# »  Solutions to puzzles in my VDARE.com monthly Diary

## December 2016

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My December diary included the following brainteaser

I found pennies, nickels and dimes for a total of 18 coins. The total monetary value was 73 cents. How many of each denomination did I find?

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• Solution

I noted that: "As usual in math there is an elegant solution and a merely enumerative one." This is true; but as one reader pointed out, the elegant solution is actually longer than the rock-breaking one.

Writing them out on the page here, in fact, I'm not sure which is which.

The odd 3 cents have to be pennies. Any further pennies must come in fives. So there can be

• 3 cents (with 15 nickels and dimes adding to 70 cents); or
• 8 cents (with 10 nickels and dimes adding to 65 cents); or
• 13 cents (with 5 nickels and dimes adding to 60 cents).

The first and last don't work because 15 nickels > 70 cents while 5 dimes < 60 cents. So there are 8 pennies.

That leaves 65 cents in ten other coins. That means an odd number of nickels.

One nickel … no. Three nickels … no. Five nickels … no. Seven nickels … there you go, 7 nickels and three dimes.

Alternatively:

• p+n+d = 18,
• p+5n+10d = 73.

Subtracting the first from the second:  4n+9d = 55.

So 55-9d is a multiple of 4; so d is odd, say d = 2k+1; then 46-18k is a multiple of 4, i.e. 23-9k is even.

That's only true when k = 1; so d = 3, giving n = 7. Then p must be 8.