Difference between revisions of "1967 AHSME Problems/Problem 24"
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| − | {{AHSME box|year=1967|num-b=23|num-a=25}} | + | {{AHSME 40p box|year=1967|num-b=23|num-a=25}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 00:40, 16 August 2023
Problem
The number of solution-pairs in the positive integers of the equation 
is:
Solution
We have 
. Thus, 
must be a positive multiple of 
. If 
, we find our first positive multiple of . From there, we note that 
will always return a multiple of for . Our first solution happens at 
.
We now want to find the smallest multiple of that will work. If , then we have 
, or 
. When 
, the expression is equal to 
, and when 
, the expression is equal to 
, which will no longer work.
Thus, all integers from 
to will generate an that will be a positive integer, and which will in turn generate a 
that is also a positive integer. So, the answer is 
.
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
