2003 AMC 12B Problems/Problem 18

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 $a+b+c+d=7+11+5+8=\boxed{\text{(D)}31}$ .

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