AMC 10 · 2009 · #4
Grade 2 number-theoryPick an answer.
The question is 'which could NOT be,' a finite five-choice list, so the natural move is Tool #3: build every total you can and cross it off, leaving the one that resists. Four of the five fall in one line of adding. For the survivor, Tool #8 (Analyze the Units) reads the coins by their multiples of 5 to show pennies would be needed, and Tool #14 (Extreme Principle) pins the smallest total four non-penny coins can make, proving the survivor is out of reach.
List the coin values
Writing down the four values is the whole setup.
Knowing each coin's cent value turns the coin puzzle into plain adding.
2.MD.C.8Analyze The UnitsBuild the other four totals
Four of the totals can be built directly.
If you can actually build a total, it is not the impossible one, so eliminate it.
2.NBT.B.5Eliminate Possibilities15 needs zero pennies
The last one's remainder forces zero of the smallest coin.
Pennies are the only coin that can break the by-5 rhythm, so a by-5 total must use a by-5 number of them.
Pennies are the only coin that can break the count-by-five rhythm, so a by-five total needs a by-five number of them.
▸ Why?
Every other coin is a whole number of fives, so together they can only ever reach multiples of five.
▸ Why?
The pennies are exactly the leftover after the fives are counted out, so their count is the remainder.
Four nickels already beat 15
Then even the cheapest four coins already overshoot it, choice (A).
Once pennies are banned, four coins cannot dip below 20 cents, so anything under 20 is unreachable.
2.OA.A.1Extreme PrincipleThe nickel, dime, and quarter only move in jumps of 5, so four of them start at 20 cents — and you can't drop to 15 without pennies, which would knock you off a count-by-5 total.
- List the coin values
- Build the other four totals
- 15 needs zero pennies
- Four nickels already beat 15