Difference between revisions of "2010 AMC 12B Problems/Problem 9"
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− | == Problem | + | == Problem == |
− | Let <math>n</math> be the smallest positive integer such that <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>? |
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math> | <math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math> | ||
== Solution == | == Solution == | ||
+ | We know that <math>n^2 = k^3</math> and <math> n^3 = m^2 </math>. Cubing and squaring the equalities respectively gives <math> n^6 = k^9 = m^4 </math>. Let <math>a = n^6</math>. Now we know <math>a</math> must be a perfect <math>36</math>-th power because <math>lcm(9,4) = 36</math>, which means that <math>n</math> must be a perfect <math>6</math>-th power. The smallest number whose sixth power is a multiple of <math>20</math> is <math>10</math>, because the only prime factors of <math>20</math> are <math>2</math> and <math>5</math>, and <math>10 = 2 \times 5</math>. Therefore our is equal to number <math>10^6 = 1000000</math>, with <math>7</math> digits <math>\Rightarrow \boxed {E}</math>. | ||
== See also == | == See also == | ||
− | {{AMC12 box|year=2010|num-b= | + | {{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}} |
+ | {{MAA Notice}} |
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 |
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