Difference between revisions of "2001 JBMO Problems/Problem 1"
Rockmanex3 (talk | contribs) (Solution to Problem 1 — mods are a game changer) |
Rockmanex3 (talk | contribs) (Completely FIXED the solution) |
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==Solution== | ==Solution== | ||
− | Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math> Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a,b,c \equiv 1 \pmod{9}.</math> | + | Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math> Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a^3,b^3,c^3 \equiv 1 \pmod{9}.</math> Thus, <math>a,b,c \equiv 1 \pmod{3},</math> and the only numbers congruent to <math>1</math> modulo <math>3</math> are <math>1,4,7,10.</math> |
<br> | <br> | ||
− | + | [[WLOG]], let <math>a \ge b \ge c.</math> That means <math>a^3 \ge b^3, c^3</math> and <math>3a^3 \ge 2001.</math> Thus, <math>a^3 \ge 667,</math> so <math>a = 10.</math> | |
+ | |||
+ | <br> | ||
+ | Now <math>b^3 + c^3 = 1001.</math> Since <math>b^3 \ge c^3,</math> we find that <math>2b^3 \ge 1001.</math> Thus, <math>b = 10</math> and <math>c = 1.</math> | ||
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+ | <br> | ||
+ | In summary, the only solutions are <math>\boxed{(10,10,1),(10,1,10),(1,10,10)}.</math> | ||
==See Also== | ==See Also== |
Revision as of 21:47, 11 August 2018
Problem
Solve the equation in positive integers.
Solution
Note that for all positive integers the value
is congruent to
modulo
Since
we find that
Thus,
and the only numbers congruent to
modulo
are
WLOG, let That means
and
Thus,
so
Now Since
we find that
Thus,
and
In summary, the only solutions are
See Also
2001 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |