Find two integers with Chinese remainder theorem

Question

Find two intergers having remainders 3, 11, 15 when divided by 10, 13, 17, respectively.

I found one is $1103$. But I'm confused about the other one.

Answer 1

You want a number $1103+a$ such that

$$\dfrac{1103+a}{10} = \dfrac{1103}{10}+\dfrac{a}{10}$$

has remainder 3,

$$\dfrac{1103+a}{13} = \dfrac{1103}{13}+\dfrac{a}{13}$$

has remainder 11, and

$$\dfrac{1103+a}{17} = \dfrac{1103}{17}+\dfrac{a}{17}$$

has remainder 15.

Since you know the remainders of $\dfrac{1103}{10}, \dfrac{1103}{13}, \dfrac{1103}{17}$ already match the remainders you desire, it must be that $10$ divides $a$, $13$ divides $a$, and $17$ divides $a$. The least common multiple of those divisors is $10\cdot 13\cdot 17$. So, the set of numbers with the remainders you are looking for would be:

$$1103+10\cdot 13\cdot 17k, k\in \mathbb{Z}$$

Choose any two.