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CAT 2023 Slot 3 — QA Question 2

FactorsEasy

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

Answer & solution

  • A

    44

  • B

    14

  • 16

  • D

    42

Solution

Easy

Everything is a power of 22: 8m=23m8^m=2^{3m} and 8n=23n8^n=2^{3n}. So we just need to count the integer exponents xx with 3m<x<3n3m\lt x\lt 3n and make that count exactly 4141, while keeping m+nm+n as small as possible.

1

Rewrite both bounds as powers of 22:

8m=23m,8n=23n8^m=2^{3m},\qquad 8^n=2^{3n}

A power of 22 strictly between them is 2x2^x with 3m<x<3n3m\lt x\lt 3n.

2

Count the integer exponents. The integers strictly between 3m3m and 3n3n are x=3m+1,…,3n−1x=3m+1,\dots,3n-1, and there are

(3n−1)−(3m+1)+1=3n−3m−1(3n-1)-(3m+1)+1=3n-3m-1

of them. We need this to equal 4141:

3n−3m−1=41 ⇒ 3(n−m)=42 ⇒ n−m=143n-3m-1=41\ \Rightarrow\ 3(n-m)=42\ \Rightarrow\ n-m=14
3

Minimise m+nm+n. Since n=m+14n=m+14,

m+n=m+(m+14)=2m+14m+n=m+(m+14)=2m+14

This is smallest when mm is smallest. As mm is a positive integer, take m=1m=1, giving n=15n=15:

m+n=1+15=16m+n=1+15=16

Smallest possible value of n+m=16n+m=\mathbf{16}.

The gap count depends only on n−mn-m, not on mm itself: 3(n−m)−1=413(n-m)-1=41 fixes n−m=14n-m=14 instantly. Then push mm to its minimum 11 to minimise the sum.

CAT 2023 Slot 3 QA Q2: Let n and m be two positive integers such that there are exactly 41 integers greater than 8 m and less than 8 — Solution | TheCATExam