Q: What is the smallest number which leaves remainders 2 and 12 when divided by 16 and 18 respectively?

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64 and 76

3

It is: 3647

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.

0.25

119

449 is the smallest number that satisfies those requirements.

23

They are the whole numbers that when divided by 2 leaves no remainders

Try 30

85

234568

41

61

46235462354623546235

12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.

121.

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.

duckweed

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