



 
Hi Veronica, I don't see $60$ as being the answer either. Concentrate on the twohundreds. 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 twohundreds with a least one digit being a $9.$ The same is true for the threehundreds, fourhundreds, fivehundreds and sixhundreds. That's $5 \times 19 = 95$ numbers with at least one digit being a $9$ and I haven't looked at the sevenhundreds yet. Penny  


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