Here's the story. is a positive integer. When is divided by 3, the quotient is with remainder 1. When is divided by 4, the quotient is with remainder 1. When is divided by 5, the quotient is with remainder 1. All three of the numbers, are positive integers.
Find the smallest possible value for that makes all this work.
Solution
All four of the numbers, have a minimal property. is smallest such that when divided by 3 the quotient is and remainder 1. has a similar property: the divisor is 4 but the remainder is also 1. Ditto for except the divisor is 5 and the quotient is . When we come to there is no division requirement. Good deal! will be the easiest of all the numbers to discover. All that has to be is a positive integer. So we can let , the smallest positive number period.
Now, all that has to do is have quotient with remainder 1. Start with . Increasing by steps of 5, so as not to disturb the "when divided by 5" part, we reach in one step and so arrive at the smallest number that when divided by 5 gives and has remainder 1. This is .
To find Q, we start again with 1, but add 4's this time until we reach a number which, when divided by 4, gives as quotient and still has remainder 1. This would be Thus .
Finally, starting again with 1, we add 3's this time to reach a number that gives when divided by 3 with remainder 1. This would be . This is the smallest number satisfying all the requirements of the problem.
The answer is .
Check: 76/3 = 25, remainder 1. 25/4 = 6, remainder 1. 6/5 = 1, remainder 1.
If you're wondering what this has to do with China, this problem is an example of the Chinese Remainder Theorem, which can be explored on Google.