CAT 2025 Slot 1 QA Q19 Logarithms & Exponents

Question 19 of the Quantitative Ability (QA) section, CAT 2025 Slot 1, with CAT's official answer and a worked solution.

For any natural number \(k\), let \(a_k=3^k\). The smallest natural number \(m\) for which \(\left\{\left(a_1\right)^1 \times\left(a_2\right)^2 \times \ldots \times\left(a_{20}\right)^{20}\right\}<\left\{a_{21} \times a_{22} \times \ldots \times a_{(20+m)}\right\}\), is
  1. A
    59
  2. B
    56
  3. C
    58
  4. D
    57

Answer: (C) 58 Logarithms & Exponents MCQ

Solution

  1. LHS \(= 3^{\sum k \cdot k} = 3^{1^2 + \dots + 20^2} = 3^{2870}\).
  2. RHS \(= 3^{21 + 22 + \dots + (20+m)}\); the exponent is \(\frac{(20+m)(21+m)}{2} - 210\).
  3. We need \((20+m)(21+m) > 2(2870 + 210) = 6160\). Since \(77 \times 78 = 6006\) and \(78 \times 79 = 6162\), \(20 + m = 78\).
  4. So \(m = 58\).

Same concept in other years

See every question on Compare powers by summing exponents or taking logs.

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