Question

Consider the following numbers 3, 6, 9, 12,...,75. Show that if we pick 15 arbitrary numbers from them, then we will find two that have sum equal to 81.

Answer 1

Answer:

See below.

Step-by-step explanation:

There are a total of (75-3) / 3 + 1 =  25 numbers in the sequence. Now 24  of them can be paired so as to get a total of 81:

6 and 75, 9 and 72, 12 and 69............. 45 and 36, 42 + 39. That is 12 pairs with the number 3 the odd one out.

If we pick 15 random numbers there is no way we cannot pick one of these pairs which add up to 81.  For example, if we pick  12 of the first numbers in the pairs and the number 3, there are 2 numbers left to pick and these 2 would be bound to match up with  2 of the numbers of the first 12.

If we pick only 6 of the first numbers in the pairs and the 3, there are another  8 numbers left to pick so at least 2 of these are bound to match up

with some of the the first 6. If we do not pick the 3 then at least 3 will match.

Whatever combination we pick will have  1 or more pairs which add up to 81.