Math Aunty

I have a large bag of gold coins, and, still, some are worth more than others.  Only now my bag contains an arbitrary number of coins, call it .  Again, one coin is worth one dollar, one is worth two dollars, and so on (making the last one worth  dollars).  What is the largest number of charities to which I can donate given that I again want to send in an equal amount of dollars to each one?  How much money will each charity receive?  Justify your answers.

Finally, if you want some help on this week’s problem, I will HAPPILY provide some, especially if you actually make a charitable donation.  The more you donate, the more help I’ll provide.”

Using 10 as an example, Casey would be able to donate to 5 charities $11 a piece because there are 5 pairs of coins and if x = 1 and y = 10, (x+n) + (y – n) = 11 where n is any integer less than or equal to 5.

Going farther and using 11, Casey would have 5 pairs of coins which if paired correctly can equal 11 , and then another coin worth 11… so he could donate $11 to  5+1 charities = 6.

So to think about it in the more general sense:

if x = 1 and y equals the number of coins in the bag,

(x + n) + (y – n) = z (where z is the amount of coins that Casey can donate)

x + n + y – n = z

x + y = z

1 + 2n = z So casey can donate one more than the number of coins he has evenly. He can donate to the number of pairs of charities if the number of coins is even so he can donate to n/2 charities $(n+1) if n is the number of coins.

If the number of coins are odd, there are n+1/2 pairs of coins. Those pairs of coins will add up to the value of the odd coin, but he can also donate the odd coin to another charity because it by itself is another donation equal to the value of the others.

So he will be able to donate to n+1/2 charities, $n if n is the number of coins and it is odd.

And now for fun!