Question

Find the smallest interger such that if dividied by 2 leaves a remainder of 1 , if divided by 3 leaves a remainder of 2, if divided by 4 leaves a remainder of 3 and if divided by 5 leaves a remainder of 4

Answer 1

Instinctively look for an odd prime as there is a remainder when divided by odd or even numbers.
So choose the smallest prime greater than the greates divisor, in this case 7 and continue.
Not 5 as divised by 3 leaves a remainder of 1 not 2.
Now try 11

Not 11 as leaves remainder 1 when divided by 5
The answer is 59

This method is called trial and improvement but there is a better way using modulus mathematics.

Answer 2

this number must be of the form 5n + 4 where n is odd eg 29, 39, 49 - it must also end in 9.
59 - yep 59