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Let's assume $a+b+c+d$ is an integer multiple of $5$, that is $$a+b+c+d=5m-------(1)$$ For all integer $m,$ where $5$ does not divide any of $a,b,c$ and $d$. What are all the sets of values of $a,b,c,d$?
The sets I could find are:

$$5w+1, 5x+1, 5y+1, 5z+7.$$ $$5w+1, 5x+1, 5y+4, 5z+4.$$ $$5w+1, 5x+2, 5y+3, 5z+4.$$ $$5w+2, 5x+2, 5y+2, 5z+4.$$ $$5w+2, 5x+2, 5y+3, 5z+3.$$ $$5w+3, 5x+3, 5y+3, 5z+1.$$ $$5w+4, 5x+4, 5y+4, 5z+3.$$ For all integers $w,x,y,z.$

Are the sets I have listed above sufficient for (1)? If not, please give a counter example.

Shuaib Lateef
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  • Does the order matter? And what is the difference between $5z+7$ and $5z+2$? – lulu Jan 14 '22 at 11:45
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    Should say: written this way, it's very hard to check the list. As you only care about the four remainders, I'd just list them. Much cleaner. Any way, you left off ${4,4,4,3}$ in whatever order. – lulu Jan 14 '22 at 11:46
  • The OP asks for all the "sets". So order would not matter. – Oscar Lanzi Jan 14 '22 at 12:06

1 Answers1

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You missed $5w+3,5x+4,5y+4,5z+4$. Also your first set can be rendered with the last term as $5z+2$.

We may render each term as $\in\{1,2,3,4\}\bmod 5$ and require four such numbers to add up to a multiple of $5$. The sum must then be $\in\{5,10,15\}$.

Thus we now have a restricted partitioning problem. Clearly only $1+1+1+2=5$ and $3+4+4+4=15$ meet the restrictions with their respecive sums. The remaining sums must equal $10$ with the minimum term $\in\{1,2\}$ (i e. less than the average which would be $5/2$) and the maximum term $\in\{3,4\}$ (greater than the average). By pairing each possible minimum with each possible maximum we do indeed generate just the last five partitions indicated in the question ($1+1+4+4,1+2+3+4,...$).

Oscar Lanzi
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