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Old 05 April 2007, 03:53 PM
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paddy_t
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If a book contains the digit 3 exactly 99 times (e.g. 3, 13, 23, 30...33-this counts as two digits) what would the final page number be?

How could you find this out without writing every page number out?
Old 05 April 2007, 04:18 PM
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KiwiGTI
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Use your head?

Probably a formula but it's too complex for Primary school.

20 occurrences per 100, but then you get the 300's

0-99 - 20
100 -199 - 20
200 - 299 - 20
300 - 399 - 110
etc

Think it's 335 give or take 1-2

Last edited by KiwiGTI; 05 April 2007 at 04:20 PM.
Old 05 April 2007, 08:36 PM
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SWRTWannabe
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As KiwiGTI says, 20 per 100 for the first three hundreds, so

0-299 = 60

Then 11 per 10 for 300-329

300-329 = 33 (total so far = 93)

330 = 2 (total so far = 95)
331 = 2 (total so far = 97)
332 = 2 (total so far = 99)

So the answer is 332
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