## Friday, May 14, 2010

### Sam Loyd's Cashier Puzzle

This puzzle comes from one of the greatest puzzle makers, Sam Loyd, who lived over 100 years ago. It tells of an old man who stepped into a bank with a check for \$200 and asked the cashier, "Give me some one-dollar bills, ten times as many twos, and the balance in fives!"

How did the cashier perform this transaction?

Submit your answer on our Family Puzzle Marathon Be first to solve three puzzles and get a prize!

Kim said...

Has to be 5 1s, 50 2s and 19 5s... the 1's and 2s gets you \$105 and the \$95 balance can be paid out with 19 5's.

If you try any other number 6 through 9, you don't get a multiple of 5 from your 1s and 2s that give you a nice remainder. And 10 is too high - 10's means 100 2s, and that's more than \$200.

If you try lower numbers (1-4) you also don't get nice remainders. If zero 1s are acceptable, you could do 0 1s, 0 2s and 20 5s, but I don't think zero fits the criteria of "some one-dollar bills".

Anonymous said...

You know the ones have to be a multiply of 5. Using the Guess and Check method, I knew 10 ones would be \$200 in twos so it had to be 5 ones. That makes \$100 in two dollar bills, plus the \$5 in dollar bills, leaving \$95 in five dollar bills which is 19.

kj said...

My first thought was that in Sam Loyd's day there was no two dollar bill (I can remember when the U.S, Mint introduced it in 1976, with much fanfare). So the teller gave out zero two dollar bills, zero one dollar bills and 40 five dollar bills.

But...research on Wikipedia says that the two dollar bill was first circulated in 1862, and remained in circulation until 1966. The U.S. Mint brought it back in 1976, and it remains in circulation today.

So the teller must have given out 5 ones, 50 twos, and 19 fives.

Maria said...

You all are correct. Kim was the first one and she gets her 38th puzzle point. I should come up with some fancy celebration as Kim is approaching fifty (puzzle points :)

Kim said...

Can I be in another puzzle? :-)

Lynnet said...

n= #1 dollar bills gives n dollars
10n= #2 dollar bills gives 20n dollars
+ ___________________
21n dollars

to give balance in \$5 bills, 21n must be a multiple of 5. since 21's factors are 3 and 7 you need a multiple of five to make the product a multiple of 5.

1 dollar bills=5
2 dollar bills=50
5 dollar bills=19

ten ones is to high because then you end up with 210 dollars and no 5s.

Maria said...

Lynnet - you are absolutely correct!

Kim - I am so happy you liked the "put you in a puzzle" trick. Perhaps we should start doing this for every ten puzzles and not only the first ten.