Difference between revisions of "1991 AIME Problems/Problem 1"
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=== Solution 2 === | === Solution 2 === | ||
− | Since <math>xy + x + y + 1 = 72</math>, this can be factored to <math>(x + 1)(y + 1) = 72</math>. As <math>x</math> and <math>y</math> are [[integer]]s, the possible sets for <math>(x,y)</math> (ignoring cases where <math>x > y</math> since it is symmetrical) are <math>(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)</math>. The second equation factors to <math>(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11</math>. The only set with a factor of <math>11</math> is <math>(5,11)</math>, and checking shows that it is | + | Since <math>xy + x + y + 1 = 72</math>, this can be factored to <math>(x + 1)(y + 1) = 72</math>. As <math>x</math> and <math>y</math> are [[integer]]s, the possible sets for <math>(x,y)</math> (ignoring cases where <math>x > y</math> since it is symmetrical) are <math>(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)</math>. The second equation factors to <math>(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11</math>. The only set with a factor of <math>11</math> is <math>(5,11)</math>, and checking shows that it is correct. |
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=== Solution 3 === | === Solution 3 === |
Revision as of 08:15, 21 July 2020
Problem
Find if
and
are positive integers such that


Contents
[hide]Solution
Solution 1
Define and
. Then
and
. Solving these two equations yields a quadratic:
, which factors to
. Either
and
or
and
. For the first case, it is easy to see that
can be
(or vice versa). In the second case, since all factors of
must be
, no two factors of
can sum greater than
, and so there are no integral solutions for
. The solution is
.
Solution 2
Since , this can be factored to
. As
and
are integers, the possible sets for
(ignoring cases where
since it is symmetrical) are
. The second equation factors to
. The only set with a factor of
is
, and checking shows that it is correct.
Solution 3
Let ,
then we get the equations
After finding the prime factorization of
, it's easy to obtain the solution
. Thus
Note that if
, the answer would exceed
which is invalid for an AIME answer.
~ Nafer
Solution 4
From the first equation, we know . We factor the second equation as
. Let
and rearranging we get
. We have two cases: (1)
and
OR (2)
and
. We find the former is true for
.
.
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.