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Problem 484

AMC 10/12, early questions

Combinatorics Difficulty 3.0 Prove it CEMC Hypatia · Canada · 2013

At the JK Mall grand opening, some lucky shoppers are able to participate in a money giveaway. A large box has been filled with many $5,$ 10, $20, and$ 50 bills. The lucky shopper reaches into the box and is allowed to pull out one handful of bills.

Rad pulls out at least two bills of each type and his total sum of money is $175. What is the total number of bills that Rad pulled out?
Sandy pulls out exactly five bills and notices that she has at least one bill of each type. What are the possible sums of money that Sandy could have?
Lino pulls out six or fewer bills and his total sum of money is $160. There are exactly four possibilities for the number of each type of bill that Lino could have. Determine these four possibilities.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Pulling out two of each type of bill gives Rad, $2\times(\$5+\$10+\$20+\$50)=2\times(\$85)=\$170$ .

Since his total sum of money is 175, the only other bill that Rad pulled out must have a value of \$175-\$170=\$5$.

That is, Rad pulled out three $5 bills, two$ 10 bills, two $20 bills, and two$ 50 bills, for a total of $3+2+2+2=9$ bills.
Sandy pulls out at least one of each type of bill and so she must have at least $\$5+\$10+\$20+\$50=\$85$ .

Thus, we know four of the five bills that Sandy pulls out and that these four bills total $85. The fifth bill that Sandy pulls could be any one of the four different types of bills.

If this fifth bill is a 5 bill, then Sandy’s total sum of money is \$85+\$5=\$90$.

If this fifth bill is a 10 bill, then Sandy’s total sum of money is \$85+\$10=\$95$.

If this fifth bill is a 20 bill, then Sandy’s total sum of money is \$85+\$20=\$105$.

Finally, if this fifth bill is a 50 bill, then Sandy’s total sum of money is \$85+\$50=\$135$.

Therefore, the sums of money that Sandy could have are $90,$ 95, $105, and$ 135.
Lino could have at most three $50 bills since four$ 50 bills exceeds his total sum of money ( $4\times \$50=\$200>\$160$ ).

If Lino had no 50 bills, then the bills (6 bills at most) each have value at most 20 and would total $\$160$ .

However, this is not possible since $6\times\$20=\$120$ which is less than the required $160.

If Lino had one 50 bill, then the remaining bills (5 bills at most) would total \$160-\$50=\$110$.

However, this is not possible since the largest denomination of the remaining bills is $20 and$ $5\times\$20=\$100$ which is less than the required 110.

Therefore, we proceed by considering the cases where Lino has two or three $50 bills.

These two cases are summarized in the table below.

Number of $\$50$ s
Money in $\$50$ s
Money remaining
Number of $\$20$ s
Number of $\$10$ s
Number of $\$5$ s
Number of bills used

3
$150
$10
0
1
0
4

3
$150
$10
0
0
2
5

2
$100
$60
3
0
0
5

2
$100
$60
2
2
0
6

In each case from the table above, Lino has a total sum of $160 and has pulled out 6 or fewer bills.

Since we are given that there are only four possibilities, then we have found them all.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.