Difference between revisions of "1960 AHSME Problems/Problem 25"
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Revision as of 18:10, 17 May 2018
Problem
Let and be any two odd numbers, with less than . The largest integer which divides all possible numbers of the form is:
Solution
First, factor the difference of squares. Since and are odd numbers, let and , where and can be any integer. Factor the resulting expression. If and are both even, then is even. If and are both odd, then is even as well. If is odd and is even (or vise versa), then is even. Therefore, in all cases, can be divided into all numbers with the form .
This can be confirmed by setting and , making . Since is not a multiple of and is less than , we can confirm that the answer is
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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 |