Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

The smallest such number is 1055.

We want to find such that

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

Taking everything together, we end up with the system

Now the moduli are coprime and we can apply the CRT.

We start with

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since , the inverse of 2 is 4.

So, we have to adjust to

and from the CRT we find

so that the general solution for all integers .

We want a 4 digit solution, so we want

which gives .

ok

step-by-step explanation:

4444444444444 i think