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Difference between revisions of "2003 AMC 12B Problems/Problem 18"

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== Solution ==
 
== Solution ==
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Substitute <math>a^cb^d</math> into <math>x</math>. We then have <math>7(a^{5c}b^{5d}) = 11y^{13}</math>. Divide both sides by <math>7</math>, and it follows that:
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<cmath>(a^{5c}b^{5d}) = \[\frac{11y^{13}}{7}.</cmath>
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Note that because <math>11</math> and <math>7</math> are prime, the minimum value of <math>x</math> must involve factors of <math>7</math> and <math>11</math> only. Thus, we try to look for the lowest power <math>p</math> of <math>11</math> such that <math>13p + 1 \equiv 0 \pmod{5}</math>, so that we can take <math>11^{13p + 1}</math> to the 5th root. Similarly, we want to look for the lowest power <math>n</math> of <math>7</math> such that <math>13n - 1 \equiv 0 \pmod{5}</math>. Again, this allows us to take the fifth root of <math>7^{13n - 1}</math>. With these conditions satisfied, we can simply multiply <math>11^{13p + 1}</math> and <math>7^{13n - 1}</math> and substitute this quantity into <math>y</math> to attain our answer.
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We can simply look for suitable values for <math>p</math> and <math>n</math>. We find that the lowest <math>p</math>, in this case, would be <math>3</math> because <math>13(3) + 1 \equiv 0 \pmod{5}</math>. Moreover, the lowest <math>q</math> should be <math>2</math> because <math>13(2) - 1 \equiv 0 \pmod{5}</math>. Hence, we can substitute the quantity <math>11^{3} \cdot 7^{2}</math> into <math>y</math>. Doing so gets us:
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<cmath>(a^{5c}b^{5d}) = \[\frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.</cmath>
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Taking the fifth root of both sides, we are left with <math>a^cb^d = 11^{8} \cdot 7^{5}</math>. <math>a + b + c + d = 11 + 8 + 7 + 5 = \boxed{\textbf{(B)}\ 31}</math>
  
 
== See Also==
 
== See Also==
 
{{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:44, 9 June 2014

Problem

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution

Substitute $a^cb^d$ into $x$. We then have $7(a^{5c}b^{5d}) = 11y^{13}$. Divide both sides by $7$, and it follows that: 

(a^{5c}b^{5d}) = \[\frac{11y^{13}}{7}. (Error compiling LaTeX. Unknown error_msg)

Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$, so that we can take $11^{13p + 1}$ to the 5th root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$. Again, this allows us to take the fifth root of $7^{13n - 1}$. With these conditions satisfied, we can simply multiply $11^{13p + 1}$ and $7^{13n - 1}$ and substitute this quantity into $y$ to attain our answer.


We can simply look for suitable values for $p$ and $n$. We find that the lowest $p$, in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$. Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$. Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$. Doing so gets us: 

(a^{5c}b^{5d}) = \[\frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}. (Error compiling LaTeX. Unknown error_msg)

Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$. $a + b + c + d = 11 + 8 + 7 + 5 = \boxed{\textbf{(B)}\ 31}$

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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