Difference between revisions of "1998 AJHSME Problems/Problem 10"
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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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== See also == | == See also == | ||
− | {{AJHSME box|year=1998| | + | {{AJHSME box|year=1998|num-b=9|num-a=11}} |
* [[AJHSME]] | * [[AJHSME]] | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:29, 4 July 2013
Problem
Each of the letters , , , and represents a different integer in the set , but not necessarily in that order. If , then the sum of and is
Solution
There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set .
has factor .
has factors and
has factors and
has factors , , and .
From here, we note that even though all numbers have the factor , only has another factor other than in the set (ie. )
We could therefore have one fraction be and another .
The sum of the numerators is
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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