Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I couldn't find any online source that validated this, so I thought I would put this up here.

Is it true that in order to determine the remainder of a very large number $n$ when divided with a particular number we could simply take the digit at the units of the large number and divide it by the divisor and that would give us the remainder ?

A simple example $\frac{1676}{5}$ means $\frac{6}{5}$ this means we get a remainder 1. Is this method valid ? Or is this for only certain cases and unsafe to use generally ?

share|improve this question
1  
11/4 has remainder 3, not 1. –  chris Jul 16 '12 at 4:27

2 Answers 2

up vote 3 down vote accepted

It does not work in general; it only works then the divisor is $1$, $2$, $5$, or $10$ (the reason being that these numbers divide $10$).

For an example where it does not work, consider $27$ divided by $7$. If we only looked at $7$ on $27$, we would conclude that the remainder is $0$. But the remainder is $6$.

The reason it works for $5$ (and it would work for $1$, $2$, or $10$), is that if you divide by $10$, the remainder is precisely the units digit (the rightmost, or least-significant digit). And dividing by $5$ is the same as first dividing by $10$ and then dividing by $2$, so you can just look at the last digit. Similarly with division by $2$ and by $1$.

Similarly, to find the remainder when dividing something by $4$, you could just look at the last two digits, because $4$ divides $100$, but the last digit would not be enough (for example, $15$ leaves a remainder of $3$, but $5$ leaves a remainder of $1$).

In general, if the divisor $q$ divides $10^n$, then the remainder is the same as the remainder of the last $n$ digits of the original number.

All of this becomes very clear when you learn about modular arithmetic.

share|improve this answer
    
For 1,2,5,10 It depends on last digit. And for 4,20,25,50 and 100 the remainder depends on last two digits of the numerator. The explanation for this same as above. –  Manty Jul 16 '12 at 6:16
    
Hello @Arturo while reviewing this post ,I cant figure the part where you said In general, if the divisor q divides $10^n$, then the remainder is the same as the remainder of the last n digits of the original number How do we get n=1 when divisor is 5 and n=2 when divisor is 4 –  MistyD Jul 30 '12 at 16:16

Divisibility of divisor depends on UNIT DIGIT - for factors of $10$.

Divisibility of divisor depends on LAST $2$ DIGITs - for factors of $10^2$.

Divisibility of divisor depends on LAST $3$ DIGITs - for factors of $10^3$.

and so on..

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.