Difference between revisions of "2010 AMC 12B Problems/Problem 9"
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Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>? | Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>? | ||
Latest revision as of 16:41, 15 February 2021
Problem
Let 
be the smallest positive integer such that is divisible by 
, 
is a perfect cube, and 
is a perfect square. What is the number of digits of ?
Solution
We know that 
and 
. Cubing and squaring the equalities respectively gives 
. Let 
. Now we know 
must be a perfect 
-th power because 
, which means that must be a perfect 
-th power. The smallest number whose sixth power is a multiple of is 
, because the only prime factors of are 
and 
, and 
. Therefore our is equal to number 
, with 
digits 
.
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
