Difference between revisions of "1969 AHSME Problems/Problem 19"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | Let <math>(xy)^2=a</math>. The expression given is equal to <math>a^2-10a+9=0</math>, which can be factored as <math>(a-9)(a-1)=0</math>. Thus, we have <math>a=(xy)^2=9</math> and <math>a=(xy)^2=1</math>. Because <math>x</math> and <math>y</math> are positive, we can eliminate the possibilities where <math>xy</math> is negative. From here it is easy to see that the only integral pairs of <math>x</math> and <math>y</math> are <math>(3, 1), (1, 3)</math>, and <math>(1, 1)</math>. The answer is <math>\fbox{B}</math>. |
== See also == | == See also == | ||
{{AHSME 35p box|year=1969|num-b=18|num-a=20}} | {{AHSME 35p box|year=1969|num-b=18|num-a=20}} | ||
| − | [[Category: | + | [[Category: Introductory Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 15:14, 10 July 2015
Problem
The number of distinct ordered pairs 
where 
and 
have positive integral values satisfying the equation 
is:
Solution
Let 
. The expression given is equal to 
, which can be factored as 
. Thus, we have 
and 
. Because and are positive, we can eliminate the possibilities where 
is negative. From here it is easy to see that the only integral pairs of and are 
, and 
. The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
