AMC 10 · 2015 · #3
Easy mode Grade 4Isaac writes one number two times and a different number three times. That gives five numbers in all. The five numbers add up to 100. One of the numbers is 28. What is the other number?
Pick an answer.
AMC 10 2015 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: Isaac writes one integer down two times and a different integer down three times, giving five numbers in all. The five numbers add up to $100$, and one of the two integers is $28$. Find the value of the other integer.
Givens: One integer is written $2$ times and a different integer is written $3$ times — five numbers in total.; The five numbers add up to $100$.; One of the two integers is $28$.; Answer choices: (A) $8$, (B) $11$, (C) $14$, (D) $15$, (E) $18$
Unknowns: The value of the other integer — the one that is not $28$.
Understand
Restated: Isaac writes one integer down two times and a different integer down three times, giving five numbers in all. The five numbers add up to $100$, and one of the two integers is $28$. Find the value of the other integer.
Givens: One integer is written $2$ times and a different integer is written $3$ times — five numbers in total.; The five numbers add up to $100$.; One of the two integers is $28$.; Answer choices: (A) $8$, (B) $11$, (C) $14$, (D) $15$, (E) $18$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #4 Introduce a Variable
We are not told whether $28$ is the number repeated twice or the number repeated three times, so there are only two possible setups. Tool #3 (Eliminate Possibilities) tests each setup and throws out the one that cannot produce a whole number, leaving exactly one valid answer. Tool #4 (Introduce a Variable) names the unknown integer $x$ so we can track what each case forces it to be. The decisive clue is that the other integer must come out whole — only one of the two cases survives that test.
Execute — Answer: A
4.OA.A.3 Step 1 Name the unknown, list the two cases
- Call the other integer $x$.
- The five numbers are made of $28$s and $x$s.
- Since $28$ is repeated either twice or three times, there are exactly two cases: Case A — $28$ appears three times and $x$ appears twice; Case B — $28$ appears twice and $x$ appears three times.
- In both cases the five numbers must add to $100$.
💡 There are only two ways to share the five slots between $28$ and $x$, so checking both covers everything.
4.NBT.B.5 Step 2 Test Case A: 28 three times
- If $28$ appears three times, those three numbers add to $3\cdot 28 = 84$.
- The remaining two numbers are both $x$ and must make up the rest: $100 - 84 = 16$.
- Two equal numbers adding to $16$ means each is $16 \div 2 = 8$, a whole number.
- This case works and gives $x = 8$.
💡 If three of the numbers are pinned down, the leftover total has to be split evenly between the last two.
4.NBT.B.6 Step 3 Test Case B, eliminate it, read the answer
- If $28$ appears twice, those two numbers add to $2\cdot 28 = 56$, leaving $100 - 56 = 44$ for the three equal $x$ values.
- But $44 \div 3$ is not a whole number ($3\cdot 14 = 42$, $3\cdot 15 = 45$), so $x$ could not be an integer.
- Case B is impossible and gets thrown out.
- Only Case A survives, so the other integer is $8$.
- So the answer is (A).
💡 If splitting the leftover into equal whole pieces is impossible, that case can't happen.
4.OA.A.3 Call the other integer $x$. The five numbers are made of $28$s and $x$s. Since $ 4.NBT.B.5 If $28$ appears three times, those three numbers add to $3\cdot 28 = 84$. The re 4.NBT.B.6 If $28$ appears twice, those two numbers add to $2\cdot 28 = 56$, leaving $100 - Review
Reasonableness: Check the winning case directly: the five numbers are $28, 28, 28, 8, 8$, and $28+28+28+8+8 = 84 + 16 = 100$. That matches the given total, one of the numbers is $28$ as required, and both repeated values are integers. The answer $8$ is also one of the listed choices.
Alternative: Set it up with algebra in one line. Since $28$ must be the value repeated three times (the other role gives no integer), write $3(28) + 2x = 100$. Then $2x = 100 - 84 = 16$, so $x = 8$ — the same answer (A).
CCSS standards used (min grade 4)
4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Translating the word problem into the two possible arrangements of $28$s and $x$s that must total $100$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $3\cdot 28 = 84$ and then splitting the leftover $16$ to find $x = 8$ in the valid case.)4.NBT.B.6Find whole-number quotients and remainders (Dividing $44 \div 3$ to see it leaves a remainder, proving the second case cannot give an integer and must be eliminated.)
⭐ When you don't know which slot a number fills, try every option and keep only the one that comes out as a whole number.
⭐ When you don't know which slot a number fills, try every option and keep only the one that comes out as a whole number.
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