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?



Answers 2


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
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
HVxvejjw
October 10, 2019 07:16 AM

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