Difference between revisions of "1979 USAMO Problems/Problem 1"
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We notice that the unit digits of the LHS of this equation should equal to <math>F_{16}</math>. In base <math>16</math>, the only unit digits of fourth powers are <math>0</math> and <math>1</math>. Thus, the maximum of these <math>14</math> terms is 14 <math>1's</math> or <math>E_{16}</math>. Since <math>E_{16}</math> is less than <math>F_{16}</math>, there are no integral solutions for this equation. | We notice that the unit digits of the LHS of this equation should equal to <math>F_{16}</math>. In base <math>16</math>, the only unit digits of fourth powers are <math>0</math> and <math>1</math>. Thus, the maximum of these <math>14</math> terms is 14 <math>1's</math> or <math>E_{16}</math>. Since <math>E_{16}</math> is less than <math>F_{16}</math>, there are no integral solutions for this equation. | ||
| − | == Video Solution == | + | == Video Solution by OmegaLearn== |
https://youtu.be/zfChnbMGLVQ?t=4778 | https://youtu.be/zfChnbMGLVQ?t=4778 | ||
Latest revision as of 03:45, 21 January 2023
Problem
Determine all non-negative integral solutions 
if any, apart from permutations, of the Diophantine Equation 
.
Solution 1
Recall that 
for all integers 
. Thus the sum we have is anything from 0 to 14 modulo 16. But 
, and thus there are no integral solutions to the given Diophantine equation.
Solution 2
In base 
, this equation would look like:
![\[n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}\]](http://latex.artofproblemsolving.com/4/f/7/4f781cb87004eba8361ae0e450dbdd301bc29832.png)
We notice that the unit digits of the LHS of this equation should equal to 
. In base , the only unit digits of fourth powers are 
and 
. Thus, the maximum of these 
terms is 14 
or 
. Since is less than , there are no integral solutions for this equation.
Video Solution by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=4778
~ pi_is_3.14
See Also
| 1979 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
