1960 AHSME Problems/Problem 19

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 $\boxed{\textbf{(C)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1960 AHSME Problems/Problem 19

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