AMC 10 · 2009 · #2
Grade 2 arithmeticFour coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
Pick an answer.
AMC 10 2009 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.
Toolkit + CCSS Solution
Understand
Restated: Four coins are drawn from a piggy bank holding pennies ($1$c), nickels ($5$c), dimes ($10$c), and quarters ($25$c). Repeats are allowed. Among $15$, $25$, $35$, $45$, and $55$ cents, decide which total four coins could never add up to.
Givens: Exactly four coins are picked; Each coin is a penny $1$c, nickel $5$c, dime $10$c, or quarter $25$c; The same coin type may be picked more than once; Answer choices: (A) $15$, (B) $25$, (C) $35$, (D) $45$, (E) $55$
Unknowns: Which listed total is impossible to form with four coins
Understand
Restated: Four coins are drawn from a piggy bank holding pennies ($1$c), nickels ($5$c), dimes ($10$c), and quarters ($25$c). Repeats are allowed. Among $15$, $25$, $35$, $45$, and $55$ cents, decide which total four coins could never add up to.
Givens: Exactly four coins are picked; Each coin is a penny $1$c, nickel $5$c, dime $10$c, or quarter $25$c; The same coin type may be picked more than once; Answer choices: (A) $15$, (B) $25$, (C) $35$, (D) $45$, (E) $55$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #8 Analyze the Units, #14 Extreme Principle
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.
Execute — Answer: A
2.MD.C.8 Step 1 List the coin values
- Write what each coin is worth in cents: a penny is $1$, a nickel is $5$, a dime is $10$, and a quarter is $25$.
- Picking four coins means adding four of these values, and the same coin may repeat.
- The task is to find the one listed total that four coins can never make.
💡 Knowing each coin's cent value turns the coin puzzle into plain adding.
2.NBT.B.5 Step 2 Build the other four totals
- Try to reach each choice with four coins.
- $25=10+5+5+5$, $35=10+10+10+5$, $45=25+10+5+5$, and $55=25+10+10+10$.
- Each one works with exactly four coins, so (B), (C), (D), and (E) are all possible and can be crossed off.
💡 If you can actually build a total, it is not the impossible one, so eliminate it.
2.NBT.A.2 Step 3 15 needs zero pennies
- Only $15$ is left.
- The nickel, dime, and quarter are all skip-counts of $5$ ($5$, $10$, $25$), so any pile of just these lands on a count-by-$5$ total.
- The penny is the only coin that is not a multiple of $5$.
- Since $15$ is itself a count-by-$5$ number, the pennies used must also add to a multiple of $5$; with only four coins the only way that happens is to use no pennies at all.
💡 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.
2.OA.A.1 Step 4 Four nickels already beat 15
- With no pennies, every one of the four coins is worth at least a nickel, $5$ cents.
- So the smallest total you can build is four nickels: $5+5+5+5=20$ cents.
- That is already more than $15$, so $15$ can never be reached with four coins.
- The total that could not happen is $15$, which is (A).
💡 Once pennies are banned, four coins cannot dip below $20$ cents, so anything under $20$ is unreachable.
2.MD.C.8 Write what each coin is worth in cents: a penny is $1$, a nickel is $5$, a dime 2.NBT.B.5 Try to reach each choice with four coins. $25=10+5+5+5$, $35=10+10+10+5$, $45=25 2.NBT.A.2 Only $15$ is left. The nickel, dime, and quarter are all skip-counts of $5$ ($5$ 2.OA.A.1 With no pennies, every one of the four coins is worth at least a nickel, $5$ cen Review
Reasonableness: The four totals we built — $25$, $35$, $45$, $55$ — are all at least $20$ and are all counts-by-$5$, matching the two rules these coins obey. Only $15$ breaks a rule: being a multiple of $5$ it cannot use pennies, yet four non-penny coins start at $20$ cents, above $15$. So $15$ is the odd one out, confirming (A).
Alternative: Split by how many pennies are used. $0$ pennies: four coins of $5$c or more total at least $20$. $1$ penny: the other three multiples of $5$ total at least $15$, so the sum is at least $16$. $2$, $3$, or $4$ pennies: the leftover coins are multiples of $5$, so the total sits $2$, $3$, or $4$ above a multiple of $5$ and can never equal $15$. No case reaches $15$, so it is impossible.
CCSS standards used (min grade 2)
2.MD.C.8Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (Reading each coin as its cent value ($1$, $5$, $10$, $25$) and treating four picked coins as a sum of those values.)2.NBT.B.5Fluently add and subtract within 100 (Adding four coin values to build $25$, $35$, $45$, and $55$ and eliminate them as possible totals.)2.NBT.A.2Count within 1000, skip-count by 5s, 10s, and 100s (Recognizing nickels, dimes, and quarters as multiples of $5$, so a multiple-of-$5$ total like $15$ must use zero pennies.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Finding the smallest four-coin total without pennies, $5+5+5+5=20$, and comparing it to $15$ to prove $15$ is impossible.)
⭐ The 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.
⭐ The 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.
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