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CAT QuanthardNumber System

Number System: Remainders, Factors and Divisibility

Updated 22 October 2026

Five questions on remainders, factors, divisibility, HCF and LCM, and last digits.

Question 1 · Type in the answer (+3 / 0)

What are the last two digits of 3²⁰²⁶ + 7²⁰²⁶? Enter the two-digit number.

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Answer: 78

  1. 7⁴ = 2,401 ≡ 1 (mod 100). 2026 = 4 × 506 + 2, so 7²⁰²⁶ ≡ 7² = 49 (mod 100).
  2. 3¹⁰ = 59,049 ≡ 49 (mod 100), so 3²⁰ ≡ 49² = 2,401 ≡ 1 (mod 100).
  3. 2026 = 20 × 101 + 6, so 3²⁰²⁶ ≡ 3⁶ = 729 ≡ 29 (mod 100).
  4. Sum ≡ 49 + 29 = 78 (mod 100). The last two digits are 78.
Question 2 · Multiple choice (+3 / −1)

How many divisors of 10,800 are multiples of 15 but are not perfect squares?

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Answer: B. 27

  1. 10,800 = 2⁴ × 3³ × 5².
  2. A divisor that is a multiple of 15 has the form 2ᵃ × 3ᵇ × 5ᶜ with a = 0–4, b = 1–3, c = 1–2: 5 × 3 × 2 = 30 divisors.
  3. Of these, perfect squares need a even (0, 2, 4), b even (2 only) and c even (2 only): 3 × 1 × 1 = 3 divisors (225, 900 and 3,600).
  4. Required count = 30 − 3 = 27.
Question 3 · Type in the answer (+3 / 0)

What is the smallest positive integer that leaves remainders 3, 4 and 5 when divided by 5, 6 and 7 respectively, and is also divisible by 11? Enter the number.

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Answer: 418

  1. In each case the remainder is 2 less than the divisor, so N + 2 is divisible by 5, 6 and 7.
  2. LCM(5, 6, 7) = 210, so N = 210k − 2.
  3. 210 = 11 × 19 + 1, so 210k − 2 ≡ k − 2 (mod 11). For divisibility by 11, k ≡ 2 (mod 11).
  4. The smallest such k is 2, giving N = 420 − 2 = 418.
  5. Check: 418 = 11 × 38; 418 leaves 3, 4 and 5 on division by 5, 6 and 7.
Question 4 · Multiple choice (+3 / −1)

What is the largest integer n such that 12ⁿ divides 50!?

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Answer: B. 22

  1. 12ⁿ = 2²ⁿ × 3ⁿ, so we need the powers of 2 and 3 in 50!.
  2. Power of 2: 25 + 12 + 6 + 3 + 1 = 47. Power of 3: 16 + 5 + 1 = 22.
  3. From 2: 2n ≤ 47, so n ≤ 23. From 3: n ≤ 22.
  4. The tighter limit wins: n = 22.
  5. Taking only the factor 2 into account gives 23; checking every prime of the base is essential.
Question 5 · Multiple choice (+3 / −1)

Two numbers have HCF 12 and LCM 2,520. How many unordered pairs of such numbers exist in which both numbers have three digits?

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Answer: B. 2

  1. Write the numbers as 12x and 12y with x and y co-prime. Then 12xy = 2,520, so xy = 210 = 2 × 3 × 5 × 7.
  2. Each prime goes wholly to x or to y, giving 2⁴ ÷ 2 = 8 unordered co-prime pairs: (1, 210), (2, 105), (3, 70), (5, 42), (6, 35), (7, 30), (10, 21), (14, 15).
  3. Both numbers have three digits when 100 ≤ 12x, 12y ≤ 999, i.e. 9 ≤ x, y ≤ 83.
  4. Only (10, 21) and (14, 15) qualify, giving (120, 252) and (168, 180).
  5. The answer is 2. Forgetting the three-digit condition gives 8.

Full worked solution

  1. Last two digits: find a power that is ≡ 1 (mod 100), e.g. 7⁴ and 3²⁰, then reduce the exponent.
  2. Number of divisors of p^a × q^b × r^c is (a + 1)(b + 1)(c + 1); restrict exponent ranges for 'multiple of' or 'perfect square' conditions.
  3. Same deficit remainders: if N leaves (d − k) for each divisor d, then N = LCM × m − k.
  4. Power of a prime p in n! = ⌊n/p⌋ + ⌊n/p²⌋ + …; for a composite base, check each prime and take the minimum.
  5. HCF h and LCM l: numbers are hx and hy with x, y co-prime and xy = l/h; the number of co-prime splits is 2^(k − 1) for k distinct primes.