KA-0075Complementary counting
5 points, difficulty 3 of 3Level 5–6about 135sHow many three-digit whole numbers contain at least one digit 7?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Count the ones with no 7 at all.
2The strategy
At least one is the signal for complementary counting. Count every three-digit number, then count those that avoid 7 entirely, and subtract. Remember the first digit cannot be zero.
3The full solution
There are 900 three-digit numbers. With no 7: 8 choices for the first digit, 9 for each of the others, giving 648. So 900 - 648 = 252.
Solution— no shortcut on this one
Count everything, then remove what you do not want. Three-digit numbers run from 100 to 999, so there are 900. Numbers with no 7 anywhere have 8 options for the first digit (1 to 9, excluding 7), and 9 for each of the other two (0 to 9, excluding 7): 8 x 9 x 9 = 648. So numbers with at least one 7 number 900 - 648 = 252. The transferable idea: at least one means count the complement, and the leading digit needs its own count because it cannot be zero.
Why this is on the test: Counting directly means three overlapping cases; the complement is one line, and the leading-zero trap catches students who forget the first digit is special.