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.
Related exercises:
- Let(xi,yi),i = 1,2,3,4,5, be a set of five distinct points with integer coordinates in the xy plane. Show that the midpoint of the line joining at least one pair of these points has integer coordinates.
- Show that if f is a function from S to T, where S and T are finite sets with |S| > |T |, then there are elements s1 and s2 in S such that f(s1)=f(s2), or in other words, f is not one-to-one
- Let d be a positive integer. Show that among any group of d + 1 (not necessarily consecutive) integers there are two with exactly the same remainder when they are divided by d
- A bowl contains 10 red balls and 10 blue balls. A woman selects balls at random without looking at them
- A drawer contains a dozen brown socks and a dozen black socks, all unmatched. A man takes socks out at random in the dark.
