AMC 10 · 2004 · #2
Grade 3 arithmeticPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
"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,…,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.
Count every two-digit number
Look at the whole pool first — the two-digit numbers run from 10 to 99, so there are 90 of them.
It is easier to count the whole group and remove the unwanted part than to chase the wanted part directly.
2.NBT.A.2Change Focus Count The ComplementCount the numbers with no 7
With no 7 allowed, the tens digit has 8 choices and the ones digit has 9, giving 72 numbers.
Pairing every allowed first digit with every allowed second digit makes equal groups, so the count is a product.
Pairing every allowed first digit with every allowed second digit makes equal groups, so the count is a product.
▸ Why?
The two digit places are chosen without regard to each other, so their counts multiply.
▸ Why?
Every number either has the digit or does not, so subtracting the ones without it leaves the ones with it.
Subtract to get the sevens
Take the no-7 group out of the whole pool: 90 minus 72 leaves 18, which is choice (B).
Whole minus the part with no 7 leaves precisely the part with at least one 7.
2.NBT.B.5Change Focus Count The ComplementWhen a problem asks for "at least one," count everything and subtract the cases that have none — it dodges the double-counting trap.
- Count every two-digit number
- Count the numbers with no 7
- Subtract to get the sevens