2003 AMC 12B Problem 18

Attempt Problem 18 of the 2003 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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18.

Let

x

and

y

be positive integers such that

7x^5 = 11y^{13}.

The minimum possible value of

x

has a prime factorization

a^c b^d.

What is

a + b + c + d?

30

31

32

33

34

The minimum

x

uses only the primes

7

and

11,

so write

x = 7^c 11^d

Match exponents of

7

and

11:

need

5c + 1 \equiv 0 \pmod{13}

and

5d \equiv 1 \pmod{13}

For the minimum

x,

neither

x

nor

y

has prime factors other than

7

and

11.

Write

x = 7^c 11^d,

7x^5 = 7^{5c+1} 11^{5d}.

Writing

y = 7^m 11^n,

we need

7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.

Matching exponents:

5c + 1 \equiv 0 \pmod{13}

gives the least

c = 5,

and

5d \equiv 1 \pmod{13}

gives the least

d = 8.

a = 7,

b = 11,

and

\begin{aligned} &a + b + c + d \\ &= 7 + 11 + 5 + 8 = 31. \end{aligned}

Thus, the correct answer is B.

Problem 18 in other years