# 1960 AHSME Problems/Problem 19

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

## Problem

Consider equation $I: x+y+z=46$ where $x, y$, and $z$ are positive integers, and equation $II: x+y+z+w=46$, where $x, y, z$, and $w$ are positive integers. Then $\textbf{(A)}\ \text{I can be solved in consecutive integers} \qquad \\ \textbf{(B)}\ \text{I can be solved in consecutive even integers} \qquad \\ \textbf{(C)}\ \text{II can be solved in consecutive integers} \qquad \\ \textbf{(D)}\ \text{II can be solved in consecutive even integers} \qquad \\ \textbf{(E)}\ \text{II can be solved in consecutive odd integers}$

## Solution

Consider each option, one at a time.

For option A, let $x=y-1$ and $z=y+1$. That means $3y=46$, so $y=\frac{46}{3}$. That is not an integer, so option A is eliminated.

For option B, let $x=y-2$ and $z=y+2$. That also means $3y=46$, so $y=\frac{46}{3}$. That is also not an integer, so option B is eliminated.

For option C, let $y=x+1$, $z=x+2$, and $w=x+3$. That means $4x+6=46$, so $x=10$. Option C works.

For options D and E, let $y=x+2$, $z=x+4$, and $w=x+6$. That means $4x+12=46$, so $x=\frac{17}{2}$. Since the result is not an integer, options D and E are eliminated.

Thus, the answer is $\boxed{\textbf{(C)}}$.

## See Also

 1960 AHSME (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 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 All AHSME Problems and Solutions
Invalid username
Login to AoPS