Difference between revisions of "1989 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
− | By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence, | + | Note that <math>n</math> is even, since the <math>RHS</math> consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence, |
<center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center> | <center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center> | ||
<center><math>4 \equiv n\pmod{5}</math></center> | <center><math>4 \equiv n\pmod{5}</math></center> | ||
Line 11: | Line 11: | ||
<center><math>0 \equiv n\pmod{3}</math></center> | <center><math>0 \equiv n\pmod{3}</math></center> | ||
− | Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5. It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n | + | Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5. It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n \geq 174</math>. It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>144</math>. |
== See also == | == See also == |
Revision as of 22:11, 30 November 2009
Problem
One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that . Find the value of
.
Solution
Note that is even, since the
consists of two odd and two even numbers. By Fermat's Little Theorem, we know
is congruent to
modulo 5. Hence,
![$3 + 0 + 4 + 7 \equiv n\pmod{5}$](http://latex.artofproblemsolving.com/9/a/e/9ae6c19669910fdcb79f9ce3947ec835f29f0733.png)
![$4 \equiv n\pmod{5}$](http://latex.artofproblemsolving.com/0/1/3/01340d89f298149cc82f56f121d194fd13476f30.png)
Continuing, we examine the equation modulo 3,
![$-1 + 1 + 0 + 0 \equiv n\pmod{3}$](http://latex.artofproblemsolving.com/c/4/c/c4ce449e7b124ba4e09da16e3c97742a85d3ad9c.png)
![$0 \equiv n\pmod{3}$](http://latex.artofproblemsolving.com/1/2/d/12dfbd3baa056d0fc22f91f6486c78285ec22d0b.png)
Thus, is divisible by three and leaves a remainder of four when divided by 5. It's obvious that
, so the only possibilities are
or
. It quickly becomes apparent that 174 is much too large, so
must be
.
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |