# Remainder of an integer divided by 5 is the same as the remainder of the division of the rightmost digit by 5 how to prove this?

by ghadah32   Last Updated October 10, 2019 07:20 AM

we have been told in arithmetic that the remainder of an integer divided by 5 is the same as the remainder of the division of the rightmost digit by 5 how to use modular properties to prove this?

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Hint: In decimal place-value notation, $$edcba = 10\times edcb + a$$. More generally, if $$a$$ is the right-most digit of a number $$n$$, then $$n=10\times m + a$$ for an appropriate number $$m$$.

Chris Culter
October 10, 2019 07:16 AM

Consider a number $$\overline{a_1a_2a_3a_4....a_n}$$. Now this can be written as $$a_1\times 10^{n-1}n +a_2\times 10^{n-2}+.....+a_{n-1}\times 10 +a_n$$ Now when you reduce this by $$5$$ you will get some remainder say $$r$$. We have $$a_1\times 10^{n-1}n +a_2\times 10^{n-2}+.....+a_{n-1}\times 10 +a_n\equiv r\pmod5$$ $$\implies a_n\equiv r\pmod5$$.

HVxvejjw
October 10, 2019 07:16 AM

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