Difference between revisions of "2010 AMC 12B Problems/Problem 9"
AuraAmbience (talk | contribs) m |
(Added CRT solution) |
||
(7 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | == Problem == |
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>? | ||
Line 5: | Line 5: | ||
== 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>. | ||
+ | |||
+ | == Solution 2 (Chinese Remainder Theorem) == | ||
+ | Let <math>n = 2^2 \cdot 5 \cdot x</math> for some integer <math>x</math>, then we know that | ||
+ | \begin{align*} | ||
+ | n^2 = 2^4 \cdot 5^2 \cdot x^2 = m_0^3\\ | ||
+ | n^3 = 2^6 \cdot 5^3 \cdot x^3 = m_1^2 | ||
+ | \end{align*} | ||
+ | Since we have <math>2^4 \cdot 5^2 \cdot x^2</math> a power of 3, that means, by the Fundamental Theorem of Arithmetic, the exponents of the primes must each be divisible by 3. This also means that there's no point in <math>p \mid x</math> if <math>p \not \in \{2,3\}</math>, since that would just serve to increase <math>n</math> and we want <math>n</math> to be minimal. | ||
+ | |||
+ | This means we can write <math>x = 2^{\alpha} \cdot 5^{\beta}</math> and thus we have | ||
+ | <math>2^4 \cdot 5^2 \cdot 2^{2\alpha} \cdot 5^{2\beta} = 2^{4 + 2\alpha} \cdot 5^{2 + 2\beta} = m_0^3</math>, therefore we know | ||
+ | \begin{align*} | ||
+ | 4 + 2\alpha \equiv 0 \pmod{3}\\ | ||
+ | 2 + 2\beta \equiv 0 \pmod{3} | ||
+ | \end{align*} | ||
+ | and doing the same with the third equation being a square, we have | ||
+ | \begin{align*} | ||
+ | 6 + 3\alpha \equiv 0 \pmod{2}\\ | ||
+ | 3 + 3\beta \equiv 0 \pmod{2} | ||
+ | \end{align*} | ||
+ | We can shift this around into two systems involving the same variables, as follows: | ||
+ | \begin{align*} | ||
+ | \alpha \equiv 0 \pmod{2}\\ | ||
+ | 1 + 2\alpha \equiv 0 \pmod{3} | ||
+ | \end{align*} | ||
+ | and | ||
+ | \begin{align*} | ||
+ | 1 + \beta \equiv 0 \pmod{2} \\ | ||
+ | 2 + 2\beta \equiv 0 \pmod{3}\\ | ||
+ | \end{align*} | ||
+ | And using [[Chinese Remainder Theorem]] to solve this, we get <math>\alpha = 4</math> and <math>\beta = 5</math>, and plugging that back in yields <math>n = 2^6 \cdot 5^6</math>. | ||
+ | |||
+ | Thus, our answer is <math>\lfloor{ \log_{10}(2^6 \cdot 5^6) + 1 \rfloor} \Rightarrow \boxed {E}</math>. | ||
+ | |||
+ | ~ <math>\color{magenta} zoomanTV</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}} |
Latest revision as of 15:52, 6 May 2024
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
.
Solution 2 (Chinese Remainder Theorem)
Let for some integer
, then we know that
\begin{align*}
n^2 = 2^4 \cdot 5^2 \cdot x^2 = m_0^3\\
n^3 = 2^6 \cdot 5^3 \cdot x^3 = m_1^2
\end{align*}
Since we have
a power of 3, that means, by the Fundamental Theorem of Arithmetic, the exponents of the primes must each be divisible by 3. This also means that there's no point in
if
, since that would just serve to increase
and we want
to be minimal.
This means we can write and thus we have
, therefore we know
\begin{align*}
4 + 2\alpha \equiv 0 \pmod{3}\\
2 + 2\beta \equiv 0 \pmod{3}
\end{align*}
and doing the same with the third equation being a square, we have
\begin{align*}
6 + 3\alpha \equiv 0 \pmod{2}\\
3 + 3\beta \equiv 0 \pmod{2}
\end{align*}
We can shift this around into two systems involving the same variables, as follows:
\begin{align*}
\alpha \equiv 0 \pmod{2}\\
1 + 2\alpha \equiv 0 \pmod{3}
\end{align*}
and
\begin{align*}
1 + \beta \equiv 0 \pmod{2} \\
2 + 2\beta \equiv 0 \pmod{3}\\
\end{align*}
And using Chinese Remainder Theorem to solve this, we get
and
, and plugging that back in yields
.
Thus, our answer is .
~
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.