Division and Remainder
Number Theory
Level
3
What the sum of the three smallest positive integers (greaters than 1), that when these numbers are divided by 1, 2, 3, 4, 5, 6 or 7 leaves remainder 1?
The answer is 2523.
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1 solution
First of all, an integer divided by 1 always has remainder of 0, so I just ignored the first "1" in the problem.
We can start by changing the "remainder 1" to "remainder 0" so that we have an easier problem - find the three smallest positive integers that are evenly divided by 2, 3, 4, 5, 6, and 7. In other words, find the smallest integers with 2, 3, 4, 5, 6, and 7 as factors. This involves finding the LCM, which is 420 . This number and all its multiples (840, 1260, ...) satisfy the conditions. It should be noted that finding the LCM of numbers doesn't necessarily mean multiplying everything.
Going back to the original problem, leaving a remainder of 1 means adding 1 to the integers we found. So the final answer would be 421 + 841 + 1261 = 2523