1963 AHSME Problems/Problem 8

Problem

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


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

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 is a multiple 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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