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Re: M09-17 [#permalink]
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deepthit wrote:
Here are the list of numbers
9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, and 99
Total - 20 9's in 1 hundred. Total = 20*7 = 140


the Q ask us about the PAGES on which 9 occurs, so 99 will be ONE and not two separately
so 19*7
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Re: M09-17 [#permalink]
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Another approach:

Pages from 1 to 705.

9 as units digit: 9- first number in set, 699-last number in set, set is inclusive. Distance between each occurrence of '9' in units digits = 10.
>> Number of occurrences = [(699-9)/10] + 1 = 70

9 as tenths digit: if hundredth digit is '0', we have 90-first number in set, 98-last number in set (omit 99 as already counted above), set is inclusive.
>> Number of occurrences: (98-90) + 1 = 9
Possible hundredth digits: 0 1 2 3 4 5 6 >> total number of occurrences = 9*7=63

Finally: total pages with '9' = 63 +70 = 133
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Re: M09-17 [#permalink]
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For those who want to solve this using combinatorics
Attachments

solving with combinatorics.png
solving with combinatorics.png [ 45.73 KiB | Viewed 21157 times ]

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Re: M09-17 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M09-17 [#permalink]
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