Difference between revisions of "1960 AHSME Problems/Problem 25"
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==See Also== | ==See Also== | ||
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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 |