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CAT 2023 Slot 2 — QA Question 4

IndicesEasy

Let a, b, m and n be natural numbers such that a > 1 and b > 1. If ambn = 144145, then the largest possible value of n - m is

Answer & solution

  • 579

  • B

    580

  • C

    289

  • D

    290

Solution

Easy

Prime-factorise 144145144^{145}. To make n−mn-m as large as possible, load almost everything into the term with the big exponent (force that exponent to be nn) and keep mm as small as 11.

1

Factorise the right side:

144=122=(24⋅32)⇒ 144145=(24⋅32)145=2580⋅3290\begin{aligned} &144=12^2=(2^4\cdot 3^2)\\ &\Rightarrow\ 144^{145}=(2^4\cdot 3^2)^{145}=2^{580}\cdot 3^{290} \end{aligned}
2

Maximise nn, minimise mm. The largest exponent available is 580580 (on the base 22). Make that the bnb^n term and bundle the rest into ama^m with m=1m=1:

a=3290, m=1b=2, n=580\begin{aligned} &a=3^{290},\ m=1\\ &b=2,\ n=580 \end{aligned}

Both a>1a\gt 1 and b>1b\gt 1, as required, and ambn=3290⋅2580=144145a^m b^n=3^{290}\cdot 2^{580}=144^{145}. ✓

3

Compute n−mn-m:

n−m=580−1=579n-m=580-1=579
n−m=579— option (a).n-m=\mathbf{579}\quad\text{— option (a).}

Don't chase n=580n=580 and mm small separately — they pull together. The trick is realising aa may itself be a prime power (32903^{290}), letting mm collapse to 11 while nn takes the biggest exponent.

CAT 2023 Slot 2 QA Q4: Let a, b, m and n be natural numbers such that a > 1 and b > 1. If a m b n = 144 145 , then the largest possib — Solution | TheCATExam