AMC 10 · 2009 · #4
Easy mode Grade 2A piggy bank holds pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents). You take out four coins. You may pick the same kind of coin more than once. Which of these totals, in cents, can the four coins never add up to?
Pick an answer.
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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