2 Left behind

lcm of 3, 4, 5, and 6 is 60 but 60 is not a 3 digit no. since 120 is a multiple of 60 120 + 2 = 122

The Least Common multiple of 3, 4, 5, and 6 is 60. This is not a 3 digit number. But the next least common multiple is 120, which is a 3 digit number. If we add 2 to 120, we get 122. Since 2 is not divisible by 3, 4, 5, and 6, 2 will be the remainder for each of these divisions.

LCM (3, 4, 5, 6) = 120 so the smallest number that leaves a remainder of 2 when divided by 3, 4, 5 or 6 is (120 + 2) = 122

I don't know if this is the right Solution. But I solved it By thinking like this.

1. the range is 100 -999.
2. number that give remainder 2 if divide by 5 is 102, 107,112.,... and so on.
3. But the number that odd, cannot give remainder 2 (even) if divide by 4.
4. So the set of the number will be : 102,112,122,...

I hope this usefull.

The Lowest Common Factor (LCM) of 3, 4, 5, 6 is 60. Take 60*2 = 120, which is the smallest 3-digit common factor of 3, 4, 5, 6. So, by adding 2 to 120, you will be able to get a remainder of 2 when the number (122) is divided by 3, 4, 5 and 6.

lcm(3, 4, 5, 6) = 60

Thus need the smallest number greater than or equal to 100 (the first 3-digit number) that is a multiple of 60 plus 2:

100 ÷ 60 = 1 r 40 → smallest multiple of 60 is 60 x 2 = 120 → required number is 120 + 2 = 122

umm...very easy in the beginning as the required number will end either in 2 or 7..if it has to leave a remainder 2..when divided by 5... moreover...if it ends in 7..then it will never leave remainder 2 when divided by 4 so...the last digit of the number is 2...now....this three digit number should be the sum of a three digit number which ends in 0 and divides 4...plus 2...! so we get the list...100+2,120+2,140+2....980+2... now...if the sum of the digits is a multiple of 3......remove all.and keep others..we get a list....120+2,140+2,180+2....,980+2 luckily...the first number in this sequence...is the required number...:D!