Three digit number divided by 5, 6 and 9

Question

A three digit number x when divided by 5, 6 and 9 gives a remainder of 1. Which is the sum of all x-es with that property?

Answer 1

So, $x$ will be of the form $1+[5,6,9]t=90t+1$ where $t$ is an integer

We need $$100\le90t+1\le999$$

Answer 2

Are you familiar with the Chinese Remainder Theorem?[1]

Well, I won't use it.

$X = 5M + 1 = 6N + 1 = 9K+1$

$5M = 6N =9K=X-1 = Y$ so

Note: This means $5|Y, 6|Y, 9|Y$ and so the least common multiple of $5,6,9= 90|Y$.

So $X = 90m + 1$.

If $X$ has $3$ digits then $100 \le 90 m + 1 < 1000$

So $99 \le 90m < 999$

$11 \le 10m < 111$ so $2 \le m \le 11$.

and the numbers are

$181,271,361,..., 721, 811, 901,991$.

If we add them together we have:

$181 + .... + 901 + 991 = $

$180 + .... + 900 + 990 + (1+1+1.....+1)=$

$90(2+3+.... + 11) + 10 =$

$90(\frac {11*12}2-1) + 10=$

$5860$

......

[1] The CRT will let you know that there is precisely one unique number between $0$ and the least common multiple of $5,6,9=90$ with remainder $1$ for $5,6,9$ and therefore that number is $1$, and therefore all possible numbers with remainder $1$ for $5,6,9$ will be in the form $90m + 1$.