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Challenges 1, Q10

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Challenges 1, Q10 [#permalink] New post 25 May 2011, 01:59
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Hi, have problem with such kinde of questions. Maybe somebody has another explanation. Or it is possible to finde a topic in Manhattan Guides.
Thank you.

How many times will the digit 7 be written when listing the integers from 1 to 1000?

(C) 2008 GMAT Club - m01#10

110
111
271
300
304
There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

The correct answer is D.
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Re: Challenges 1, Q10 [#permalink] New post 25 May 2011, 08:20
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another way:

1. The number of 7s is the same in 000-999 as in 1-1000.
2. There is 1000 numbers and 3000 digits in 000-999
3. All digits (0-9) appear with the same probability - 1/10
4. So, 7 is written 3000*1/10 = 300 times
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Re: Challenges 1, Q10 [#permalink] New post 27 May 2011, 18:07
Thanks a lot Walker.... it was quite difficult but you made it easy to understand...
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Re: Challenges 1, Q10 [#permalink] New post 02 Sep 2011, 14:33
walker wrote:
another way:

1. The number of 7s is the same in 000-999 as in 1-1000.
2. There is 1000 numbers and 3000 digits in 000-999
3. All digits (0-9) appear with the same probability - 1/10
4. So, 7 is written 3000*1/10 = 300 times



just wanted someone to clarify, how are there 3000 digits in 000-999

awaiting a reply
Thanx
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Re: Challenges 1, Q10 [#permalink] New post 03 Sep 2011, 11:04
There is 1000 numbers in 000-999 set ((999 - 0) + 1 = 1000) and each number consists with 3 digits: 3*1000 = 3000 digits
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Re: Challenges 1, Q10   [#permalink] 03 Sep 2011, 11:04
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