Let n and m be two positive integers such that there are exactly 41 integers greater than 8^(m) and less than 8ⁿ , which can be expressed as powers of 2. Then, the smallest possible value of n+m is

CAT 2023 Slot 3 · Quant · Question 2

CAT 2023 Slot 3 previous year question on Number of Factors, with the answer and a step-by-step solution.

Number of FactorsNumber SystemMultiple choiceMedium

Let n and m be two positive integers such that there are exactly 41 integers greater than {8}^{m} and less than {8}^{n}, which can be expressed as powers of 2. Then, the smallest possible value of n+m is

A
14
B
42
C
16
D
44

Pick an option, then check.

Answer & solution

Answer: Option C

Write both bounds as powers of 2: {8}^{m}={2}^{3m} and {8}^{n}={2}^{3n}.

The powers of 2 strictly between them are {2}^{3m+1}, {2}^{3m+2}, and so on up to {2}^{3n-1}.

Their exponents run from 3m+1 to 3n-1, so the count is \left( 3n-1 \right)-\left( 3m+1 \right)+1.

Simplifying, the count is 3n-3m-1.

Setting this equal to 41 gives 3n-3m=42.

Dividing by 3, n-m=14.

To make n+m as small as possible, make m as small as possible.

Since m is a positive integer, the smallest value is m=1, which forces n=15.

Then n+m=15+1=16 — the third option.

Shortcut. Counting whole numbers strictly between two exponents is just \left( \text{upper}-\text{lower}\right)-1, so 3n-3m-1=41 comes out in a line. Note the gap n-m=14 is fixed, so the sum is minimised at the smallest allowed m — the answer cannot be 14, since that would need m=0.

CAT 2023 Slot 3 · Quant · Question 2

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