Difference between revisions of "1992 AHSME Problems/Problem 13"
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== Solution 1 == | == Solution 1 == | ||
− | <math>\fbox{C}</math> We can rewrite the left-hand side as <math>\frac{a^2b+a}{b+ab^2}</math> by multiplying the numerator and denominator by <math>ab</math>. Now we can multiply | + | <math>\fbox{C}</math> We can rewrite the left-hand side as <math>\frac{a^2b+a}{b+ab^2}</math> by multiplying the numerator and denominator by <math>ab</math>. Now we can multiply the top part of the fraction by the denominator to give <math>a^2b+a = 13b + 13ab^2 \implies ab(a-13b)=13b-a = -(a-13b) \implies (ab+1)(a-13b) = 0</math>. Now, as <math>a</math> and <math>b</math> are positive, we cannot have <math>ab+1=0</math>, so we must have <math>a-13b=0 \implies a=13b</math>, so the solutions are <math>(13,1), (26,2), (39,3), (52,4), (65,5), (78,6), (91,7)</math>, which is 7 ordered pairs. |
== Solution 2 == | == Solution 2 == |
Latest revision as of 14:53, 21 October 2021
Contents
Problem
How many pairs of positive integers (a,b) with satisfy the equation
Solution 1
We can rewrite the left-hand side as by multiplying the numerator and denominator by . Now we can multiply the top part of the fraction by the denominator to give . Now, as and are positive, we cannot have , so we must have , so the solutions are , which is 7 ordered pairs.
Solution 2
The equation is equivalent to or . The s cancel out leaving us with . This is asking us for the positive multiples of up to , and we know that , so our answer is .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AHSME Problems and Solutions |
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