AMC 8 · 2010 · #22
Grade 6 arithmeticnumber-theoryProblem
The hundreds digit of a three-digit number is more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Take a three-digit number whose hundreds digit is exactly $2$ more than its units digit. Reverse its digits to make a new three-digit number. Subtract the reversed number from the original. What is the units digit of that difference?
Givens: The number has three digits — call them hundreds $h$, tens $t$, units $u$; Constraint between digits: $h = u + 2$; A new number is formed by reversing the digits (new hundreds $= u$, new units $= h$); We compute (original) $-$ (reversed); Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$
Unknowns: The units digit of the difference (original $-$ reversed)
Understand
Restated: Take a three-digit number whose hundreds digit is exactly $2$ more than its units digit. Reverse its digits to make a new three-digit number. Subtract the reversed number from the original. What is the units digit of that difference?
Givens: The number has three digits — call them hundreds $h$, tens $t$, units $u$; Constraint between digits: $h = u + 2$; A new number is formed by reversing the digits (new hundreds $= u$, new units $= h$); We compute (original) $-$ (reversed); Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #13 Convert to Algebra
The problem describes a whole family of three-digit numbers, not just one. Instead of fighting with abstract digits, Tool #9 (Easier Problem) says: pick a couple of concrete numbers that fit the rule and just do the subtraction. Tool #5 (Pattern) then checks whether the units digit comes out the same every time — if it does, that is the answer regardless of which specific number we picked. Tool #13 (Algebra) is held in reserve to confirm the pattern using place-value expansion ($100h + 10t + u$), but only after the simpler tools have already revealed the answer.
Execute — Answer: E
4.NBT.B.4 Step 1 - Pick the simplest example that fits the rule $h = u + 2$.
- Try $u = 0$, $t = 0$, $h = 2$, which gives the original number $200$.
- The reversed number is $002 = 2$.
- Subtract.
💡 Replacing the abstract digits with the smallest legal choice is the Grade 4 "use what you can compute" move — a concrete three-digit subtraction.
5.OA.B.3 Step 2 - Now try a different example to see if the units digit of the difference is stable.
- Take $u = 3$, $t = 5$, so $h = 5$, the original number is $553$.
- Reversed, it is $355$.
💡 Same answer again. Tool #5 (Pattern) suggests the difference is always $198$ whenever $h - u = 2$, no matter what the tens digit is — a Grade 5 "analyze patterns and relationships" observation.
5.OA.B.3 Step 3 - Confirm with one more example so we are not fooled by coincidence.
- Take $u = 5$, $t = 9$, so $h = 7$, original $= 795$, reversed $= 597$.
💡 Three different originals, identical difference of $198$. The pattern is real.
6.EE.A.2 Step 4 - Lock in the reason with a one-line algebra check.
- The original is $100h + 10t + u$ and the reversed is $100u + 10t + h$.
- Subtract them; the tens cancel and the rest factors.
💡 Writing each number with letters for digits is Grade 6 expression-writing, and it explains why $t$ never matters: the tens digits subtract to $0$.
4.NBT.A.2 Step 5 Read off the units digit of $198$.
💡 Identifying the ones place of a multi-digit number is a Grade 4 place-value skill.
4.NBT.B.4 Pick the simplest example that fits the rule $h = u + 2$. Try $u = 0$, $t = 0$, 5.OA.B.3 Now try a different example to see if the units digit of the difference is stabl 5.OA.B.3 Confirm with one more example so we are not fooled by coincidence. Take $u = 5$, 6.EE.A.2 Lock in the reason with a one-line algebra check. The original is $100h + 10t + 4.NBT.A.2 Read off the units digit of $198$. Review
Reasonableness: Across three very different starting numbers ($200$, $553$, $795$), the difference came out to exactly $198$ each time, and the algebra step $99(h - u)$ explains why: the constraint $h - u = 2$ forces the difference to be $99 \times 2 = 198$ no matter what the tens digit is. The units digit of $198$ is $8$, which matches choice (E) — and (E) is the only choice that could come from a multiple of $99$ ending in $8$, so the answer is consistent.
Alternative: Tool #3 (Eliminate Possibilities): the difference of a number and its reversal is always a multiple of $99$ (digits $0, 99, 198, 297, 396, 495, 594, 693, 792, 891$). The units digits of those multiples are $0, 9, 8, 7, 6, 5, 4, 3, 2, 1$. Since $h - u = 2$, the difference is $99 \times 2 = 198$, so the units digit must be $8$, eliminating (A)–(D) immediately.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals (Reading off the units digit of $198$ as $8$ — a basic place-value identification.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Performing the three-digit subtractions $200 - 2$, $553 - 355$, and $795 - 597$ on concrete examples.)5.OA.B.3Generate two numerical patterns and identify relationships (Observing that multiple examples obeying $h = u + 2$ all give the same difference $198$, so the units digit is fixed.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing the original as $100h + 10t + u$ and the reverse as $100u + 10t + h$, then simplifying the difference to $99(h - u)$.)
⭐ Trying two or three concrete examples first (Tool #9) often cracks an "abstract digits" problem before you ever need algebra!
⭐ Trying two or three concrete examples first (Tool #9) often cracks an "abstract digits" problem before you ever need algebra!