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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 .

 Jun 30, 2022
 #1
avatar+192 
+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.

 Jun 30, 2022
 #2
avatar+2275 
+1

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 \(\color{brown}\boxed{68}\)

 Jun 30, 2022

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