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> | |
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+ | <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 |