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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.

enter image description here

So , from here.

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

9+x = 9+x

But x gets cancel out.

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  • $\begingroup$ 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=$2+13k$. $\endgroup$ – Gareth Ma Mar 14 at 7:49
  • $\begingroup$ Yes @GarethMa . A 3 digit number which on dividing by 13 gives 11 as remainder. $\endgroup$ – Srijan M.T Mar 14 at 7:50
  • $\begingroup$ @GarethMa Just like 9 /2 = 4 , not 1 which is remainder . Similarly , 100+x / 13 is not equal to 11. $\endgroup$ – Srijan M.T Mar 14 at 7:53
  • $\begingroup$ $104 = 8\times 13$ so leaving remainder $11$, which is the same as $-2$, we just take $102$ so $x=2$. $\endgroup$ – Henno Brandsma Mar 14 at 9:33
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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!

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

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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 $7\times 13+11=102$.

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