## Alfred’s Tough Cash Request

Alfred is at the bank to cash his \$200 check. He tells the cashier he would like some one dollar bills, ten times as many two dollar bills and the rest in fives.

How many of each denomination does the cashier need to give Alfred?

Five \$1 bills, 50 \$2 bills and 19 \$5 bills.

We know that in order to give the rest of the amount in fives, the sum of the one and two dollar bills needs to be divisible by five (i.e. end in 0 or 5).

If we start with a single one dollar bill, we’d need ten two dollar bills to satisfy the request, making \$21. But we need a sum that is divisible by 5. So we keep going up, like so:

\$1 + \$2 * 10 = \$21
\$2 + \$2 * 20 = \$42
\$3 + \$2 * 30 = \$63
\$4 + \$2 * 40 = \$84
\$5 + \$2 * 50 = \$105 (Aha! It’s divisible by 5)
\$6 + \$2 * 60 = \$126
\$7 + \$2 * 70 = \$147
\$8 + \$2 * 80 = \$168
\$9 + \$2 * 90 = \$189

So the only option that works is 5 \$1 bills and 50 \$2 bills, leaving \$95 (95 / 5 = 19) to be paid out in 19 fives.

Alfred is one tough customer.

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