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

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<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math>
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math>
  
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== Solution ==
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== Solution 1==
  
 
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:
 
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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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 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}</math>
 
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 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}</math>
  
 +
==Solution 2==
 +
 +
A simpler way to tackle this problem without all that modding is to keep the equation as:
 +
 +
<cmath>7*a^{5b}c^{5d} = 11y^{13}</cmath>
 +
 +
As stated above, <math>a</math> and <math>c</math> must be the factors 7 and 11 in order to keep <math>x</math> at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
 +
 +
<cmath>7^{5b+1}11^{5d-1}=y^{13}</cmath>
 +
 +
The above equation means that <math>y</math> must also contain only the factors 7 and 11 (again, in order to keep <math>x</math> at a minimum), so we end up with:
 +
 +
<cmath>7^{5b+1}11^{5d-1}=7^{13h}11^{13j}</cmath>
 +
 +
(<math>h</math> and <math>j</math> are arbitrary variables placed in order to show that <math>y</math> could have more than just one 7 or one 11 as factors)
 +
 +
Since 7 and 11 are prime, we know that <math>5b+1=13h</math> and <math>5d-1=13j</math>. The smallest positive combinations that would work are <math>b=5,h=2</math> and <math>d=8,j=3</math>. Therefore, <math>a+b+c+d=7+5+11+8=31</math>. <math>\boxed{\textbf{(B)}\ 31}</math> is correct.
 +
 +
==Solution 3==
 +
 +
Another way to solve this problem solve for x. First, we can divide both sides by 7 to get:
 +
 +
<cmath>x^5 = \frac{11y^{13}}{7}</cmath>
 +
 +
Next, we take the fifth root on both sides, which gets us:
 +
 +
<cmath>x = \sqrt[5]{\frac{11y^{13}}{7}}</cmath>
 +
 +
<cmath>x = y^2 \cdot \sqrt[5]{\frac{11y^{3}}{7}}</cmath>
 +
 +
Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let <math>y = 11^3 \cdot 7^2</math>(Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in:
 +
 +
<cmath>x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}</cmath>
 +
 +
<cmath>x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot 11^9 \cdot 7^6}</cmath>
 +
 +
<cmath>x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{11^{10} \cdot 7^5}</cmath>
 +
 +
<cmath>x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1</cmath>
 +
 +
<cmath>x = 7^5 \cdot 11^ 8</cmath>
 +
 +
This gets us <math>a^cb^d = 7^5\cdot 11^8</math>, so <math>a + b + c + d = 7 + 5 + 11 + 8 = \boxed{\textbf{(B)}\ 31}</math>
 +
~lucaswujc, <math>\LaTeX</math> help from Technodoggo
 +
 +
==Solution 4 (easy)==
 +
According to the problem, we have that <math>x^5</math> and <math>y^{13}</math> must be a multiple of both <math>7</math> and <math>11</math>. Thus, in their prime factorisation, there must be <math>7</math> and <math>11</math>. Thus, we have <math>a=7</math> and <math>b=11</math>. Now, let <math>x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}</math>.
 +
We can now divide both sides by 11 in our original equation <math>7x^5=11y^{13}</math> to get <math>y^{13}=7^{5c+1}11^{5d-1}</math>. As we are only considering integers, we have <cmath>5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)</cmath> and <cmath>5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).</cmath>
 +
 +
We can apply brute force to solve for <math>c</math> and <math>d</math> as the numbers aren't big. For the first congruence, we find that <math>25</math> is the smallest number that satisfies, thus <math>c=5</math>. For the second congruence, we find that <math>40</math> is the smallest number that satisfies, thus <math>d=8</math>. Summarising, we get <math>a+b+c+d=7+11+5+8=\boxed{\textbf{(D)}~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}}

Latest revision as of 11:07, 7 September 2023

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 1

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:

\[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\]

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 fifth 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 $7^{13n - 1}$. Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$, respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ 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:

\[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\]

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

Solution 2

A simpler way to tackle this problem without all that modding is to keep the equation as:

\[7*a^{5b}c^{5d} = 11y^{13}\]

As stated above, $a$ and $c$ must be the factors 7 and 11 in order to keep $x$ at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:

\[7^{5b+1}11^{5d-1}=y^{13}\]

The above equation means that $y$ must also contain only the factors 7 and 11 (again, in order to keep $x$ at a minimum), so we end up with:

\[7^{5b+1}11^{5d-1}=7^{13h}11^{13j}\]

($h$ and $j$ are arbitrary variables placed in order to show that $y$ could have more than just one 7 or one 11 as factors)

Since 7 and 11 are prime, we know that $5b+1=13h$ and $5d-1=13j$. The smallest positive combinations that would work are $b=5,h=2$ and $d=8,j=3$. Therefore, $a+b+c+d=7+5+11+8=31$. $\boxed{\textbf{(B)}\ 31}$ is correct.

Solution 3

Another way to solve this problem solve for x. First, we can divide both sides by 7 to get:

\[x^5 = \frac{11y^{13}}{7}\]

Next, we take the fifth root on both sides, which gets us:

\[x = \sqrt[5]{\frac{11y^{13}}{7}}\]

\[x = y^2 \cdot \sqrt[5]{\frac{11y^{3}}{7}}\]

Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let $y = 11^3 \cdot 7^2$(Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in:

\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}\]

\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot 11^9 \cdot 7^6}\]

\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{11^{10} \cdot 7^5}\]

\[x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1\]

\[x = 7^5 \cdot 11^ 8\]

This gets us $a^cb^d = 7^5\cdot 11^8$, so $a + b + c + d = 7 + 5 + 11 + 8 = \boxed{\textbf{(B)}\ 31}$ ~lucaswujc, $\LaTeX$ help from Technodoggo

Solution 4 (easy)

According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$. Thus, in their prime factorisation, there must be $7$ and $11$. Thus, we have $a=7$ and $b=11$. Now, let $x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}$. We can now divide both sides by 11 in our original equation $7x^5=11y^{13}$ to get $y^{13}=7^{5c+1}11^{5d-1}$. As we are only considering integers, we have \[5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)\] and \[5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).\]

We can apply brute force to solve for $c$ and $d$ as the numbers aren't big. For the first congruence, we find that $25$ is the smallest number that satisfies, thus $c=5$. For the second congruence, we find that $40$ is the smallest number that satisfies, thus $d=8$. Summarising, we get $a+b+c+d=7+11+5+8=\boxed{\textbf{(D)}~31}$.

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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