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.
Four 20, past 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 ListFor 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.
4.OA.B.4Guess And CheckOnly two cases pass: one 2s, and three 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).