66 66/16 = 4 and a remainder of 2 66/18 = 3 and a remainder of 12
36 divided by 28 would be 1 remainder 8and44 divided by 32 would be 1 remainder 12
62 is the smallest number that leaves a remainder of 2 when divided by 3, 4, 5, and 6.
449 is the smallest number that satisfies those requirements.
They are the whole numbers that when divided by 2 leaves no remainders
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.
L.C.M of 17 and 29 is 493.493 + 3 = 496Answer = 496
If the primes are 5 or greater, then the remainders are 1 or 5.This is so trivially obvious.The remainder cannot be 0 or else the number is divisible by 6 and so not a prime.The remainder cannot be 2 or 4 or else the number is divisible by 2 and so not a prime.The remainder cannot be 3 or else the number is divisible by 3 and so not a prime.That just leaves 1 and 5: 11 leaves a remainder of 5, 13 leaves 1 for example.
6If a divisor of 51 leaves a quotient of 40, the dividend was 2040.If 2040 is divided by 17, the quotient left is 120.From superscot85: All very interesting but nothing to do with the question, which was about remainders, not quotients.Any number which leaves 40 on division by 51 must leave a remainder of 6 on division by 17, as 51 is 3 x 17 and 40 is (2 x 17) + 6.
LCM of 1 - 6 is 60 so 61 would leave a remainder of 1, but doesn't divide by 11. Try other multiples of 60 and add 1: 120 + 1 = 121 which is an exact multiple of 11, and there you have it!
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
71 : Python code that demonstrates this. N = 1 while True : if N % 8 8 : print N break N += 1