This Problem of the Week was pretty cool, and it will be even cooler if I did it right! The question is asking what would happen if you had some arbitrary number of coins, n, in a bag that each had a unique value (1,2,3….all the way up to n), how many charities could you donate to if you wanted to donate with the same money.
First you can start by modeling the problem with 10 coins. So you have 10 coins worth, 1,2,3,4,5,6,7,8,9, and 10 dollars. You can set them up into pairs, with the largest coin and the smallest coin, then the second biggest and the second smallest, and so on. That way, every pair will be the same. You get 10+1, 9+2, 8+3, 7+4, and 6+5: 5 pairs of $11!
And then all of a sudden, I remembered good old Gauss.
Gauss was a german mathematician. Legend has it, that when he was little, one of his especially cruel teachers told him to count up all of the numbers between one and one hundred. Being the clever little squirt that he was (even at that young of age), he did the same thing we just did, but on a bigger scale. He matched up 100 and 1, 99 and 2, and so on. Since he split them into pairs, he realized that there would be 100/2 pairs, or 50 pairs of 101. Therefore, 101 x 50 is 5,050, and you now have the sum of the first 100 hundred digits with pretty much no mathematics needed. Pretty cool right? (Right.)
It gets cooler. We’ll move on to n number of coins..
Okay, so we can still set up pairs. And since there are n coins, we know that there will be n/2 pairs. The least amount a coin is worth would be 1 dollar, so it (n)+1, (n – 1) +2, (n – 2) +3, (n – 3) + 4, and so on. Notice that the bigger numbers become smaller be subtracting 1, then two, then three, and so on. So when you start adding 1, then 2, then 3, you get n + 1 for every pair.
So, all in all, for n coins, you’ll be able to donate to n/2 charities, and donate n + 1 dollars to each charity!
(It even works for 10! 10/2 = 5 pairs, and 10 + 1 = $11 per charity.)
Problem of the Week: Officially Solved.