# 1963 AHSME Problems/Problem 8

## Problem

The smallest positive integer $x$ for which $1260x=N^3$, where $N$ is an integer, is:

$\textbf{(A)}\ 1050 \qquad \textbf{(B)}\ 1260 \qquad \textbf{(C)}\ 1260^2 \qquad \textbf{(D)}\ 7350 \qquad \textbf{(6)}\ 44100$

## Solution

Factoring $1260$ results in $2^2 \cdot 3^2 \cdot 5 \cdot 7$. If an integer $N$ is a perfect cube, then the exponents of all the primes in its prime factorization are multiples of 3. Thus, the smallest positive integer that can be multiplied by $1260$ to result in a perfect cube is $2 \cdot 3 \cdot 5^2 \cdot 7^2 = 7350$, which is answer choice $\boxed{\textbf{(D)}}$.

## See Also

 1963 AHSC (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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