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

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==Problem 10==
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==Problem==
  
 
Each of the letters <math>\text{W}</math>, <math>\text{X}</math>, <math>\text{Y}</math>, and <math>\text{Z}</math> represents a different integer in the set <math>\{ 1,2,3,4\}</math>, but not necessarily in that order.  If <math>\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1</math>, then the sum of <math>\text{W}</math> and <math>\text{Y}</math> is
 
Each of the letters <math>\text{W}</math>, <math>\text{X}</math>, <math>\text{Y}</math>, and <math>\text{Z}</math> represents a different integer in the set <math>\{ 1,2,3,4\}</math>, but not necessarily in that order.  If <math>\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1</math>, then the sum of <math>\text{W}</math> and <math>\text{Y}</math> is
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 23:29, 4 July 2013

Problem

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 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AJHSME/AMC 8 Problems and Solutions

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