AMC 8 · 2016 · #11
Easy mode Grade 6Problem
Pick any two-digit number. Now flip its two digits to make a new number. For example, flipped becomes .
We want pairs where the original number plus its flipped version equals . (For and : , so counts.)
How many two-digit numbers have this property?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Count the two-digit numbers $N$ with the property that $N$ plus the number you get by reversing its digits equals $132$.
Givens: $N$ is a two-digit number, so its tens digit is between $1$ and $9$ and its units digit is between $0$ and $9$; Reversing the digits gives a second number (also $1$ or $2$ digits, but written with the same two digit symbols swapped); $N + (\text{digit-reversal of } N) = 132$; Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $11$, (E) $12$
Unknowns: How many such two-digit numbers $N$ exist
Understand
Restated: Count the two-digit numbers $N$ with the property that $N$ plus the number you get by reversing its digits equals $132$.
Givens: $N$ is a two-digit number, so its tens digit is between $1$ and $9$ and its units digit is between $0$ and $9$; Reversing the digits gives a second number (also $1$ or $2$ digits, but written with the same two digit symbols swapped); $N + (\text{digit-reversal of } N) = 132$; Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $11$, (E) $12$
Plan
Primary tool: #5 Use Variables
Secondary: #13 Work Systematically
The condition is a statement about digits, so Tool #5 (Use Variables) lets us name the tens digit $t$ and the units digit $u$ and write $N = 10t + u$ in place-value form. Adding $N$ to its reversal $10u + t$ collapses neatly to $11(t+u) = 132$, which simplifies to $t + u = 12$. Once the condition reduces to a single equation in two digits, Tool #13 (Work Systematically) takes over: list every $(t,u)$ with $t + u = 12$, $1 \le t \le 9$, $0 \le u \le 9$, and count what survives.
Execute — Answer: B
4.NBT.A.1 Step 1 - Write the two-digit number in place-value form using variables.
- Let $t$ be the tens digit and $u$ be the units digit.
- Then the original number is $10t + u$ and the digit-reversed number is $10u + t$.
💡 Place value (Grade 4) says the tens digit contributes ten times its face value, and the ones digit contributes its face value.
6.EE.A.2 Step 2 - Translate the word condition into an equation.
- The sum of the number and its reversal is $132$, so add the two place-value expressions.
💡 Writing a word condition as an algebraic equation is Grade 6 expressions-and-equations work.
6.EE.A.3 Step 3 Combine like terms on the left, then divide both sides by $11$ to isolate $t + u$.
💡 Grouping $11t + 11u$ as $11(t+u)$ is the distributive property — a Grade 6 "equivalent expressions" move.
6.EE.B.5 Step 4 - Apply the digit constraints.
- The tens digit $t$ must be from $1$ to $9$ (so $N$ stays two-digit) and the units digit $u$ must be from $0$ to $9$.
- List every $t$ from $1$ to $9$, set $u = 12 - t$, and keep only the pairs where $u$ is also a valid digit.
💡 Substituting each candidate $t$ into $t + u = 12$ and checking the digit range is the Grade 6 idea of "which values make the equation true."
4.OA.C.5 Step 5 - Count the valid $(t, u)$ pairs.
- Each pair gives exactly one two-digit number $N = 10t + u$.
💡 Counting the items in a generated list is the Grade 4 "analyze a pattern" finishing step.
4.NBT.A.1 Write the two-digit number in place-value form using variables. Let $t$ be the t 6.EE.A.2 Translate the word condition into an equation. The sum of the number and its rev 6.EE.A.3 Combine like terms on the left, then divide both sides by $11$ to isolate $t + u 6.EE.B.5 Apply the digit constraints. The tens digit $t$ must be from $1$ to $9$ (so $N$ 4.OA.C.5 Count the valid $(t, u)$ pairs. Each pair gives exactly one two-digit number $N Review
Reasonableness: Sanity check one number: $39 + 93 = 132$. Yes. Try another: $66 + 66 = 132$. Yes. The valid tens digits run from $3$ to $9$, giving $9 - 3 + 1 = 7$ numbers, matching the count above. Also, the reversed numbers $\{93, 84, 75, 66, 57, 48, 39\}$ are just the original list in reverse order — no number is missed and none is double-counted, because we labeled by the tens digit of $N$.
Alternative: Tool #15 (Look for Symmetry): the digit-reversal map pairs each number $\overline{tu}$ with $\overline{ut}$, and a pair sums to $132$ exactly when $t + u = 12$. Without writing any equation, just list two-digit numbers whose digits add to $12$: $39, 48, 57, 66, 75, 84, 93$ — the tens digit only ranges over $3$–$9$ because smaller tens digits would force the units digit above $9$. That is $7$ numbers, choice (B).
CCSS standards used (min grade 6)
4.NBT.A.1Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right (Writing the two-digit number $N$ in place-value form as $10t + u$ and the reversed number as $10u + t$.)4.OA.C.5Generate and analyze a number or shape pattern that follows a given rule (Listing the generated pattern $\{39, 48, 57, 66, 75, 84, 93\}$ and counting that it has $7$ entries.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Translating "the number plus its digit-reversal equals $132$" into the algebraic equation $(10t + u) + (10u + t) = 132$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Combining $11t + 11u$ as $11(t + u)$ and dividing by $11$ to reduce the equation to $t + u = 12$.)6.EE.B.5Understand solving an equation as a process of answering which values from a specified set make the equation true (Checking each candidate digit $t \in \{1, 2, \dots, 9\}$ against the digit constraint on $u = 12 - t$ to find which pairs satisfy the equation.)
⭐ Once you write a two-digit number as $10t + u$, the whole problem turns into a Grade 6 equation $t + u = 12$ — then you just list the digit pairs that work!
⭐ Once you write a two-digit number as $10t + u$, the whole problem turns into a Grade 6 equation $t + u = 12$ — then you just list the digit pairs that work!