The sum of 11 positive integers is 65. Show that at least two of the eleven numbers are equal.

**Answer**

Some problems are easier to approach if we start by assuming that the statement we have to prove is false, and then show that this assumption leads to contradiction with the other facts in the problem, or contradiction with mathematical facts in general. The fancy mathematical name for this approach is proof by contradiction (or *reductio ad absurdum*, its Latin name).

For this problem, *let’s assume that the statement we have to prove (i.e. at least two numbers are equal) is false*, and that the eleven numbers are all different.

What do we know about the sum of the first ten positive integers?

1+2+3+4+5+6+7+8+9+10 =55

We can now assume that **the problem involves the smallest 11 consecutive positive integers**: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. Their sum is 66.

1+2+3+4+5+6+7+8+9+10+11 =66

The problem states that the sum has to be 65. We have now the contradiction with the other facts of the problem!

How can we eliminate the difference of 1 between the 65 given in the problem and the 66 we arrived at? We can replace any of the numbers (with the exception of 1^{*}) by a number that is one less than the number being replaced. For example, 8 could be replaced by a 7, but we already have a 7. After every possible replacement we do end up with two equal numbers, which is exactly the statement we had to prove.

** We are working with positive integers, so we cannot have a 0 (zero) as part of the sum. Therefore we cannot replace 1 with 0.*

* *

*Try another problem that requires a similar approach:*

The sum of 16 different whole numbers is 191. Show that at least one of the numbers is a multiple of 3.

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