To solve this exercise, we will use the fact of how many reminders of the division there can be when dividing by a certain number.
If you divide by n, you can get n possible reminders. These are 0, 1, 2, …, n-1.
If we have a certain set of n consecutive integers, we can write them as follows:
x, x+1, x+2, x+3, …, x+n-1
Every term of the sequence above is different than the rest. We make use of the premise “n consecutive integers”.
Because they are consecutive, when we divide every term of the sequence by n, we should get a different reminder.
In other words: x mod n ≠ (x+1) mod n ≠ (x+2) mod n≠ … ≠ (x+n-1) mod n
Because there are n terms of the sequence, all of them different and consecutive, when we divide all the terms by n, we can obtain exactly n reminders.
Then, exactly 1 reminder must be 0.
Therefore, there must be exactly one number in that sequence that is divisible by n.
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