Difference between revisions of "1998 AJHSME Problems/Problem 10"

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== See also ==
 
== See also ==
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
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{{AJHSME box|year=1998|num-b=9|num-a=11}}
 
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* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]

Revision as of 22:59, 9 June 2011

Problem 10

Each of the letters $\text{W}$, $\text{X}$, $\text{Y}$, and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$, but not necessarily in that order. If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$, then the sum of $\text{W}$ and $\text{Y}$ is

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution

There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{ 1,2,3,4\}$.

$1$ has factor $1$.

$2$ has factors $1$ and $2$

$3$ has factors $1$ and $3$

$4$ has factors $1$, $2$, and $4$.

From here, we note that even though all numbers have the factor $1$, only $4$ has another factor other than $1$ in the set (ie. $2$)

We could therefore have one fraction be $\frac{4}{2}$ and another $\frac{3}{1}$.

The sum of the numerators is $4+3=7=\boxed{E}$

See also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AJHSME/AMC 8 Problems and Solutions