AMC 8 · 2000 · #20
Grade 4 arithmeticnumber-theoryPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
First peel off one coin of each kind to satisfy the "at least one" rule. That fixes 41¢ in 4 coins and leaves 61¢ to share among 5 more coins. Tool #8 (Analyze the Units) gives the key restriction: nickels, dimes, and quarters all contribute multiples of 5, so the number of extra pennies must make the leftover a multiple of 5. Tool #3 (Eliminate Possibilities) then knocks out every penny count except one. With pennies pinned down, Tool #2 (Make a Systematic List) tries each possible extra-dime count and eliminates the ones that can't be completed by nickels and quarters. No equations needed — just divisibility and short bookkeeping.
Give one coin of each kind away first: those 4 coins fix 1+5+10+25=41¢, leaving 5 coins to make 102-41=61¢.
Spending the "at least one" requirement up front turns a four-variable puzzle into a smaller one with no minimum constraints.
4.MD.A.2Eliminate PossibilitiesNickels, dimes, and quarters are all multiples of 5, so the leftover's ones digit forces 1 extra penny (6 pennies won't fit in 5 coins).
Looking only at the ones digit (the unit of 1¢) forces the penny count without touching the other coins.
4.OA.B.4Analyze The UnitsPlace that penny: now 4 coins remain to make 60¢, and they can only be nickels, dimes, or quarters.
With pennies done, every remaining coin is worth a multiple of 5¢ — the rest is pure trial on a small board.
4.MD.A.2Eliminate PossibilitiesTest extra dimes d=0..4: only d=0 works, filled by 2 quarters and 2 nickels — every other d can't reach 60¢.
Five small cases, each settled by one quick check — exactly the situation a systematic list is built for.
4.OA.A.3Make A Systematic ListAdd the set-aside dime back: the final bag is 2 pennies, 3 nickels, 1 dime, 3 quarters = 9 coins totaling 102¢.
Don't forget to put back the dime that was set aside at the very start — the count of 1 comes from the original alone.
4.MD.A.2Eliminate PossibilitiesHand out one of each coin first to clear the "at least one" rule. The leftover 61¢ ends in a 1, so only pennies can supply that odd unit — and with just 5 coins left, there's room for exactly 1 extra penny. From there, a short check of dime counts 0 through 4 leaves only one working bag, with 1 dime total — answer (A).