← Sensim Lab Sensim Math

AMC 8 · 1999 · #1

Grade 6 arithmetic
Archetype Small-Domain Constrained Search
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.(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
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.

⭐ 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.

More like this

Same archetype — closest grade level first.