Difference between revisions of "2010 AMC 12B Problems/Problem 9"
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== Solution == | == Solution == | ||
| + | A number whose square root is a perfect cube and whose cube root is a perfect square will be in the form <math>a^6</math> where a is an integer, because <math>6</math> is the <math>LCM</math> of <math>2</math> and <math>3</math>. A number that is divisible by <math>20</math> obviously ends in a <math>0</math>. The only way <math>a^6</math> can end in a zero is if <math>a</math> ends in a zero. The smallest number that ends in a <math>0</math> is <math>10</math>, so our number <math>a^6=10^6=1000000</math>, with <math>\boxed {7}</math> digits. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}} | {{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}} | ||
Revision as of 14:47, 3 February 2011
Problem 9
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
A number whose square root is a perfect cube and whose cube root is a perfect square will be in the form 
where a is an integer, because 
is the 
of 
and 
. A number that is divisible by obviously ends in a 
. The only way can end in a zero is if 
ends in a zero. The smallest number that ends in a is 
, so our 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 | |
