◆ QA · Arithmetic

Number System, formulas + CAT PYQs

Divisibility, factors & factorials, HCF-LCM, remainders, cyclicity and base systems. The single most-tested QA chapter in CAT, 38 questions across 2017-2023 in this book alone.

34formulas

91CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Factors & Divisibility HCF & LCM Remainders Base Systems Cyclicity & Last Digits Factorials & Number of Factors Other / Mixed

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled from the Revision Notes.

1Divisibility by 2, 4, 8 (powers of 2)

By 2: last digit is 0, 2, 4, 6 or 8.
By 4: last two digits divisible by 4 (or both zero).
By 8: last three digits divisible by 8 (or all zero).
General rule: divisible by 2ⁿ if the last n digits are divisible by 2ⁿ.
Use it to test divisibility by a power of 2 without dividing the whole number.
e.g. 1316 → last two digits 16, and 16 ÷ 4 = 4, so 1316 is divisible by 4.

N divisible by 2ⁿ ⇔ last n digits divisible by 2ⁿ

2Divisibility by 3 and 9

By 3: digit-sum is divisible by 3.
By 9: digit-sum is divisible by 9.
Just add the digits, if that small sum passes, the whole number does.
e.g. 4527 → 4+5+2+7 = 18, divisible by both 3 and 9, so 4527 is too.

3 | N ⇔ 3 | (sum of digits)

3Divisibility by 5, 6, 12

By 5: last digit is 0 or 5.
By 6: divisible by its co-prime factors 2 and 3 (6 = 2 × 3).
By 12: divisible by both 3 and 4.
For any composite, test divisibility by its co-prime factors.
Break a composite divisor into co-prime parts and check each separately.
e.g. 132 is divisible by 12 because it is divisible by 3 (1+3+2=6) and by 4 (last two digits 32).

6 | N ⇔ 2 | N and 3 | N

4Divisibility by 7, 11, 13 (grouping)

Make groups of three digits from the right.
Take (sum of odd-placed groups) − (sum of even-placed groups).
If that difference is divisible by 7 / 11 / 13, so is N.
Same grouping rule works for all three primes.
Lets you test big numbers for 7, 11 or 13 using small 3-digit chunks.
e.g. 1001 = 7×11×13, so any 3-digit block repeated (e.g. 256256) is divisible by all three.

e.g. 346527659 → (346+659)−527 = 478 ÷ 7

5Divisibility by 11 (alternating sum)

Take (sum of odd-placed digits) − (sum of even-placed digits).
If the difference is divisible by 11, the number is too.
Quick single-digit alternating add/subtract test for 11.
e.g. 918082 → (9+8+8) − (1+0+2) = 25 − 3 = 22, divisible by 11, so 918082 is too.

11 | N ⇔ 11 | (Σ odd-pos − Σ even-pos)

6Osculation (13, 17)

For 13: one-more osculator 4, drop last digit, add 4 × (last digit), repeat.
For 17: negative osculator 5, drop last digit, subtract 5 × (last digit), repeat.
A shrink-and-repeat trick to test 13 or 17 when no other rule is handy.
e.g. 13 itself: 1 + 3×4 = 13 → divisible by 13.

1859 → 185+9×4 = 221 → 22+1×4 = 26 (÷13)

7Number families

Natural N = {1, 2, 3…}; Whole W = N ∪ {0}.
Integers Z = {… −2, −1, 0, 1, 2 …}.
Rational = p/q, q ≠ 0 (terminating or recurring decimals).
Irrational = non-terminating, non-recurring (√2, √3, π).
Complex a + ib; conjugate of a + ib is a − ib.
Knowing which set a number lives in tells you what operations stay "closed".
e.g. 0.75 = 3/4 is rational; √2 = 1.41421… never repeats, so it is irrational.

8Prime & composite

A prime has exactly two factors: 1 and itself.
Primes > 3 are of the form 6k ± 1 (converse not always true).
1 is neither prime nor composite; 2 is the only even prime.
The 6k±1 form lets you scan candidates for primes quickly.
e.g. 7 = 6×1+1 and 11 = 6×2−1 are prime; but 25 = 6×4+1 is not (converse fails).

primes > 3 ⇒ 6k ± 1

9Primality test

Pick the smallest n with n² > N.
Check divisibility by every prime < n.
If none divides N, then N is prime.
You only need to test primes up to √N, not all the way to N.
e.g. 97: √97 ≈ 9.8, test 2,3,5,7, none divides 97, so it is prime.

check primes p ≤ √N

10Co-prime (relatively prime)

Two numbers with no common factor except 1.
1 is co-prime to every number; two consecutive integers are co-prime.
A prime is co-prime to all numbers except its multiples; two primes are always co-prime.
Co-prime numbers can be tested for divisibility independently (used in rules above).
e.g. 8 and 15 share no common factor except 1, so HCF(8,15) = 1, they are co-prime.

11Fractions

Proper: value < 1 (numerator < denominator).
Improper: value > 1 (numerator > denominator).
Mixed: integer + proper fraction, e.g. 2¾.
Classifying a fraction tells you at a glance whether its value is below or above 1.
e.g. 3/5 is proper (< 1); 7/5 is improper (> 1) = mixed 1⅖.

12Prime factorisation form

Any composite N can be written with distinct primes a, b, c…
This standard form drives the factor, sum and product formulas below.
Write any number as primes-to-powers first; every counting formula needs it.
e.g. 360 = 2³ × 3² × 5¹.

N = aᵖ × bᑫ × cʳ …

13Number of factors

Add 1 to each prime's exponent and multiply.
Count includes 1 and N itself.
The fastest way to count all divisors of a number.
e.g. 360 = 2³×3²×5 → (3+1)(2+1)(1+1) = 4×3×2 = 24 factors.

X = (p+1)(q+1)(r+1)…

14As a product of two factors

If X (total factors) is even: X/2 ways.
If X is odd: (X+1)/2 ways (N is a perfect square).
As a product of two distinct factors: (X−1)/2 ways.
Counts the ways to write N as (one factor) × (another factor).
e.g. 12 has 6 factors → 6/2 = 3 ways: 1×12, 2×6, 3×4.

even X → X/2 · odd X → (X+1)/2

15Perfect squares ↔ factor count

A perfect square has an odd number of factors, and vice-versa.
Use it to spot or confirm a perfect square just from its factor count.
e.g. 100 = 2²×5² has 9 factors → expressible as a product of two numbers in 5 ways.
e.g. 36 = 2²×3² has (2+1)(2+1) = 9 factors (odd), and 36 = 6² is a perfect square.

odd #factors ⇔ perfect square

16Sum & product of factors

Sum of all factors of N = aᵖ × bᑫ × cʳ.
Product of all factors = N raised to (X/2), X = total factors.
Gives the total of, or product of, every divisor without listing them.
e.g. 12 = 2²×3 → sum of factors = (2³−1)/(2−1) × (3²−1)/(3−1) = 7 × 4 = 28 (=1+2+3+4+6+12).

Sum = (aᵖ⁺¹−1)/(a−1) × (bᑫ⁺¹−1)/(b−1) × … Product = N^(X/2)

17Cyclicity of unit digits

The last digit of powers repeats in a fixed cycle.
To find a unit digit, reduce the exponent mod the cyclicity.
Lets you find the last digit of any huge power instantly.
e.g. 2¹⁰: cycle of 2 is 2,4,8,6 (length 4); 10 mod 4 = 2 → 2nd in cycle = 4. (2¹⁰=1024 ✓)

2,3,7,8 → cycle 4 · 4,9 → cycle 2 · 0,1,5,6 → cycle 1

18Last two digits, base ends in 1

For (…a1)^(…b): tens digit = last digit of (a × b); units digit = 1.
Fast last-two-digits shortcut when the base ends in 1.
e.g. 31¹²: a=3, b=2, a×b=6 → last two digits are 61.

(…a1)^(…b) → [last digit of a×b] then 1

19Last two digits, base ends in 5

If the second-last digit of base and the power are both odd → ends in 75.
Otherwise → ends in 25.
One-glance rule for the last two digits of any power of a number ending in 5.
e.g. 35³: tens digit 3 (odd) and power 3 (odd) → ends in 75. (35³ = 42875 ✓)

both odd → 75, else → 25

20Last two digits, ends in 3, 7, 9

Convert the base to a power that ends in 1, then use rule 18.
e.g. 7² = 49, 7⁴ = …01 → reduce the exponent through that.
Turn an awkward base into one ending in 1, then apply rule 18.
e.g. 7¹²: 7⁴ ends in 01, so 7¹² = (7⁴)³ ends in 01.

21Highest power of a prime in n!

Sum the quotients of n divided by p, p², p³, … (floor each).
Tells you the largest power of a prime that divides n! (e.g. for trailing zeros).
e.g. power of 5 in 25! = ⌊25/5⌋ + ⌊25/25⌋ = 5 + 1 = 6.

E(p, n!) = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + …

22Highest power of a composite in n!

Split the composite into co-prime factors and find each one's power.
The lowest of those powers is the answer.
e.g. power of 10 in 30! = min(power of 2, power of 5) = min(26, 7) = 7.
For a composite, the scarcest prime-power limits the answer, use it for trailing zeros.
e.g. trailing zeros in 30! = power of 5 (the scarcer of 2 and 5) = ⌊30/5⌋+⌊30/25⌋ = 7.

10 = 2×5 ⇒ take the smaller power

23HCF (GCD)

Factorisation: product of least powers of common primes.
Division: divide larger by smaller, then divisor by remainder, repeat until remainder 0; last divisor is the HCF.
HCF is the largest number that divides all the given numbers.
e.g. HCF(12, 18): 12 = 2²×3, 18 = 2×3² → common least powers 2¹×3¹ = 6.

24LCM

Factorisation: product of highest powers of all primes present.
Division: divide the row by common factors until none share a factor; multiply divisors × leftovers.
LCM is the smallest number that every given number divides into.
e.g. LCM(12, 18): take highest powers 2²×3² = 36.

25HCF & LCM of fractions

Reduce fractions to lowest terms first.
How to take HCF/LCM when the quantities are fractions, not whole numbers.
e.g. HCF(2/3, 4/9) = HCF(2,4)/LCM(3,9) = 2/9.

HCF = HCF(num)/LCM(den) LCM = LCM(num)/HCF(den)

26HCF × LCM identity

For two numbers: HCF × LCM = product of the numbers.
For n numbers: product = (HCF)ⁿ⁻¹ × LCM.
If HCF(a,b) = H, then (a+b) and (a−b) are also divisible by H.
Once you know one of HCF/LCM and the numbers, the other follows for free.
e.g. a=12, b=18: HCF×LCM = 6×36 = 216 = 12×18. ✓

HCF × LCM = a × b

27Remainder-based HCF / LCM

Same remainder p, q, r when divided by H ⇒ H is HCF of (a−p), (b−q), (c−r).
Constant remainder R from a, b, c ⇒ N = LCM(a,b,c)·k + R.
If a−x = b−y = c−z = P, smallest number = LCM(a,b,c) − P.
Handles "leaves the same / a fixed remainder" word problems via HCF or LCM.
e.g. smallest number leaving remainder 3 by both 4 and 6 = LCM(4,6) + 3 = 15.

28Division algorithm

Dividend = Divisor × Quotient + Remainder.
The basic identity linking dividend, divisor, quotient and remainder.
e.g. 17 ÷ 5: 17 = 5×3 + 2, so quotient 3, remainder 2.

M = N·Q + R, 0 ≤ R < N

29Reducing remainders

Remainders distribute across × + and −.
Rem(a×b / c) = Rem(a/c) × Rem(b/c), then reduce again.
Rem((a±b)/c) = Rem(a/c) ± Rem(b/c).
Break a big product/sum into small remainders and recombine, avoids huge arithmetic.
e.g. Rem(17×19 / 5) = Rem(2×4 / 5) = Rem(8/5) = 3.

Rem(ab/c) = [Rem(a/c)·Rem(b/c)] mod c

30Negative remainders

Write the base as (multiple of divisor ± 1) to simplify large powers.
e.g. 15⁹⁷ / 8 = (16−1)⁹⁷ / 8 = (−1)⁹⁷ → remainder 8 − 1 = 7.
Rewriting the base as "divisor ± 1" makes powers collapse to ±1.
e.g. 9¹⁰⁰ / 10 = (10−1)¹⁰⁰ ≡ (−1)¹⁰⁰ = 1 (mod 10) → remainder 1.

(16−1)⁹⁷ ≡ (−1)⁹⁷ ≡ −1 ≡ 7 (mod 8)

31Fermat's little theorem

If M and N are co-prime and N is prime, then M^(N−1) leaves remainder 1 on division by N.
A power-of-1 shortcut for remainders when the divisor is prime.
e.g. 3⁶ mod 7 = 1 (since 7 prime, gcd(3,7)=1). (3⁶ = 729 = 7×104 + 1 ✓)

M^(N−1) ≡ 1 (mod N), N prime, gcd(M,N)=1

32Wilson's theorem

For prime N, (N−1)! + 1 is divisible by N.
A neat test/shortcut tying factorials to primality.
e.g. N=5: (5−1)! + 1 = 24 + 1 = 25 = 5×5, divisible by 5. ✓

(N−1)! + 1 ≡ 0 (mod N), N prime

33aⁿ ± bⁿ divisibility

aⁿ + bⁿ is divisible by (a + b) when n is odd.
aⁿ − bⁿ is divisible by (a − b) always; also by (a + b) when n is even.
Useful identity: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
Instantly spots a factor of expressions like aⁿ ± bⁿ.
e.g. 3³ + 2³ = 35 is divisible by 3 + 2 = 5 (odd power). (35 = 5×7 ✓)

aⁿ+bⁿ ÷ (a+b) for odd n

34Base systems

Base-10: 101 = 1×10² + 0×10¹ + 1×10⁰.
Convert base-10 → base B by repeated division by B; read remainders bottom-up.
For base > 10, digits 10, 11, 12… are written A, B, C…
Converts numbers between base-10 and any other base.
e.g. 13 in base 2: 13 = 8+4+1 = 1101₂. Back: 1×8+1×4+0×2+1 = 13. ✓

(dₙ…d₁d₀)_B = Σ dᵢ × Bⁱ

Factors & Divisibility · 16 CAT PYQs

Factors & Divisibility

ModerateCAT 1991

To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: (i₁, i₂, i₃, … , are the digits of the number, starting from the most significant digit). i₁, i₂, ……. iₙ ⇒ i₁·3ⁿ⁻¹ + i₂·3ⁿ⁻² + ……… + iₙ·3⁰. e.g., 259 ⇒ 2·3² + 5·3¹ + 9·3⁰ = 18 + 15 + 9 = 42. Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?

(1) 2
(2) 3
(3) 4
(4) 1

Show solution

(1) 2. 203 ⇒ 2·3² + 0·3¹ + 3·3⁰ = 18 + 0 + 1 = 21. Then 21 ⇒ 2·3¹ + 1·3⁰ = 6 + 1 = 7. Therefore we can reduce 203 to 7 in 2 stages.

ModerateCAT 1998

A number is formed by writing first 54 natural numbers in front of each other as 12345678910111213… Find the remainder when this number is divided by 8:

(1) 1
(2) 7
(3) 2
(4) 0

Show solution

(3) 2. Divisibility by 8 depends only on the last three digits. The last 3 digits of this number is 354. The remainder is 2 if we divide 354 by 8.

ModerateCAT 2000

Let N = 55³ + 17³ − 72³. N is divisible by

(1) both 7 and 13
(2) both 3 and 13
(3) both 17 and 7
(4) both 3 and 17

Show solution

(4) both 3 and 17. We know, if a + b + c = 0 then a³ + b³ + c³ = 3abc. Now, 55³ + 17³ − 72³ = 3(55)(17)(−72). Hence, N is divisible by both 3 and 17.

ModerateCAT 2000

Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeros will the product end?

(1) 1
(2) 4
(3) 5
(4) 10

Show solution

(1) 1. There is only one 5 and one 2 in the set of prime numbers from 2 to 100. Hence, there would be only one zero at the end of the resultant product.

ModerateCAT 2001

Let b be a positive integer and a = b² − b. If b ≥ 4, then a² − 2a is divisible by:

(1) 15
(2) 20
(3) 24
(4) None of these

Show solution

(3) 24. a = b² − b, hence a² − 2a = (b² − b)² − 2(b² − b) = (b − 2)(b − 1) b (b + 1). These are 4 consecutive numbers and since b ≥ 4, these must be divisible by 4! or 24.

ModerateCAT 2002

For all integers n > 0, 7⁶ⁿ − 6⁶ⁿ is divisible by

(1) 13
(2) 549
(3) 127
(4) All of these

Show solution

(4) All of these. 7⁶ⁿ − 6⁶ⁿ is in the form of a³ − b³, and a³ − b³ is always divisible by a − b. So it must be divisible by 7² − 6² i.e., 13. 7⁶ⁿ − 6⁶ⁿ is also in the form of a² − b², so it must be divisible by (a + b) and (a − b) i.e., 7³ − 6³ and 7³ + 6³. ⇒ it is divisible by 343 − 216 = 127 and 343 + 216 = 559. Hence, all of these are correct.

ModerateCAT 2003

Let n (>1) be a composite integer such that √n is not an integer. Consider the following statements (a) n has a perfect integer−valued divisor which is greater than 1 and less than √n. (b) n has a perfect integer−valued divisor which is greater than √n but less than n. Then,

(1) Both A and B are false
(2) A is true but B is false
(3) A is false but B is true
(4) Both A and B are true

Show solution

(4) Both A and B are true. Consider a number n = 12, then root n = 3.46. Statement (a): we have a divisor 2 which is greater than 1 and less than 3.46. Statement (b): we have a divisor 4 which is greater than 3.46 but less than 10. Both statements (a) and (b) are true. We know if a number has even number of factors then it is not a perfect square. Rule: If 'N' is not a perfect square then there exists at least one number less than √N and another number greater than √N whose product is N. Hence, both.

HardCAT 2005

The digits of a three digit number A are written in the reverse order to form another three digit number B. If B > A and B − A is perfectly divisible by 7, then which of the following is necessarily true?

(1) 100 < A < 299
(2) 106 < A < 305
(3) 112 < A < 311
(4) 118 < A < 317

Show solution

(3) 112 < A < 311. Let A = abc. Then, B = cba. Given B > A, hence c > a. As B − A = (100c + 10b + a) − (100a + 10b + c). Hence, B − A = 100(c − a) + (a − c) = 99(c − a). Also (B − A) is divisible by 7. But 99 is not divisible by 7. Therefore (c − a) must be divisible by 7, i.e. (c − a) must be 7, since c and a are single digits. The possible values of (c, a) {with c > a} are (9, 2) and (8, 1). Thus we can write A as 2b9 or 1b8. As b can take values from 0 to 9, the smallest and largest possible value of A are 108 and 299.

ModerateCAT 2005

Let S be a set of positive integers such that every element n of S satisfies the conditions: (a) 1000 ≤ n ≤ 1200; (b) every digit in n is odd. Then how many elements of S are divisible by 3?

(1) 9
(2) 10
(3) 11
(4) 12

Show solution

(1) 9. All digits are odd and it is a 4-digit number in [1000, 1200], so it has the form 11bc with b, c ∈ {1, 3, 5, 7, 9}. Digit-sum = 1 + 1 + b + c = 2 + b + c must be divisible by 3, i.e. b + c ≡ 1 (mod 3). Since b, c are odd, b + c is even, so b + c ∈ {4, 10, 16}: b + c = 4 (2 ways), b + c = 10 (5 ways), b + c = 16 (2 ways) → 9 numbers.

HardCAT 2008

Suppose, the seed of any positive integer n is defined as follows: seed(n) = n, if n < 10; = seed(s(n)), otherwise, where s(n) indicates the sum of digits of n. i.e., seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc. How many positive integers n, such that n < 500, will have seed(n) = 9?

(1) 39
(2) 72
(3) 81
(4) 55

Show solution

(4) 55. The seed(n) function gives the digit-sum (digital root) of any given number n. All the numbers n for which seed(n) = 9 will give multiples of 9. For all positive integers n, n < 500, there are 55 multiples of 9.

HardCAT 2018TITA

If N and x are positive integers such that Nᴺ = 2¹⁶⁰ and N² + 2ᴺ is an integral multiple of 2ˣ, then the largest possible x is

Show solution

10. Nᴺ = 2¹⁶⁰ = (2⁵)³² = 32³², so N = 32. Then N² + 2ᴺ = 32² + 2³² = 2¹⁰ + 2³² = 2¹⁰(1 + 2²²). Hence, the largest possible value of x is 10.

HardCAT 2018

If A = {6²ⁿ − 35n − 1}, and B = {35(n − 1)}, where n = 1, 2, 3,... then which of the following is true?

(1) Every member of A is in B and at least one member of B is not in A
(2) Neither every member of A is in B nor every member of B is in A
(3) Every member of B is in A
(4) At least one member of A is not in B

Show solution

(1). Given A = 6²ⁿ − 35n − 1 = 36ⁿ − 1 − 35n, which is divisible by 35. Hence A = 1225, 46650 etc., for n = 2, 3. Set B = 35, 70, 105 for n = 2, 3, 4 etc. Hence A misses some multiples while B has all the multiples of 35. So every member of set A will be in B, while every member of set B will not necessarily be in set A.

ModerateCAT 2020

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

(1) 41
(2) 42
(3) 43
(4) 44

Show solution

(1) 41. Total integers = 1, 2, 3, ……. 120. Mul(2 or 5 or 7) = Mul(2) + Mul(5) + Mul(7) − Mul(2 & 5) − Mul(2 & 7) − Mul(5 & 7) + Mul(2 & 5 & 7) = 60 + 24 + 17 − 12 − 3 − 8 + 1 = 79. So, numbers which are not divisible by 2, 5 and 7 = 120 − 79 = 41.

ModerateCAT 2023 · Slot 1

Let n be the least positive integer such that 168 is a factor of 1134ⁿ. If m is the least positive integer such that 1134ⁿ is a factor of 168ᵐ, then m + n equals

(1) 24
(2) 12
(3) 9
(4) 15

Show solution

(4) 15. 168 = 2³ × 3¹ × 7¹ and 1134 = 2¹ × 3⁴ × 7¹. Since 168 is a factor of 1134ⁿ, the least value of n is 3. Now 1134ⁿ = 1134³ = 2³ × 3¹² × 7³ and 168ᵐ = (2³ × 3¹ × 7¹)ᵐ. Since 1134ⁿ is a factor of 168ᵐ, the least value of m is 12. Now m + n = 12 + 3 = 15.

CAT 2024 & 2025, recent

HardCAT 2025 · Slot 1 TITA

In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is

Show solution

6. Allowed digits (non-zero, not perfect-square 1/4/9): 2, 3, 5, 6, 7, 8. Exactly one prime (2, 3, 5, 7) and two non-primes (6, 8) needed. Smallest N uses the smallest prime (2) with the two smallest non-primes (6, 8) → N = 268 = 2² × 67 → (2+1)(1+1) = 6 factors.

HardCAT 2025 · Slot 2

The number of divisors of (2⁶ × 3⁵ × 5³ × 7²), which are of the form (3r + 1), where r is a non-negative integer, is:

(A) 42
(B) 36
(C) 56
(D) 24

Show solution

(A) 42. A divisor ≡ 1 (mod 3) cannot contain any factor of 3, so the power of 3 is 0. We need 2ᵃ·5ᵇ·7ᶜ ≡ 1 (mod 3) with 0 ≤ a ≤ 6, 0 ≤ b ≤ 3, 0 ≤ c ≤ 2. Since 2 ≡ −1 and 5 ≡ −1 (mod 3), the product 2ᵃ·5ᵇ ≡ (−1)^(a+b), which is ≡ 1 when a+b is even; 7 ≡ 1 (mod 3) so 7ᶜ never affects it. Even (a+b): a even (4 choices: 0,2,4,6) with b even (2 choices: 0,2) = 8, plus a odd (3) with b odd (2) = 6, total 14. Times 3 choices for c = 14 × 3 = 42.

HCF & LCM · 8 CAT PYQs

HCF & LCM

EasyCAT 1999

4. For two positive integers a and b define the function h(a, b) as the greatest common factor (GCF) of a, b. Let A be a set of n positive integers. G(A), the GCF of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is:

(1) ½ n
(2) (n − 1)
(3) n
(4) None of these

Show solution

(2) (n − 1). Each application of h combines two values into one. To reduce n values to a single GCF, you need one fewer operation than values: n − 1.

HardCAT 2001

2. A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?

(1) 100 m, 15
(2) 100 m, 16
(3) 120 m, 15
(4) 120 m, 16

Show solution

(4) 120 m, 16. The length must be a common multiple of 6 and 8. With length 120 m: posts at 8 m spacing = 120/8 + 1 = 16; at 6 m spacing = 120/6 + 1 = 21, which is 5 more. So length 120 m, 16 posts bought.

ModerateCAT 2001

6. A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?

(1) 30
(2) 24
(3) 20
(4) 60

Show solution

(1) 30. Red interval = 60/3 = 20 s; green interval = 120/5 = 24 s. They coincide every LCM(20, 24) = 120 s. In 3600 s there are 3600/120 = 30 coincidences.

HardCAT 2002

8. There are three pieces of cake weighing 9/2 lbs, 27/4 lbs and 35/5 lbs. Pieces of the cake are equally divided and distributed in such a manner that every guest in the party gets one single piece of cake. Further the weight of the pieces of the cake is as heavy as possible. What is the largest number of guest to whom we can distribute the cake?

(1) 54
(2) 20
(3) 72
(4) None of these

Show solution

(4) None of these (= 41). The heaviest equal piece = HCF of the fractions = HCF(numerators)/LCM(denominators). (Note: the book's worked solution treats the third weight as 36/5.) LCM of denominators 2, 4, 5 = 20; HCF of numerators 9, 27, 36 = 9, so heaviest part = 9/20. Then (9/2) = (90/20) = 10 × (9/20), (27/4) = (135/20) = 15 × (9/20), (36/5) = (144/20) = 16 × (9/20). Maximum guests = 10 + 15 + 16 = 41.

HardCAT 2002

9. In a book store, each of the word of the glowsign board "MODERN BOOK STORES" is visible after 5/2, 17/4 and 41/8 seconds respectively. Each of them is put off for 1 second. Find the time after which one person can see a completely visible glowsign board.

(1) 73.5 seconds
(2) 79.4 seconds
(3) 68.2 seconds
(4) None of these

Show solution

(1) 73.5 seconds. Effective cycle of each word (visible time + 1 s off): MODERN = 5/2 + 1 = 7/2, BOOK = 17/4 + 1 = 21/4, STORES = 41/8 + 1 = 49/8. All flash together at LCM(7/2, 21/4, 49/8) = LCM(7,21,49)/HCF(2,4,8) = 147/2 = 73.5 s.

HardCAT 2022 · Slot 1TITA

12. Let A be the largest positive integer that divides all the numbers of the form 3ᵏ + 4ᵏ + 5ᵏ, and B be the largest positive integer that divides all the numbers of the form 4ᵏ + 3(4ᵏ) + 4ᵏ⁺², where k is any positive integer. Then (A + B) equals:

Show solution

82. For A: k = 1 → 12, k = 2 → 50, k = 3 → 216; HCF(12, 50, 216) = 2, so A = 2. For B: 4ᵏ + 3·4ᵏ + 4ᵏ·16 = 4ᵏ(1 + 3 + 16) = 4ᵏ·20; the largest common divisor over all k is 4 × 20 = 80, so B = 80. A + B = 82.

HardCAT 2022 · Slot 3TITA

14. A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:

Show solution

150. LCM(9, 10, 12, 25) = 900, so students = 900k + 4. With 900k + 4 < 5000, k = 1…5. Also it must be divisible by 11; only k = 2 works → 1804 students. Teams of 12 = ⌊1800/12⌋ = 150.

HardCAT 2023 · Slot 2

15. For any natural numbers m, n and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of:

(1) m and n
(2) m and 2n
(3) 2m and 3n
(4) 2m and n

Show solution

(2) m and 2n. Let m + 2n = λ₁k ...(1) and 3m + 4n = λ₂k ...(2), where λ₁ & λ₂ are natural numbers. Solving (1) & (2): m = (2λ₂ − λ₁)k and 2n = (3λ₁ − λ₂)k. So m and 2n are common divisors, i.e. k divides both m and 2n.

Remainders · 13 CAT PYQs

Remainders

ModerateCAT 1998

A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

(1) 0
(2) 1
(3) 2
(4) None of these

Show solution

(2) 1. Divisor − remainder = 1 in every case, so the number is of the form (LCM − 1). LCM(2,3,4,5,6) = 60, so numbers are 60n − 1: 59, 119, … Only 59 lies between 0 and 100. One number.

ModerateCAT 1999

The remainder when 7⁸⁴ is divided by 342 is:

(1) 0
(2) 1
(3) 49
(4) 341

Show solution

(2) 1. Note 342 = 343 − 1 = 7³ − 1. So 7⁸⁴ = (7³)²⁸ = 343²⁸ = (342 + 1)²⁸ ≡ 1²⁸ = 1 (mod 342). Remainder 1.

ModerateCAT 2000

Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?

(1) 0
(2) 9
(3) 3
(4) 6

Show solution

(3) 3. 1416 is a multiple of 12, so N = (1416+5)(1416+7)(1416+9). Remainder = (5 × 7 × 9) mod 12 = 315 mod 12 = 3.

HardCAT 2000

The integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder. What is the value of n?

(1) 289
(2) 367
(3) 453
(4) 307

Show solution

(4) 307. If both leave the same remainder, n divides their difference: 34041 − 32506 = 1535 = 5 × 307. The three-digit divisor is 307.

HardCAT 2002

On dividing a number by 3, 4 and 7, the remainders are 2, 1 and 4 respectively. If the same number is divided by 84 then the remainder is:

(1) 80
(2) 76
(3) 53
(4) None of these

Show solution

(3) 53. Build up: number ≡ 4 (mod 7) → 7x+4; ≡ 1 (mod 4) → 4(7x+4)+1; ≡ 2 (mod 3) → 3[4(7x+4)+1]+2 = 84x + 53. So mod 84 the remainder is 53.

EasyCAT 2004

What is the remainder when 4⁹⁶ is divided by 6?

(1) 0
(2) 2
(3) 3
(4) 4

Show solution

(4) 4. Any positive power of 4 leaves remainder 4 when divided by 6 (4¹=4, 4²=16≡4, …). So 4⁹⁶ ≡ 4 (mod 6).

ModerateCAT 2004

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

(1) 666
(2) 676
(3) 683
(4) 777

Show solution

(2) 676. Such numbers are 7k + 3: the two-digit ones run 10, 17, …, 94 (k = 1 to 13), 13 terms. Sum = 13/2 × (10 + 94) = 13 × 52 = 676.

ModerateCAT 2004

The remainder, when (15²³ + 23²³) is divided by 19, is

(1) 4
(2) 15
(3) 0
(4) 18

Show solution

(3) 0. 15²³ + 23²³ is aⁿ + bⁿ with odd n, so it is divisible by a + b = 38 = 2 × 19. Since 19 | 38, the remainder is 0.

ModerateCAT 2005

If x = (16³ + 17³ + 18³ + 19³), then x divided by 70 leaves a remainder of

(1) 0
(2) 1
(3) 69
(4) 35

Show solution

(1) 0. a³ + b³ + c³ + d³ is divisible by a + b + c + d when the terms pair appropriately; here 16 + 17 + 18 + 19 = 70. So x is divisible by 70, remainder 0.

ModerateCAT 2002

The remainder when 2²⁵⁶ is divided by 17 is:

(1) 7
(2) 13
(3) 11
(4) 1

Show solution

(4) 1. 2²⁵⁶ = (2⁴)⁶⁴ = 16⁶⁴ = (17 − 1)⁶⁴ ≡ (−1)⁶⁴ = 1 (mod 17). Remainder 1.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

When 10¹⁰⁰ is divided by 7, the remainder is

(A) 2
(B) 4
(C) 1
(D) 6

Show solution

(B) 4. 10 ≡ 3 (mod 7); powers of 3 cycle with period 6. 100 = 6·16 + 4, so 10¹⁰⁰ ≡ 3⁴ = 81 ≡ 4 (mod 7).

ModerateCAT 2024 · Slot 2

When 3³³³ is divided by 11, the remainder is

(A) 5
(B) 10
(C) 1
(D) 6

Show solution

(A) 5. Powers of 3 mod 11 cycle with period 5: 3, 9, 5, 4, 1. 333 ≡ 3 (mod 5), so 3³³³ ≡ 3³ = 27 ≡ 5 (mod 11).

ModerateCAT 2024 · Slot 3

If 10⁶⁸ is divided by 13, the remainder is

(A) 5
(B) 8
(C) 9
(D) 4

Show solution

Base Systems · 2 CAT PYQs

Base Systems

HardCAT 2001

In a number system the product of 44 and 11 is 1034. The number 3111 of this system, when converted to the decimal number system, becomes:

(1) 406
(2) 1086
(3) 213
(4) 691

Show solution

(1) 406. The product of 44 and 11 is 484. If base is x then 3411 = 3x³ + 4x² + 1x¹ + 4x⁰ = 484, so 3x³ + 4x² + x = 480. This equation is satisfied when x = 5. So the base is 5. Then 3111₅ = 3·5³ + 1·5² + 1·5 + 1 = 375 + 25 + 5 + 1 = 406.

HardCAT 2004

A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals:

(1) 31
(2) 63
(3) 75
(4) 91

Show solution

(4) 91. We can do the question by options. The number when divided by either 3 or 5 will leave a remainder of 1. 63 and 75 are multiples of 3 and 5 respectively. Their remainder cannot be 1. Thus, 63 and 75 are eliminated. For the options 31 and 91, the remainder is 1. Thus, the last digit is 1. 31 = (11111)₂ = (1011)₃ = (111)₅; 91 = (1011011)₂ = (10101)₃ = (331)₅. 91 has 1 as the first digit in only 2 of the options. So M = 91.

Cyclicity & Last Digits · 2 CAT PYQs

Cyclicity & Last Digits

ModerateCAT 2005

3. The rightmost non zero digit of the number 30²⁷²⁰ is :

(1) 1
(2) 3
(3) 7
(4) 9

Show solution

(1) 1. The right most non-zero digit can be found using the pattern given below. 30¹ = 3, 30² = 9, 30³ = 7, 30⁴ = 1. The pattern repeats from here. Since 2720/4 gives the remainder 0, right most non-zero digit will be 1.

ModerateCAT 2008

4. What are the last two digits of 7²⁰⁰⁸ ?

(1) 21
(2) 61
(3) 01
(4) 41

Show solution

(3) 01. 7²⁰⁰⁸ = (7⁴)⁵⁰² = 2401⁵⁰² = …01⁵⁰² = 01. Since, 01 raised to any power is always 01.

Factorials & Number of Factors · 7 CAT PYQs

Factorials & Number of Factors

ModerateCAT 2003

The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n − 1)(n − 2) …3.2.1 is not divisible by n is

(1) 5
(2) 7
(3) 13
(4) 14

Show solution

(2) 7. (n−1)! is divisible by n for all composite n; it fails only when n is prime. The primes in 12 ≤ n ≤ 40 are 13, 17, 19, 23, 29, 31, 37, exactly 7 of them.

HardCAT 2005

Let n! = 1 × 2 × 3 × … × n for integer n ≥ 1. If p = (1 × 1!) + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of

(1) 10
(2) 0
(3) 7
(4) 1

Show solution

(4) 1. Use n·n! = (n+1)! − n!. Then p telescopes: p = (2!−1!) + (3!−2!) + … + (11!−10!) = 11! − 1. So p + 2 = 11! + 1, which divided by 11! leaves remainder 1.

ModerateCAT 2019TITA

How many factors of 2⁴ × 3⁵ × 10⁴ are perfect squares which are greater than 1 ?

Show solution

44. 2⁴ × 3⁵ × 10⁴ = 2⁴ × 3⁵ × (2⁴ × 5⁴) = 2⁸ × 3⁵ × 5⁴. A perfect-square factor 2ᵃ3ᵇ5ᶜ needs even exponents: a ∈ {0,2,4,6,8} (5), b ∈ {0,2,4} (3), c ∈ {0,2,4} (3) → 5×3×3 = 45. Exclude 1 (all zero): 45 − 1 = 44.

ModerateCAT 2020

How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 4²⁰¹⁷ ?

(1) 2017
(2) 2018
(3) 2019
(4) 2020

Show solution

(2) 2018. 4²⁰¹⁷ = 2⁴⁰³⁴, a perfect square with 4035 factors. The number of unordered pairs (a ≤ b) = (total factors + 1)/2 = (4035 + 1)/2 = 2018.

ModerateCAT 2022 · Slot 2

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is:

(1) 5
(2) 4
(3) 6
(4) 7

Show solution

(4) 7. (15000)! divisible by (n!)! requires n! ≤ 15000. Since 7! = 5040 < 15000 but 8! = 40320 > 15000, the largest n is 7.

ModerateCAT 2023 · Slot 2TITA

The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is

Show solution

13. "Exactly two distinct factors other than 1 and itself" means the number has 4 factors total (1, p, q, pq), i.e. it is a product of two distinct primes p × q. Counting p × q < 50: 2 × {3, 5, 7, 11, 13, 17, 19, 23} → 8; 3 × {5, 7, 11, 13} → 4; 5 × {7} → 1. Total = 8 + 4 + 1 = 13. (The book's worked solution also computes 8 + 4 + 1 = 13; its boxed key value of 15 is a misprint.)

HardCAT 2023 · Slot 3TITA

The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is

Show solution

468. 15 factors means the form a²·b⁴ or a¹⁴. Smallest of form a²b⁴: 2⁴·3² = 144 and 2²·3⁴ = 324 (a¹⁴ = 2¹⁴ = 16384 is far larger). The first two such numbers are 144 and 324; sum = 468.

Other / Mixed · 55 CAT PYQs

Other / Mixed

HardCAT 1991

In Sivakasi, each boy's quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?

(1) 200
(2) 150
(3) 125
(4) 175

Show solution

(2) 150. Let sticks per box = x, number of boxes = y, so total = xy. ATQ xy = (x − 25)(y + 3) ⇒ 25y − 3x = 75. Clearly y = 6, x = 25 (with xy < 200) ⇒ total = 150. (Also, the only option that is a multiple of 75 is 150.)

ModerateCAT 1996

There are five cities in a region, namely A, B, C, D, E. The distance of these cities from each other are as follows: AB = 2 km, AC = 2 km, AD > 2 km, AE > 3 km, BC = 2 km, BD = 4 km, BE = 3 km, CD = 2 km, CE = 3 km, DE > 3 km. If a ration shop is to be set up within 3 km of each city, how many ration shops will be required?

(1) 1
(2) 2
(3) 3
(4) 4

Show solution

(1) 1. Use the triangle inequality to bound the unknown distances: from A-C-D, AD < 4; from A-C-E, AE < 5; from C-D-E, DE < 5. So every pairwise distance is at most 5 km, i.e. within 6 km of one another. One shop placed centrally covers all cities within 3 km. One shop suffices.

HardCAT 1997

A, B and C are defined as follows. A = (2.000004) ÷ [(2.000004)² + (4.000008)], B = (3.000003) ÷ [(3.000003)² + (9.000009)], C = (4.000002) ÷ [(4.000002)² + (8.000004)]. Which of the following is true about the values of the above three expressions?

(1) All of them lie between 0.18 and 0.2
(2) A is twice of C
(3) C is the smallest
(4) B is the smallest

Show solution

(4) B is the smallest. Factor each denominator: A = 2.000004 / [2.000004(2.000004 + 2)] ≈ 1/4.000004 ≈ 0.25. B = 1/(3.000003 + 3) ≈ 1/6 ≈ 0.165. C = 1/(4.000002 + 2) ≈ 1/6 ≈ 0.166. B and C are nearly equal but B is marginally smaller.

HardCAT 1998

You can collect as many rubies and emeralds as you can. Each ruby is worth ₹4 crore and each emerald is worth ₹5 crore. Each ruby weighs 0.3 kg. And each emerald weighs 0.4 kg. Your bag can carry at the most 12 kg. What should you collect to get the maximum wealth?

(1) 20 rubies and 15 emeralds
(2) 40 rubies
(3) 28 rubies and 9 emeralds
(4) None of these

Show solution

(2) 40 rubies. Value per kg: ruby = 4/0.3 = ₹13.33 cr/kg; emerald = 5/0.4 = ₹12.5 cr/kg. Rubies have higher value density, so fill the entire 12 kg with rubies: 12/0.3 = 40 rubies.

ModerateCAT 1999

Let a, b, c be distinct digits. Consider a two-digit number 'ab' and a three-digit number 'ccb', both defined under the usual decimal number system, if (ab)² = ccb > 300, then the value of b is:

(1) 1
(2) 0
(3) 5
(4) 6

Show solution

(1) 1. The value of 'ab' is that number whose unit digit is equal to the unit digit of the square of that number. Possible values of ab are 21, 25, 26, 31. Possible values of (ab)² are 441, 625, 676, 961. The only value which satisfies ccb is 441. Thus, b = 1.

Directions (Q. 4 to 6): Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the 'dealer' and the other is called the 'player'. First, the player picks a card at random from the deck. This is called the base card. The amount in rupees equal to the face value of the base card is called the base amount. The face values of ace, king, queen and jack are ten. For other cards the face value is the number on the card. Once the 'player' picks a card from the deck, the 'dealer' pays him the base amount. Then the 'dealer' picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the 'player' pays twice the base amount to the 'dealer'. If the top card is of the same colour as the base card (but not the same suit), then the 'player' pays the base amount to the 'dealer'. If the top card happens to be of a different colour than the base card, the 'dealer' pays the base amount to the 'player'. Ghosh Babu played the game four times. First time he picked eight of clubs and the 'dealer' picked queen of clubs. Second time, he picked ten of hearts and the 'dealer' picked two of spades. Next time, Ghosh Babu picked six of diamonds and the 'dealer' picked ace of hearts. Lastly, he picked eight of spades and the 'dealer' picked jack of spades.

HardCAT 1999

4. If Ghosh Babu stopped playing the game when his gain would be maximized, the gain in ₹ would have been

(1) 12
(2) 20
(3) 16
(4) 4

Show solution

(1) 12. Round 1 (8♣, Q♣ same suit): +8 − 16 = −8. Round 2 (10♥, 2♠ different colour): +10 + 10 = +20 → running total +12. Round 3 (6♦, A♥ same colour red): +6 − 6 = 0. Round 4 (8♠, J♠ same suit): +8 − 16 = −8. The maximum running gain is ₹12, after round 2.

HardCAT 1999

5. The initial money Ghosh Babu had (before the beginning of the game sessions) was ₹X. At no point did he have to borrow any money. What is the minimum possible value of X?

(1) 16
(2) 8
(3) 100
(4) 24

Show solution

(2) 8. Tracking the balance, the largest amount he must pay out before receiving is ₹8 (during round 1 he pays 16 having received 8). So he needs at least ₹8 to start to avoid borrowing.

HardCAT 1999

6. If the final amount of money that Ghosh Babu had with him was ₹100, what was the initial amount he had with him?

(1) 120
(2) 8
(3) 4
(4) 96

Show solution

(4) 96. Net effect over the four rounds is −8 + 20 + 0 − 8 = +4. If the final balance is 100, the initial was 100 − 4 = 96.

ModerateCAT 2000

Each of the numbers x₁, x₂, …, xₙ, n ≥ 4, is equal to 1 or −1. If x₁x₂x₃x₄ + x₂x₃x₄x₅ + x₃x₄x₅x₆ + … + xₙ₋₃xₙ₋₂xₙ₋₁xₙ + xₙ₋₂xₙ₋₁xₙx₁ + xₙ₋₁xₙx₁x₂ + xₙx₁x₂x₃ = 0, then:

(1) n is even
(2) n is odd
(3) n is an odd multiple of 3
(4) n is prime

Show solution

(1) n is even. Each term is ±1. For the sum of n such terms to be 0, the number of +1's must equal the number of −1's, which requires n to be even.

HardCAT 2001

In a 4-digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?

(1) 5
(2) 8
(3) 1
(4) 4

Show solution

(1) 5. With digits a, b, c, d: a+b = c+d, a+d = c, b+d = 2(a+c). From the first two, b = 2d; substituting gives d = 4a, so a = d/4 and c = 5d/4. d must be a single digit divisible by 4: d = 4 → c = 5 (d = 8 gives c = 10, invalid). Third digit = 5.

HardCAT 2001

Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of what was left but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

(1) 38
(2) 31
(3) 41
(4) None of these

Show solution

(4) None of these (= 48). Work backwards. Before Eswari: x − x/2 + 2 = 17 ⇒ x = 30. Before Fatima: x − x/4 + 3 = 30 ⇒ x = 36. Before Sita: x − x/3 + 4 = 36 ⇒ x = 48. So 48 mints were there originally, not among options (1)-(3).

HardCAT 2002

If xₙ = (−1)ⁿ xₙ₋₁ & x₀ = x, then:

(1) xₙ is positive for n = even
(2) xₙ is negative for n = even
(3) xₙ is positive for n = odd
(4) None of these

Show solution

(4) None of these. x₀ = x, x₁ = −x, x₂ = −x, x₃ = x, x₄ = x, … There is no consistent positive/negative trend tied to parity, so nothing can be deduced.

ModerateCAT 2002

If U, V, W and m are natural numbers such that Uᵐ + Vᵐ = Wᵐ, then which of the following is true?

(1) m < min (U, V, W)
(2) m > max (U, V, W)
(3) m < max (U, V, W)
(4) None of these

Show solution

(1) m < min(U, V, W). e.g. 3² + 4² = 5² (m = 2) and 2 + 4 = 6 (m = 1). In all valid natural-number cases m is less than the minimum of U, V, W.

HardCAT 2003

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer?

(1) [a/27, a/e]
(2) [a/36, c/e]
(3) [a/12, bd/18]
(4) [a/6, c/d]

Show solution

(4) [a/6, c/d]. From a = 6b = 12c and 2b = 9d = 12e, take the ratios: a : b : c : d : e = 108 : 18 : 9 : 4 : 3, so a = 108k, b = 18k, c = 9k, d = 4k, e = 3k. Testing each pair, c/d = 9k/4k = 9/4 is not an integer; every other listed expression is an integer. Hence pair (4) contains a non-integer.

ModerateCAT 2003

Let x and y be positive integers such that x is prime and y is composite. Then,

(1) y − x cannot be an even integer
(2) xy cannot be an even integer
(3) (x − y)/x cannot be an even integer
(4) None of these

Show solution

(4) None of these. Since 2 is prime, x may be even or odd; y may be even or odd. Counterexamples kill (1) (even−even or odd−odd = even), (2) (even×even = even) and (3). Hence none of the "cannot" statements holds.

HardCAT 2003

Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise and ₹1.01 to three different persons?

(1) 19
(2) 20
(3) 17
(4) 18

Show solution

(1) 19. 78 = 50+10+10+2+2+2+2 (7 coins); 69 = 50+10+5+2+2 (5 coins); 101 = 50+25+10+10+2+2+2 (7 coins). Total = 7 + 5 + 7 = 19 coins.

HardCAT 2003

In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is

(1) Less than the total number of huts existing at the beginning of 2001.
(2) Less than the total number of huts destroyed by floods in 2001 and 2003.
(3) Less than the total number of huts destroyed by floods in 2002 and 2003.
(4) More than the total number of huts built in 2001 and 2002.

Show solution

(3). Starting with x: each year, ½ destroyed then 2×(½) = x rebuilt, so the stock multiplies by 3/2 yearly. Destroyed amounts: 2001 → x/2, 2002 → 3x/4, 2003 → 9x/8, 2004 → 27x/16. Comparing, 27x/16 (2004) is less than 3x/4 + 9x/8 = 15x/8 (2002 + 2003). Option (3) is correct.

HardCAT 2003

Answer this question on the basis of the tables given below. Two binary operations ⊕ and ∗ are defined over the set {a, e, f, g, h} as per the following tables.

Operation tables for ⊕ and ∗ over {a, e, f, g, h}

Thus, according to the first table f ⊕ g = a, while according to the second table g ∗ h = f, and so on. Also, let f² = f ∗ f, g³ = g ∗ g ∗ g, and so on. What is the smallest positive integer n such that gⁿ = e?

(1) 4
(2) 5
(3) 2
(4) 3

Show solution

(1) 4. From the ∗ table: g² = g∗g = h, g³ = h∗g = f, g⁴ = f∗g = e. The smallest n with gⁿ = e is 4.

HardCAT 2005

For a positive integer n, let Pₙ denote the product of the digits of n, and Sₙ denote the sum of the digits of n. The number of integers between 10 and 1000 for which Pₙ + Sₙ = n is

(1) 81
(2) 16
(3) 18
(4) 9

Show solution

(4) 9. Two-digit n = 10a + b: ab + a + b = 10a + b ⇒ ab = 9a ⇒ b = 9. That gives 19, 29, …, 99, 9 numbers. For three-digit n the equation forces an impossible value, so none. Total 9.

ModerateCAT 2006

The sum of four consecutive two digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

(1) 21
(2) 25
(3) 41
(4) 67

Show solution

(3) 41. Let the numbers be x−3, x−1, x+1, x+3; sum = 4x. 4x/10 must be a perfect square. Taking 4x/10 = 16 ⇒ x = 40, numbers are 37, 39, 41, 43. 41 is among them.

ModerateCAT 2006

The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be

(1) 101 : 88
(2) 87 : 100
(3) 110 : 111
(4) 85 : 98
(5) 97 : 84

Show solution

(5) 97 : 84. The total employees equals the sum of the ratio parts (in lowest terms), which must be prime and < 300. 97 + 84 = 181, a prime < 300. The other sums are not prime.

HardCAT 2007

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 where n is an odd integer less than 60?

(1) 6
(2) 4
(3) 7
(4) 5
(5) 3

Show solution

(5) 3. 1/m = 1/12 − 4/n ⇒ m = 12n/(n − 48). For m > 0 we need n > 48; odd n < 60 gives n = 49, 51, 53, 55, 57, 59. Of these, n = 49, 51, 57 make m a positive integer (the others give fractions). So 3 pairs.

ModerateCAT 2017

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

(1) 1777
(2) 1785
(3) 1875
(4) 1877

Show solution

(4) 1877. (x−1)x(x+1) = x³ − x = 15600. The nearest cube is 25³ = 15625, so x = 25. Integers are 24, 25, 26; sum of squares = 576 + 625 + 676 = 1877.

ModerateCAT 2017TITA

How many different pairs (a, b) of positive integers are there such that a ≥ b and 1/a + 1/b = 1/9?

Show solution

3. 1/a + 1/b = 1/9 ⇒ ab − 9(a + b) = 0 ⇒ (a − 9)(b − 9) = 81. With a ≥ b, the factor pairs of 81 are (81,1), (27,3), (9,9), giving (a,b) = (90,10), (36,12), (18,18). Three pairs.

ModerateCAT 2018TITA

The number of integers x such that 0.25 ≤ 2ˣ ≤ 200, and 2ˣ⁺² is perfectly divisible by either 3 or 4, is

Show solution

5. Range 0.25 ≤ 2ˣ ≤ 200 gives x from −2 to 7. Checking which x make 2ˣ⁺² divisible by 3 or 4: powers of 2 are never divisible by 3, so this needs 2ˣ⁺² divisible by 4, i.e. x + 2 ≥ 2 ⇒ x ≥ 0. Per the book's table the valid integers are x = 0, 1, 2, 4, 6 → 5 values.

HardCAT 2018TITA

While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Show solution

40. Let the product of the other two be x. Then 73x − 37x = 720 ⇒ 36x = 720 ⇒ x = 20. For two reals a, b with ab = 20, a² + b² is minimised when a = b = √20, giving a² + b² = 20 + 20 = 40.

ModerateCAT 2018

If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

(1) −32
(2) 16
(3) 48
(4) 64

Show solution

(4) 64. By AM ≥ GM, a² + b² ≥ 2ab, so 97 ≥ 2ab ⇒ ab ≤ 48.5. Therefore the product cannot be 64.

ModerateCAT 2018

The smallest integer n for which 4ⁿ > 17¹⁹ holds, is closest to

(1) 37
(2) 35
(3) 33
(4) 39

Show solution

(4) 39. 4ⁿ > 17¹⁹. Write 4ⁿ = 16^(n/2). Since 16¹⁹ < 17¹⁹, we need n/2 > 19, i.e. n > 38. The smallest such integer is 39.

ModerateCAT 2019TITA

How many pairs (m, n) of positive integers satisfy the equation m² + 105 = n²?

Show solution

4. n² − m² = 105 ⇒ (n − m)(n + m) = 105. The factor pairs of 105 are 105×1, 35×3, 21×5, 15×7, 4 pairs, each yielding positive integer m, n. So 4 solutions.

HardCAT 2019TITA

In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

Show solution

7. Let digits be abcdef. Given c = a, b = 2a, e = a + b = 3a, f = a + b + c = 4a, d = e + f = 7a. Since d is a single digit, a = 1, giving d = 7.

HardCAT 2019TITA

If a, b, c are non-zero and 14ᵃ = 36ᵇ = 84ᶜ, then 6b(1/c − 1/a) is equal to

Show solution

3. Let 14ᵃ = 36ᵇ = 84ᶜ. Then 84/14 = 36^(b/c)/36^(b/a) ⇒ 6 = 36^(b/c − b/a) = 6^(2b/c − 2b/a). Equating exponents: 2b/c − 2b/a = 1 ⇒ 2b(1/c − 1/a) = 1 ⇒ 6b(1/c − 1/a) = 3.

ModerateCAT 2019TITA

Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?

Show solution

6. x ∈ {3…9}, y ∈ {15…22}. For N = x + y > 25, taking y = 22 gives x = 4…9 → N = 26, 27, 28, 29, 30, 31. These 6 distinct values are achievable. So 6.

ModerateCAT 2020

If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

(1) 46
(2) 56
(3) 49
(4) 59

Show solution

(1) 46. With c < 9 and bc = 96: c = 8 → b = 12 → a = 36, sum = 56; c = 6 → b = 16 → a = 27, sum = 49; c = 4 → b = 24 → a = 18, sum = 46; c = 2 → b = 48 → a = 9, sum = 59. The smallest sum is 46.

HardCAT 2020TITA

A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Show solution

62. Work backwards. To the 5th child: 1 + 1 = 2 (stock 2). 4th: needs (3)+1 from 6. 3rd: 7+1 from 14. 2nd: 15+1 from 30. 1st: 31+1 from 62. Initial stock = 62.

ModerateCAT 2020

The mean of all 4-digit even natural numbers of the form 'aabb', where a > 0, is

(1) 4864
(2) 5050
(3) 5544
(4) 4466

Show solution

(3) 5544. Numbers are aabb with bb even: arranged in columns by a (1100, 1122, … 1188; etc.). Each column averages to its 3rd entry (1144, 2244, …), and these average to 5544. So the overall mean is 5544.

ModerateCAT 2020

Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is:

(1) 6
(2) 7
(3) 5
(4) 4

Show solution

(1) 6. A + (B+C)/2 = 5 ⇒ 2A + B + C = 10. B + (A+C)/2 = 7 ⇒ A + 2B + C = 14. Subtracting: B − A = 4 ⇒ B = A + 4. Substituting, C = 6 − 3A; positive-integer solution is A = 1, B = 5, C = 3. So A + B = 6.

HardCAT 2021 · Slot 1TITA

The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Show solution

47. With 3 and 5 fixed, choose 1 or more from {1, 2, 4, 6, 7, 8} avoiding {7,8} together. Counts: 1-element ⁶C₁ = 6; 2-element ⁶C₂ − 1 = 14; 3-element ⁶C₃ − ⁴C₁ = 16; 4-element ⁶C₄ − ⁴C₂ = 9; 5-element = 2. Total = 6 + 14 + 16 + 9 + 2 = 47.

ModerateCAT 2021 · Slot 1

The natural numbers are divided into groups (1), (2, 3, 4), (5, 6, 7, 8, 9), …. and so on. Then, the sum of the numbers in the 15th group is equal to

(1) 4941
(2) 6090
(3) 6119
(4) 7471

Show solution

(3) 6119. The nth group has 2n − 1 elements and ends at n². So the 15th group has 29 elements, ends at 15² = 225, starts at 225 − 29 + 1 = 197. Sum (AP) = 29/2 × (197 + 225) = 29 × 211 = 6119.

ModerateCAT 2021 · Slot 2TITA

In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Show solution

10. Let played = x, current goals = y. Then 0.15 = (y+1)/(x+10) and 0.2 = (y+2)/(x+10). Subtracting: 0.05 = 1/(x+10) ⇒ x + 10 = 20 ⇒ x = 10.

ModerateCAT 2022 · Slot 1

Let a and b be natural numbers. If a² + ab + a = 14 and b² + ab + b = 28, then (2a + b) equals:

(1) 8
(2) 9
(3) 7
(4) 10

Show solution

(1) 8. a(a + b + 1) = 14 and b(a + b + 1) = 28. Dividing: a/b = 1/2 ⇒ b = 2a. Substituting: a(3a + 1) = 14 ⇒ 3a² + a − 14 = 0 ⇒ (3a + 7)(a − 2) = 0 ⇒ a = 2 (natural). Then b = 4, so 2a + b = 8.

ModerateCAT 2022 · Slot 2

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

(1) 1480
(2) 1440
(3) 1200
(4) 1420

Show solution

(2) 1440. 4-digit (> 2000): first digit ∈ {2,3,4,5} → 4 × 5 × 4 × 3 = 240. 5-digit: 5 × 5 × 4 × 3 × 2 = 600. 6-digit: 5 × 5 × 4 × 3 × 2 × 1 = 600. Total = 240 + 600 + 600 = 1440.

HardCAT 2022 · Slot 2TITA

For any real number x, let [x] be the largest integer less than or equal to x. If Σ (n = 1 to N) ⌊1/5 + n/25⌋ = 25, then N is

Show solution

44. Note that ⌊1/5 + n/25⌋ = ⌊(5 + n)/25⌋. This equals 0 for n = 1…19 (since 5 + n < 25) and 1 for n = 20…44 (since 25 ≤ 5 + n < 50). To get a total of 25 ones we need n running from 20 to 44, i.e. N = 19 + 25 = 44.

ModerateCAT 2023 · Slot 1

If x and y are real numbers such that x² + (x − 2y − 1)² = −4y(x + y), then the value x − 2y is

(1) 1
(2) 2
(3) −1
(4) 0

Show solution

(1) 1. x² + (x − 2y − 1)² = −4y(x + y) ⇒ x² + 4xy + 4y² + (x − 2y − 1)² = 0 ⇒ (x + 2y)² + (x − 2y − 1)² = 0. Both squares zero ⇒ x − 2y − 1 = 0 ⇒ x − 2y = 1.

HardCAT 2023 · Slot 1

If √(5x + 9) + √(5x − 9) = 3(2 + √2), then √(10x + 9) =

(1) 3√31
(2) 2√7
(3) 3√7
(4) 4√5

Show solution

(3) 3√7. The book equates 5x + 9 = 36 and 5x − 9 = 18, giving 5x = 27. Then 10x + 9 = 54 + 9 = 63, so √(10x + 9) = √63 = 3√7.

ModerateCAT 2023 · Slot 1

The number of all natural numbers up to 1000 with non-repeating digits is

(1) 738
(2) 648
(3) 504
(4) 585

Show solution

(1) 738. 1-digit: 9. 2-digit: 9 × 9 = 81. 3-digit (non-repeating): 9 × 9 × 8 = 648. Total = 9 + 81 + 648 = 738.

HardCAT 2023 · Slot 3

The population of a town in 2020 was 1,00,000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was

(1) 73,000
(2) 75,000
(3) 74,000
(4) 72,000

Show solution

(1) 73,000. Overall change positive with x = y + 10: net % ≈ (x − y) − xy/100 = 10 − y(y+10)/100 > 0 ⇒ y(y + 10) < 1000. The largest y satisfying this is 27. Lowest 2021 population = 100000 × (1 − 27/100) = 73,000.

HardCAT 2023 · Slot 2

Any non-zero real numbers x, y such that y ≠ 3 and x/y < (x + 3)/(y − 3), will satisfy the condition

(1) If y > 10, then −x > y
(2) x/y < y/x
(3) If x < 0, then −x < y
(4) If y < 0, then −x < y

Show solution

(4) If y < 0, then −x < y. x/y < (x+3)/(y−3) ⇒ x/y − (x+3)/(y−3) < 0 ⇒ [xy − 3x − xy − 3y] / [y(y−3)] < 0 ⇒ −3(x + y) / [y(y−3)] < 0. If y < 0 then y(y − 3) > 0, so the inequality forces (x + y) > 0 ⇒ y > −x, i.e. −x < y.

HardCAT 2023 · Slot 3

The value of 1 + (1 + 1/3)·(1/4) + (1 + 1/3 + 1/9)·(1/16) + (1 + 1/3 + 1/9 + 1/27)·(1/64) + …, is

(1) 15/8
(2) 15/13
(3) 16/11
(4) 27/12

Show solution

(3) 16/11. Regroup by collecting like powers of 1/4 (= 1/4, 1/16, 1/64 …) against the geometric pieces 1, 1/3, 1/9, 1/27 …: S = [1 + 1/4 + 1/16 + …] + (1/3)[1/4 + 1/16 + …] + (1/9)[1/16 + …] + … Each inner bracket is a geometric series with ratio 1/4. Summing the outer geometric series of ratio (1/3·1/4) and the leading 4/3 factor, the book simplifies S = 1 + (1/3)·(1/(1 − 1/12)) ·… = 48/33 = 16/11.

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 1 TITA

The sum of all four-digit numbers that can be formed with the distinct non-zero digits a, b, c, and d, with each digit appearing exactly once in every number, is 153310 + n, where n is a single digit natural number. Then, the value of (a + b + c + d + n) is

Show solution

31. With 4 distinct digits, 24 numbers are formed and each digit appears 24/4 = 6 times in each place ⇒ sum = 6 × 1111 × (a + b + c + d) = 6666 × S. Since 6666 × 23 = 153318, we get S = 23 and n = 8, so a + b + c + d + n = 23 + 8 = 31.

ModerateCAT 2024 · Slot 1 TITA

For any natural number n, let aₙ be the largest integer not exceeding √n. Then the value of a₁ + a₂ + … + a₅₀ is

Show solution

217. aₙ = k when k² ≤ n < (k+1)². Count by value: k=1 for n=1-3 (3 terms), k=2 for 4-8 (5), k=3 for 9-15 (7), k=4 for 16-24 (9), k=5 for 25-35 (11), k=6 for 36-48 (13), k=7 for 49-50 (2). Sum = 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217.

ModerateCAT 2024 · Slot 1

A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is

(A) 50
(B) 36
(C) 42
(D) 18

Show solution

(C) 42. Work backwards from empty stock. Before the 3rd customer: x/2 − 3 = 0 ⇒ x = 6; before the 2nd: x/2 − 3 = 6 ⇒ x = 18; before the 1st: x/2 − 3 = 18 ⇒ x = 42. Initial quantity = 42 kg.

HardCAT 2025 · Slot 2

Find the sum of digits of the number 625⁶⁵ × 128³⁶.

(A) 20
(B) 25
(C) 30
(D) 35
(E) 40

Show solution

(B) 25. 625⁶⁵ = 5²⁶⁰ and 128³⁶ = 2²⁵². Product = 2²⁵² × 5²⁵² × 5⁸ = 10²⁵² × 390625, i.e. 390625 followed by 252 zeros. Digit sum = 3 + 9 + 0 + 6 + 2 + 5 = 25.

HardCAT 2025 · Slot 3 TITA

If 12¹²ˣ × 4²⁴ˣ⁺¹² × 5²ʸ = 8⁴ᶻ × 20¹²ˣ × 243³ˣ⁻⁶, where x, y and z are natural numbers, then x + y + z equals

Show solution

112. Write every base in primes 2, 3, 5 and equate exponents on both sides. Solving the resulting system for natural numbers gives x = 10, y = 60, z = 42, so x + y + z = 112.

ModerateCAT 2025 · Slot 3

The sum of all the digits of the number (10⁵⁰ + 10²⁵ − 123), is

(A) 21
(B) 221
(C) 324
(D) 255

Show solution

(B) 221. 10²⁵ − 123 = 22 nines followed by 877 (a 25-digit number). Adding 10⁵⁰ gives 1, then 25 zeros, then 22 nines, then 877. Digit sum = 1 + 9 × 22 + 8 + 7 + 7 = 1 + 198 + 22 = 221.

ModerateCAT 2025 · Slot 3

For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is:

(A) 811
(B) 735
(C) 3289
(D) 4078

Show solution

(A) 811. Let digits be a (thousands), b, c (tens), d (units). c = d + 6; b + c + d = 16 ⇒ b = 10 − 2d; a + b + c = 15 ⇒ a = d + 1. Single-digit constraints give d = 1 (number 1682) or d = 2 (number 2493). Difference = 2493 − 1682 = 811.