SOLUTION: What is the smallest positive integer which, when divided by each of 2, 3, 4, 5, 6 and 7, will give in each case a remainder that is one less than the divisor?

Question 1143478: What is the smallest positive integer which, when divided by each of 2, 3, 4, 5, 6
and 7, will give in each case a remainder that is one less than the divisor?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

You are looking for the number N which, when divided by any of 2, 3, 4, 5, 6, or 7, leaves a remainder that is 1 less than the divisor.

With those conditions, the number N+1 will be divisible by each of those divisors.

So the number you are looking for is the smallest positive integer that is 1 less than the least common multiple of 2, 3, 4, 5, 6, and 7.

The LCM of those divisors is 2*2*3*5*7 = 420; the number you are looking for is 419.

CHECK:
419/2 = 209 remainder 1
419/3 = 139 remainder 2
419/4 = 104 remainder 3
419/5 = 83 remainder 4
419/6 = 69 remainder 5
419/7 = 59 remainder 6
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