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 Question from Veronica, a student: How many numbers between 200 and 750 have a 9 for at least one of the digits? The answer in the textbook is 60 but I don't see how that is correct, that each 100 has 90,91,92,93 etc.

Hi Veronica,

I don't see $60$ as being the answer either.

Concentrate on the two-hundreds. The nine numbers $209, 219, ..., 289$ each contain at least one $9$ and so do the ten numbers $290, 291, ..., 299$. That gives me $9 + 10 = 19$ numbers in the two-hundreds with a least one digit being a $9.$ The same is true for the three-hundreds, four-hundreds, five-hundreds and six-hundreds. That's $5 \times 19 = 95$ numbers with at least one digit being a $9$ and I haven't looked at the seven-hundreds yet.

Penny

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