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

Question

kinoNoidae1wj · Accepted Answer

Firstly, the correct notation for remainder

R E M (n \div p) = r

is

n = p q + r \Rightarrow 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 \Rightarrow 102 \equiv 11 mod 13

, but other examples are

n = 15, 28, 41, 54, \dots

. These give 115, 128, 141, 154, ... which all give remainder of 11 when divided by 13.

cinereod3am · Accepted Answer

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 \cdot 6 = 91

we have remainder

= 0 . (\frac{91}{13})

.
Then,

\frac{92}{13}

gives remainder 1. Therefore, 102 when we reach, we get remainder as 11.

Find the value of this 3 digit number