AMC 10 · 2004 · #2
Grade 3 arithmeticHow many two-digit positive integers have at least one 7 as a digit?
Pick an answer.
AMC 10 2004 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: Among the two-digit positive integers, count how many have the digit $7$ in the tens place, the ones place, or both.
Givens: The numbers considered are the two-digit positive integers, which run from $10$ to $99$.; A number counts if at least one of its two digits is a $7$.; Answer choices: (A) $10$, (B) $18$, (C) $19$, (D) $20$, (E) $30$
Unknowns: How many two-digit integers have a $7$ as the tens digit, the ones digit, or both.
Understand
Restated: Among the two-digit positive integers, count how many have the digit $7$ in the tens place, the ones place, or both.
Givens: The numbers considered are the two-digit positive integers, which run from $10$ to $99$.; A number counts if at least one of its two digits is a $7$.; Answer choices: (A) $10$, (B) $18$, (C) $19$, (D) $20$, (E) $30$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List
"At least one $7$" is the classic trigger for Tool #16 (Count the Complement). Chasing the numbers that contain a $7$ directly means adding the ten numbers $70$–$79$ to the nine numbers $17,27,\dots,97$, but $77$ sits in both lists and is easy to double-count. Flipping the question is cleaner: count every two-digit number, then subtract the ones with no $7$ at all, because a number either has a $7$ or it does not. Tool #2 (Make a Systematic List) supports this by organizing the digit choices — how many tens digits and how many ones digits avoid $7$ — so the no-$7$ group is counted without missing or repeating any case.
Execute — Answer: B
2.NBT.A.2 Step 1 Count every two-digit number
- Instead of hunting for the sevens, look at the whole pool first.
- The two-digit positive integers are $10, 11, 12, \dots, 99$.
- Counting a run that includes both ends gives $99-10+1=90$ numbers in all.
- Every one of them either contains a $7$ or contains no $7$, so once we know how many have no $7$, subtraction gives the rest.
💡 It is easier to count the whole group and remove the unwanted part than to chase the wanted part directly.
3.OA.A.1 Step 2 Count the numbers with no 7
- Now count the two-digit numbers that avoid $7$ completely.
- The tens digit can be $1,2,3,4,5,6,8,9$ — that is $8$ choices, since it cannot be $0$ and cannot be $7$.
- The ones digit can be $0,1,2,3,4,5,6,8,9$ — that is $9$ choices, since it just cannot be $7$.
- Each of the $8$ allowed tens digits pairs with each of the $9$ allowed ones digits, giving $8$ groups of $9$: $8\times9=72$ numbers with no $7$.
💡 Pairing every allowed first digit with every allowed second digit makes equal groups, so the count is a product.
2.NBT.B.5 Step 3 Subtract to get the sevens
- The numbers that have at least one $7$ are exactly the ones left over after removing the no-$7$ group from the whole pool.
- Subtract: $90-72=18$.
- This lands on choice (B).
- No double-counting is possible here, because each number was placed in exactly one of the two groups — has a $7$ or has no $7$.
💡 Whole minus the part with no $7$ leaves precisely the part with at least one $7$.
2.NBT.A.2 Instead of hunting for the sevens, look at the whole pool first. The two-digit p 3.OA.A.1 Now count the two-digit numbers that avoid $7$ completely. The tens digit can be 2.NBT.B.5 The numbers that have at least one $7$ are exactly the ones left over after remo Review
Reasonableness: The answer $18$ should be a bit under the count you would guess from adding $70$–$79$ (ten numbers) to $\_7$ endings (nine numbers), and indeed $18$ is one less than that raw $19$ — exactly the correction for counting $77$ twice. That rules out (C) $19$, which is the classic double-count trap, and (D) $20$. It also comfortably exceeds (A) $10$, since numbers ending in $7$ add several more beyond the $70$s. So $18$ sits right where it should.
Alternative: Count directly with a Venn view (Tool #12). Numbers with a $7$ in the tens place: $70,71,\dots,79$ — that is $10$ numbers. Numbers with a $7$ in the ones place: $17,27,37,47,57,67,77,87,97$ — that is $9$ numbers. The single number $77$ appears in both lists, so by inclusion–exclusion the total is $10+9-1=18$, confirming (B).
CCSS standards used (min grade 3)
2.NBT.A.2Count within 1000, skip-count by 5s, 10s, and 100s (Counting the whole pool of two-digit integers from 10 to 99 as 90 numbers.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Counting the no-7 numbers as 8 allowed tens digits times 9 allowed ones digits, 8x9=72.)2.NBT.B.5Fluently add and subtract within 100 (Subtracting the no-7 numbers from the whole pool: 90-72=18.)
⭐ When a problem asks for "at least one," count everything and subtract the cases that have none — it dodges the double-counting trap.
⭐ When a problem asks for "at least one," count everything and subtract the cases that have none — it dodges the double-counting trap.
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