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Difference between revisions of "2010 AMC 12B Problems/Problem 9"

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== Solution ==
 
== Solution ==
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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>7</math> digits <math>\Rightarrow \boxed {E}</math>.
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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>. <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 sixth 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=8|num-a=10|ab=B}}
 
{{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}}

Revision as of 16:29, 18 February 2012

Problem 9

Let $n$ be the smallest positive integer such that $n$ is divisible by $20$, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution

We know that $n^2 = k^3$ and $n^3 = m^2$. Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$. Let $a = n^6$. $a$ must be a perfect $36$-th power because $lcm(9,4) = 36$, which means that $n$ must be a perfect sixth power. The smallest number whose sixth power is a multiple of $20$ is $10$, because the only prime factors of $20$ are $2$ and $5$, and $10 = 2 \times 5$. Therefore our is equal to number $10^6 = 1000000$, with $7$ digits $\Rightarrow \boxed {E}$.

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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