AMC 8 · 2002 · #2

Grade 4 number-theorycounting

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

linear-diophantinesystematic-enumerationparitybound-inequality-then-enumerate systematic-enumerationcasework ↑ Prerequisites: multi-digit-arithmeticparity

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Problem

How many different combinations of $5 bills and$ 2 bills can be used to make a total of $17? Order does not matter in this problem.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Count how many different ways you can pay exactly $\$17$ using only $\$5$ bills and $\$2$ bills. Two ways are the same if they use the same number of each bill (order does not matter).

Givens: Bills available: $\$5$ and $\$2$; Total to make: $\$17$; Each bill count is a whole number $\ge 0$; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$

Unknowns: The number of $(\text{fives}, \text{twos})$ pairs whose total equals $\$17$

Understand

Restated: Count how many different ways you can pay exactly $\$17$ using only $\$5$ bills and $\$2$ bills. Two ways are the same if they use the same number of each bill (order does not matter).

Givens: Bills available: $\$5$ and $\$2$; Total to make: $\$17$; Each bill count is a whole number $\ge 0$; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #6 Guess and Check

The number of $\$5$ bills can only be $0, 1, 2,$ or $3$, because $4 \times 5 = 20$ already passes $17$. That is a tiny list, so Tool #2 (Make a Systematic List) walks through every case in seconds. For each case, Tool #6 (Guess and Check) asks the same yes/no question: after the fives, is the leftover an exact multiple of $\$2$? Counting how many cases answer yes gives the final answer, with no algebra or number-theory machinery needed.

Execute — Answer: A

#2 Make a Systematic List 4.OA.A.3 Step 1

Find the largest number of $\$5$ bills that could fit. Each $\$5$ bill takes $\$5$ out of $\$17$.
Four fives would be $\$20$, already over. So the number of fives is one of $0, 1, 2, 3$ — a list of just four cases to check.

$$4 \times 5 = 20 > 17 \;\Rightarrow\; \text{fives} \in \{0, 1, 2, 3\}$$

💡 Whenever a problem says "how many combinations," first bound the smaller list — here the $\$5$ bills — so the search ends quickly.

#6 Guess and Check 4.OA.B.4 Step 2

For each case, compute the leftover and check whether $\$2$ bills can pay it exactly. A leftover works only if it is even (a multiple of $2$).

$$\begin{array}{c|c|c} \text{fives} & \text{leftover} & \text{twos needed} \\ \hline 0 & 17 & 17 \div 2 = 8.5 \;\text{(no)} \\ 1 & 12 & 12 \div 2 = 6 \;\text{(yes)} \\ 2 & 7 & 7 \div 2 = 3.5 \;\text{(no)} \\ 3 & 2 & 2 \div 2 = 1 \;\text{(yes)} \end{array}$$

💡 The leftover must be even — an even number of dollars can always be paid in $\$2$ bills, an odd amount never can.

#2 Make a Systematic List 4.OA.A.3 Step 3

Count the rows marked "yes." Only two cases work: one $\$5$ plus six $\$2$s, and three $\$5$s plus one $\$2$.
That gives $2$ different combinations.

$$(1, 6): 5 + 12 = 17 \;\checkmark \quad (3, 1): 15 + 2 = 17 \;\checkmark \;\Rightarrow\; 2 \text{ combinations} \;\Rightarrow\; \textbf{(A)}$$

💡 Once the table is built, just count the "yes" rows — that is the answer.

[1] #2 4.OA.A.3 Find the largest number of $\$5$ bills that could fit. Each $\$5$ bill takes $\$

[2] #6 4.OA.B.4 For each case, compute the leftover and check whether $\$2$ bills can pay it exa

[3] #2 4.OA.A.3 Count the rows marked "yes." Only two cases work: one $\$5$ plus six $\$2$s, and

Review

Reasonableness: Verify both winning combinations add to $\$17$: $1 \times 5 + 6 \times 2 = 5 + 12 = 17$, and $3 \times 5 + 1 \times 2 = 15 + 2 = 17$. Both check out. Also notice the pattern: the leftover after the fives must be even, and $17$ is odd, so the number of fives must be odd (because odd $-$ odd $=$ even). The odd choices from ${0,1,2,3}$ are $1$ and $3$ — exactly two cases, matching answer (A).

Alternative: Tool #5 (Look for a Pattern): the total $\$17$ is odd, and each $\$2$ bill contributes an even amount. So the $\$5$ bills together must contribute an odd amount, which means the count of $\$5$ bills must be odd. From $\{0, 1, 2, 3\}$ the odd counts are $1$ and $3$, and both leave an even leftover that $\$2$ bills can cover ($\$12 = 6 \times \$2$ and $\$2 = 1 \times \$2$). That gives $2$ combinations, confirming (A).

CCSS standards used (min grade 4)

4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Setting up the case list $\text{fives} \in \{0,1,2,3\}$, computing each leftover, and counting how many cases pay $\$17$ exactly.)
4.OA.B.4 Find all factor pairs for a whole number; recognize multiples (Checking each leftover for the question "is this a multiple of $2$?" — the test that decides whether $\$2$ bills can cover it exactly.)

⭐ When a problem asks "how many combinations," pin down the smaller list first — here, the $\$5$ bills can only be $0, 1, 2,$ or $3$. Walk that short list, check each leftover, and the answer pops out: $2$ combinations, choice (A).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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