# Find the value of this 3 digit number such that dividing by it leaves REM = 11

REM = Remainder.

Now , It is easy to think of doing it by dividing the numbers starting from 100 till whatever until you get remainder as 11. But isthere a way to solve it by formula.

Like , what I have doing is and not getting the right answer for it.

$$\frac{100 + x}{13}$$= y (for y it is Some value but we don’t know)and (100+x because it has to be a number either greater or or equal to 100)

Then , 100+x - 13(y) = 11.

I’m not getting how to solve further than this.

One way I tried is this but got wrong ans.

So , from here.

100 + x - 13(6) = 9 + x.

9+x = 9+x

But x gets cancel out.

• So given a divisor, say $k=13$ in your example, you want to find the minimal number (3-digit) such that $REM(x \div k)=11$? (Or more "formally", $x \equiv 11 \mod k$). Your approach is correct that $REM(100+x \div 13)=11 \implies REM(9+x \div 13) = 11 \implies x=2$ will work. There are more answers, for example x=15,28,41,... and in general x=$2+13k$. – Gareth Ma Mar 14 at 7:49
• Yes @GarethMa . A 3 digit number which on dividing by 13 gives 11 as remainder. – Srijan M.T Mar 14 at 7:50
• @GarethMa Just like 9 /2 = 4 , not 1 which is remainder . Similarly , 100+x / 13 is not equal to 11. – Srijan M.T Mar 14 at 7:53
• $104 = 8\times 13$ so leaving remainder $11$, which is the same as $-2$, we just take $102$ so $x=2$. – Henno Brandsma Mar 14 at 9:33

Firstly, the correct notation for remainder $$REM(n \div p) = r$$ is

$$n = pq + r \implies n\equiv r \mod p$$

Examples include $$13\equiv 3\mod 5$$ and $$444 \equiv 3 \mod 7$$

So you want to find the minimum $$n\geq 100$$ such that $$n\equiv 11\mod 13$$

$$n\equiv 11\mod 13, n\geq 100$$

$$n + 100 \equiv 11 \mod 13, n \geq 0$$

$$n+9+13 \cdot 7 \equiv 11\mod 13,n\geq 0$$

Since adding 13 won't change the remainder, we can ignore the $$13\cdot 7$$ term:

$$n+9\equiv 11\mod 13, n\geq 0$$

$$n\equiv 2\mod 13$$

In english terms, this means $$n$$ (the two-digit part) has remainder 2 when divided by $$13$$. The trivial example is $$n=2 \implies 102\equiv 11\mod 13$$, but other examples are $$n=15,28,41,54,\cdots$$. These give $$115,128,141,154,\cdots$$ which all give remainder of $$11$$ when divided by $$13$$.

Hope this helps!

• Thank you very much. – Srijan M.T Mar 14 at 8:36

Actually , I already solved the answer in Q itself but did a silly mistake.

As you can see , I did write 9+x as remainder. So , 9+x = 11 gives x = 2. Therefore , 102 is ans.

Also , one thing to notice is that if at 13*6 = 91 we have remainder = 0.(91/13).

Then , 92/13 gives remainder 1. Therefore , 102 when we reach , we get remainder as 11.

We have:

$$N\equiv 11\bmod 13\equiv -2 \bmod 13$$

Now we consider the least odd prime of 2 greater than 100 which is 7 and we can write:

$$2^7=128=117+11=7\times 13+11+2\times 13$$

$$26=2\times 13$$ can be ignored, so the smallest 3 digit number can be $$7\times 13+11=102$$.