Difference between revisions of "1963 AHSME Problems/Problem 8"
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Keerthanaa (talk | contribs) (→Problem) |
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\textbf{(C)}\ 1260^2 \qquad | \textbf{(C)}\ 1260^2 \qquad | ||
\textbf{(D)}\ 7350 \qquad | \textbf{(D)}\ 7350 \qquad | ||
| − | \textbf{( | + | \textbf{(6)}\ 44100 </math> |
==Solution== | ==Solution== | ||
Revision as of 17:47, 3 January 2024
Problem
The smallest positive integer 
for which 
, where 
is an integer, is:
Solution
Factoring 
results in 
. If an integer 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 to result in a perfect cube is 
, which is answer choice 
.
See Also
| 1963 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
