Competition · AMC preparation · step 4 of 4
AMC 8 · 2002 · #2
Grade 4 number-theorycountingPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The number of 5 bills can only be 0, 1, 2, or 3, because 4 × 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 of2? Counting how many cases answer yes gives the final answer, with no algebra or number-theory machinery needed.
Bound the number of fives
Four $5 bills already make $20, past $17, so the number of $5 bills is one of 0, 1, 2, 3 — just four cases.
Whenever a problem says "how many combinations," first bound the smaller list — here the $5 bills — so the search ends quickly.
4.OA.A.3Make A Systematic ListCheck each leftover
For each case, the leftover must be paid by $2 bills, which works only when the leftover is even.
The leftover must be even — an even number of dollars can always be paid in $2 bills, an odd amount never can.
After the $5 bills are set aside, the dollars left over can be paid exactly with $2 bills only when that leftover is an even number of dollars.
▸ Why?
Paying a leftover with only $2 bills means it must equal some number of $2 bills added together, and that sum is always even, so an odd leftover can never be covered while an even leftover always can.
▸ Why?
A pile of n two-dollar bills is worth 2 × n, and two times any whole number is even, so the reachable totals are exactly the even amounts 0, 2, 4, 6, … and never an odd one.
Count the working cases
Only two cases pass: one $5 with six $2s, and three $5s with one $2 — giving 2 combinations, choice (A).
Once the table is built, just count the "yes" rows — that is the answer.
4.OA.A.3Make A Systematic ListWhen 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).
- Bound the number of fives
- Check each leftover
- Count the working cases
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