AMC 8 · 1999 · #1

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

order-of-operationsmental-arithmeticsystematic-enumeration guess-and-checksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticorder-of-operations

📏 Short solution 💡 2 insights

Problem

$(6?3) + 4 - (2 - 1) = 5.$ To make this statement true, the question mark between the 6 and the 3 should be replaced by

Pick an answer.

(A)

$\div$

(B)

$\times$

(C)

(D)

(E)

None of these

AMC 8 1999 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: In the equation $(6\,?\,3) + 4 - (2 - 1) = 5$, which of $\div$, $\times$, $+$, $-$, or "None of these" should replace the question mark to make the statement true?

Givens: The equation is $(6\,?\,3) + 4 - (2 - 1) = 5$; The unknown is the operator between $6$ and $3$; Answer choices: (A) $\div$, (B) $\times$, (C) $+$, (D) $-$, (E) None of these

Unknowns: The operator that makes the equation true

Understand

Restated: In the equation $(6\,?\,3) + 4 - (2 - 1) = 5$, which of $\div$, $\times$, $+$, $-$, or "None of these" should replace the question mark to make the statement true?

Givens: The equation is $(6\,?\,3) + 4 - (2 - 1) = 5$; The unknown is the operator between $6$ and $3$; Answer choices: (A) $\div$, (B) $\times$, (C) $+$, (D) $-$, (E) None of these

Plan

Primary tool: #11 Work Backwards

Secondary: #6 Guess and Check

The equation already tells us the final value is $5$. Tool #11 (Work Backwards) lets us undo the known pieces — first simplify $(2-1)$, then peel off the $+4 - 1$ — until only the mystery expression $(6\,?\,3)$ is left equal to a number. Once we know what $(6\,?\,3)$ must equal, Tool #6 (Guess and Check) tests each of the four operators on the small numbers $6$ and $3$ — a one-line check per option.

Execute — Answer: A

#11 Work Backwards 5.OA.A.1 Step 1

Simplify the parenthesis on the right side of the equation first.
$(2 - 1) = 1$, so the equation becomes $(6\,?\,3) + 4 - 1 = 5$.

$$(6\,?\,3) + 4 - 1 = 5$$

💡 Order of operations says do the inside of parentheses first — a Grade 5 skill.

#11 Work Backwards 3.NBT.A.2 Step 2

Combine the known numbers on the left: $4 - 1 = 3$.
The equation now reads $(6\,?\,3) + 3 = 5$.

$$(6\,?\,3) + 3 = 5$$

💡 $4 - 1 = 3$ is a Grade 3 within-$100$ subtraction fact.

#11 Work Backwards 6.EE.B.7 Step 3

Undo the $+3$ by subtracting $3$ from both sides.
This isolates the mystery expression.

$$(6\,?\,3) = 5 - 3 = 2$$

💡 Subtracting the same value from both sides keeps the equation balanced — the Grade 6 inverse-operation move.

#6 Guess and Check 3.OA.C.7 Step 4

Test each operator on $6$ and $3$ to see which one gives $2$.
Only division works: $6 \div 3 = 2$.
The others give $18$, $9$, and $3$.

$6 \div 3 = 2$ \;\checkmark, \;\; $6 \times 3 = 18$, \;\; $6 + 3 = 9$, \;\; $6 - 3 = 3 \;\Rightarrow\; \textbf{(A)}$

💡 All four checks use one-digit basic facts — Grade 3 fluency with multiplication and division within $100$.

[1] #11 5.OA.A.1 Simplify the parenthesis on the right side of the equation first. $(2 - 1) = 1$,

[2] #11 3.NBT.A.2 Combine the known numbers on the left: $4 - 1 = 3$. The equation now reads $(6\,

[3] #11 6.EE.B.7 Undo the $+3$ by subtracting $3$ from both sides. This isolates the mystery expr

[4] #6 3.OA.C.7 Test each operator on $6$ and $3$ to see which one gives $2$. Only division work

Review

Reasonableness: Plug $\div$ back into the original equation: $(6 \div 3) + 4 - (2 - 1) = 2 + 4 - 1 = 5$. The left side equals the right side, so (A) is confirmed. The other three operators all fail the same check: $(6 \times 3) + 4 - 1 = 21$, $(6 + 3) + 4 - 1 = 12$, $(6 - 3) + 4 - 1 = 6$. None equals $5$, so (E) "None of these" is also ruled out.

Alternative: Tool #3 (Eliminate Possibilities): scan the four operators with a quick size estimate. Since $6 + 3 = 9$ and $6 \times 3 = 18$ are both much larger than the $2$ we need, addition and multiplication are out. Between subtraction ($6 - 3 = 3$) and division ($6 \div 3 = 2$), only division hits $2$ exactly. That immediately gives (A) without restoring the full equation.

CCSS standards used (min grade 6)

5.OA.A.1 Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols (Simplifying $(2-1)$ before the surrounding addition and subtraction, in line with the order of operations.)
3.OA.C.7 Fluently multiply and divide within 100 (Testing each candidate operator on $6$ and $3$ — basic multiplication and division facts.)
3.NBT.A.2 Fluently add and subtract within 1000 (Combining $4 - 1 = 3$ and $5 - 3 = 2$ while simplifying the equation.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Subtracting $3$ from both sides to isolate $(6\,?\,3)$ and find that it must equal $2$.)

⭐ Clean up the parts you already know, see what the mystery piece has to equal, then try each operator. AMC 8 #1 is a Grade 6 balance-the-equation problem with a Grade 3 fact check at the end.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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