# 2006 iTest Problems/Problem 13

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
Invalid username
Login to AoPS