Difference between revisions of "1979 USAMO Problems/Problem 1"

(Solution 1)
(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 $(n_1,n_2,\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\cdots +n_{14}^4=1599$.

Solution 1

Recall that $n_i^4\equiv 0,1\bmod{16}$ for all integers $n_i$. Thus the sum we have is anything from 0 to 14 modulo 16. But $1599\equiv 15\bmod{16}$, and thus there are no integral solutions to the given Diophantine equation.

Solution 2

In base $16$, this equation would look like: \[n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}\]

We notice that the unit digits of the LHS of this equation should equal to $F_{16}$. In base $16$, the only unit digits of fourth powers are $0$ and $1$. Thus, the maximum of these $14$ terms is 14 $1's$ or $E_{16}$. Since $E_{16}$ is less than $F_{16}$, there are no integral solutions for this equation.

Video Solution

https://youtu.be/zfChnbMGLVQ?t=4778

~ pi_is_3.14

See Also

1979 USAMO (ProblemsResources)
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. AMC logo.png

Invalid username
Login to AoPS