Question 223183: what is the lowest number possible that when divided by 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 always has one left over, or a remainder of one? thanks a bunch.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Well first off, if you divide ANY number by one, you'll ALWAYS get a remainder of 0 (since 1 goes into every number evenly). So I'm assuming that the set of numbers should be 2, 3, 4, 5, 6, 7, 8, 9, and 10 (otherwise, there is no answer).
Note: I'm going to ignore the obvious answer of 1 and assume that this number is larger than each value in the group.
Take note that 2, 3, 4, 5, 6, 7, 8, 9, and 10 all go into the LCM which is 2520. So 2520 divided by any number in this group leaves a remainder of 0. So just add 1 to this value to get 2521. This new number will leave a remainder of 1 if you divide it by any number in that group. So 2521 is the smallest number (except for 1) that leaves a remainder of 1 when dividing by these given numbers.
Note: Other values that leave a remainder of 1 can be generated by adding on multiples of 2520.
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