2006 iTest Problems/Problem 13

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Problem

Suppose that $x,  y,  z$ are three distinct prime numbers such that $x  +  y  +  z  =  49$. Find the maximum possible value for the product $xyz$.

$\text{(A) } 615 \quad \text{(B) } 1295 \quad \text{(C) } 2387 \quad \text{(D) } 1772 \quad \text{(E) } 715 \quad \text{(F) } 442 \quad \text{(G) } 1479 \quad \\ \text{(H) } 2639 \quad \text{(I) } 3059 \quad \text{(J) } 3821 \quad \text{(K) } 3145 \quad \text{(L) } 1715 \quad \text{(M) } \text{none of the above} \quad$

Solution

The prime numbers less than 49 are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$. The largest must be at least $\lceil \tfrac{49}{3} \rceil = 17$. Afterward, we can perform casework on the largest number. One of the cases ais $19 \cdot 17 \cdot 13 = 4199$, which is larger than all of the given numeric answer choices. Thus, the answer is $\boxed{\textbf{(M)}}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 12
Followed by:
Problem 14
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