Difference between revisions of "1960 AHSME Problems/Problem 19"

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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=18|num-a=20}}
 
{{AHSME 40p box|year=1960|num-b=18|num-a=20}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:03, 17 May 2018

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 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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