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 $\boxed{\textbf{(D)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1963 AHSME Problems/Problem 8

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