MBApyq

Powers of Three Consecutive Integers

CAT 2008 Slot 1 · Quantitative Ability · Medium · Number System

This is a medium Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Number System. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

All CAT 2008 Slot 1 questionsMore Number System questionsAll CAT QA

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say mm, of these three integers?

  1. A.

    1≤m≤31 \le m \le 3

  2. B.

    4≤m≤64 \le m \le 6

  3. C.

    7≤m≤97 \le m \le 9

  4. D.

    10≤m≤1210 \le m \le 12

  5. E.

    13≤m≤1513 \le m \le 15

Answer

A

Explanation

Let the three consecutive positive integers be n−1,n,n+1n - 1, n, n + 1. Sum of powers: (n−1)1+n2+(n+1)3(n - 1)^1 + n^2 + (n + 1)^3.

Sum of integers =(n−1)+n+(n+1)=3n= (n - 1) + n + (n + 1) = 3n.

Given: (n−1)+n2+(n+1)3=(3n)2(n - 1) + n^2 + (n + 1)^3 = (3n)^2 n−1+n2+n3+3n2+3n+1=9n2n - 1 + n^2 + n^3 + 3n^2 + 3n + 1 = 9n^2 n3−5n2+4n=0  ⟹  n(n−1)(n−4)=0n^3 - 5n^2 + 4n = 0 \implies n(n - 1)(n - 4) = 0

Since nn is a positive integer greater than 1, n=4n = 4.

The three integers are 3,4,53, 4, 5. The minimum is m=3m = 3. This satisfies 1≤m≤31 \le m \le 3.

Related Number System questions

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace