Difference between revisions of "1963 AHSME Problems/Problem 8"
Rockmanex3 (talk | contribs) (Solution to Problem 8) |
m (→Problem: fixed letter E) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
| − | Factoring <math>1260</math> results in <math>2^2 \cdot 3^2 \cdot 5 \cdot 7</math>. If an integer <math>N</math> is a perfect cube, then the exponents of all the primes in its [[prime factorization]] | + | Factoring <math>1260</math> results in <math>2^2 \cdot 3^2 \cdot 5 \cdot 7</math>. If an integer <math>N</math> 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 <math>1260</math> to result in a perfect cube is <math>2 \cdot 3 \cdot 5^2 \cdot 7^2 = 7350</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 08:16, 17 October 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. 
