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> | ||
| + | |||
| + | <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 | ||
