Let n be a positive integer. Click and drag the steps to their corresponding step numbers to show that in any set of n consecutive integers there is exactly one divisible by n. Step 1 Therefore exactly one of the numbers in our collection is divisible by n. Subtract x+) from x+i to get j-i which is divisible by n. Step 2 Check if the integers cover all the possible remainders or not. Step 3 But j-i is a positive integer strictly greater than n. Denote the n consecutive integers as x + 1,x+ 2,. . .,x+n, where x is some integer Step 4 But j- i is a positive integer strictly less than n. Step 5 If they do not, apply pigeonhole principle, so that at least one remainder must occur twice. Let x+i and x+j have the same remainder Step 6 Subtract x+ti from x+j to get j-i which is divisible by n