AMC 10 · 2005 · #16
Grade 6 arithmeticThe sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is 6. How many two-digit numbers have this property?
Pick an answer.
AMC 10 2005 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: Take a two-digit number and subtract the sum of its two digits from it. Count how many two-digit numbers give a result whose units (ones) digit is $6$.
Givens: The number has two digits, so it runs from $10$ to $99$; From the number you subtract the sum of its digits; The units digit of that result must be $6$; Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $10$, (E) $19$
Unknowns: How many two-digit numbers produce a result ending in $6$
Understand
Restated: Take a two-digit number and subtract the sum of its two digits from it. Count how many two-digit numbers give a result whose units (ones) digit is $6$.
Givens: The number has two digits, so it runs from $10$ to $99$; From the number you subtract the sum of its digits; The units digit of that result must be $6$; Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $10$, (E) $19$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List
Testing all $90$ two-digit numbers by hand is slow and easy to botch. Tool #4 names the tens digit $a$ and units digit $b$ and writes the number as $10a+b$; then "subtract the digit sum" becomes $(10a+b)-(a+b)$, which collapses to $9a$. That single simplification is the whole problem: the result never depends on $b$ at all, only on the tens digit. Tool #2 then just lists the nine multiples of $9$ to see which tens digit makes the result end in $6$, and the count of matching numbers falls out immediately.
Execute — Answer: D
6.EE.B.6 Step 1 Name the digits
- Let $a$ be the tens digit and $b$ be the units digit, so the two-digit number is $10a+b$.
- Here $a$ runs from $1$ to $9$ and $b$ runs from $0$ to $9$.
- The sum of the digits is simply $a+b$.
💡 A two-digit number is worth ten times its tens digit plus its units digit.
6.EE.A.3 Step 2 Subtract the digit sum
- Subtract the digit sum from the number and combine like terms.
- The $b$ terms cancel and the $a$ terms give $10a-a=9a$.
- So no matter what the units digit is, the result is exactly $9a$ — it depends only on the tens digit.
💡 The units digit you subtract is the same one that was in the number, so it disappears and only $9a$ survives.
4.OA.B.4 Step 3 Which tens digit ends in 6
- We need the units digit of $9a$ to be $6$.
- List the multiples of $9$ for $a=1$ through $9$ and read off their units digits: $9,18,27,36,45,54,63,72,81$.
- Only $36$ ends in $6$, and that happens when $a=4$.
💡 The result is a multiple of $9$, and among the one-digit-tens multiples of $9$ only $36$ ends in $6$.
3.OA.A.1 Step 4 Count the numbers
- So the tens digit must be $4$, while the units digit $b$ can be any of the ten values $0,1,2,\dots,9$ (it never affected the result).
- That gives the ten numbers $40,41,\dots,49$ — each of them yields $36$, ending in $6$.
- So there are $10$ such numbers, choice (D).
💡 The tens digit is pinned to $4$, but the units digit is free, so an entire decade of ten numbers works.
6.EE.B.6 Let $a$ be the tens digit and $b$ be the units digit, so the two-digit number is 6.EE.A.3 Subtract the digit sum from the number and combine like terms. The $b$ terms can 4.OA.B.4 We need the units digit of $9a$ to be $6$. List the multiples of $9$ for $a=1$ t 3.OA.A.1 So the tens digit must be $4$, while the units digit $b$ can be any of the ten v Review
Reasonableness: Spot-check both ends of the decade: $40 \to 4+0=4,\ 40-4=36$ (ends in $6$), and $49 \to 4+9=13,\ 49-13=36$ (ends in $6$). Both land on $36$, matching the claim that the result is always $9a=36$. A different tens digit fails, e.g. $50 \to 50-5=45$ ends in $5$. Because the result ignores the units digit, the answer must be a whole block of ten, so $10$ (D) is exactly right; the tempting $9$ (C) would come from forgetting one of the ten units digits, and $19$ (E) from wrongly thinking two tens digits qualify.
Alternative: Tool #3 (Eliminate Possibilities): once you see the result is $9a$ and never depends on the units digit, the numbers that work come in complete decades, so the count has to be a multiple of $10$. Among the choices only $10$ (D) is a multiple of $10$, which settles it without even finding that $a=4$.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Writing the two-digit number as $10a+b$ and its digit sum as $a+b$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Simplifying $(10a+b)-(a+b)$ to $9a$, showing the result depends only on the tens digit.)4.OA.B.4Find factor pairs and recognize multiples of a whole number in the range 1-100 (Listing the multiples of $9$ to find which tens digit makes the result end in $6$.)3.OA.A.1Interpret products of whole numbers as equal groups (Counting $1$ tens choice times $10$ units choices to get the $10$ qualifying numbers.)
⭐ Subtracting a two-digit number's digit sum always leaves $9$ times the tens digit, so the units digit is forgotten — find the one tens digit that works and every number in that decade of ten counts.
⭐ Subtracting a two-digit number's digit sum always leaves $9$ times the tens digit, so the units digit is forgotten — find the one tens digit that works and every number in that decade of ten counts.
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