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Problem 152

AMC 10/12, early questions

Number theory Difficulty 3.7 Find the answer

Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$

Your answer

Official solutions — 2

Solution 1

The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{\textbf{(D) }160}$

Solution 2

1. Prime Factorization of 27000:
$27000 = 27 \times 1000 = 3^3 \times 10^3 = 3^3 \times (2 \times 5)^3 = 3^3 \times 2^3 \times 5^3$
The prime factorization of 27000 is $2^3 \times 3^3 \times 5^3$ .

2. Finding Three Pairwise Relatively Prime Numbers:
Since the three numbers must be pairwise relatively prime, each number must be composed of different prime factors. We can assign the prime factors to three different numbers:
$a = 2^3 = 8, \quad b = 3^3 = 27, \quad c = 5^3 = 125$

3. Verification:
- The product of these numbers is:
$8 \times 27 \times 125 = 27000$
- They are pairwise relatively prime because:
$\gcd(8, 27) = 1, \quad \gcd(8, 125) = 1, \quad \gcd(27, 125) = 1$

4. Sum of the Numbers:
$8 + 27 + 125 = 160$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.