Find the smallest number such that

(i) it leaves a remainder 2 when divided by 3 ;

(ii) it leaves a remainder 3 when divided by 5 ;

(iii) it leaves a remainder 5 when divided by 7 .

Guest Jun 30, 2022

#1**+1 **

By trial and error find the smallest number to satisfy (i) and (ii) is 8.

Then go on by trialing numbers of 8 + 15n since 15 is the least common multiple of 3 and 5.

You will find the answer to be 68 within just a few tries.

There should be a better and more official way by modulo arithmetic though so it would be great if someone can also add that to this thread.

OrangeJuicy Jun 30, 2022

#2**0 **

We have:

\(x = 2 (\mod 3)\)

\(x = 3( \mod 5)\)

\(x = 5 ( \mod 7)\)

Using the Chinese Remainder Theorem, we have \(x = 68 ( \mod 105)\), so the smallest value that works is \(\color{brown}\boxed{68}\)

BuilderBoi Jun 30, 2022