An easy four-stuff problem

Number Theory Level 3

There are four numbers $a$ , $b$ , $c$ and $d$ , such that when they are each divided by 5, distinct nonzero remainders are obtained. Is it true that the sum of the numbers is not divisible by 5?

No. Yes.

1 solution

X X
Nov 21, 2018

If $a=1,b=2,c=3,d=4$ , then $a+b+c+d=10$ which is a multiple of 5, so it sometimes is a multiple of five. Hence, the answer is no.

Let $a=1, b=2, c=3, b=5$ , the sum is $1+2+3+5=11$ , which is not a multiple of 5.

Sometimes it is a multiple, sometimes not, generally it works in $\frac 15$ of all cases; exactly when the numbers have remainders $\{1,2,3,4\}$ but not for remainders $\{0,1,2,3\}, \{0,1,2,4\}, \{0,1,3,4\}, \{0,2,3,4\}$ . You cannot say something general about the divisibility.

Henry U - 2 years, 6 months ago

Thanks for telling me, I have editted the solution.

X X - 2 years, 6 months ago

An easy four-stuff problem

1 solution

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