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nuox95
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
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.
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