Difference between revisions of "1979 USAMO Problems/Problem 1"
Jerry122805 (talk | contribs) (→Solution 1) |
Pi is 3.14 (talk | contribs) (→Solution 2) |
||
Line 9: | Line 9: | ||
<cmath>n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}</cmath> | <cmath>n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}</cmath> | ||
− | 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 == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=4778 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Latest revision as of 01:27, 17 January 2021
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:
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
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.