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# How many number of times will the digit 7' be written when

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How many number of times will the digit 7' be written when [#permalink]

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29 Oct 2005, 10:07
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How many number of times will the digit â€˜7' be written when listing the integers from 1 to 1000?

(1) 271
(2) 300
(3) 252
(4) 304
(5) None of the above

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Manager
Joined: 01 Aug 2005
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29 Oct 2005, 10:18
2) 300?

I think they are trying to fool you to choose 271, but then you are ignoring some 7's in 7xx.

In every 100 (non 7xx) you have:

7 17 27 37 47 57 67 70 71 72 73 74 75 76 77 78 79 87 97 = 20 sevens.
so 20x10 = 200 sevens.

Then you must add the 100 sevens in all 700 numbers. You are up to 300 sevens give or take.

I am doing this quite fast so it is possible I added up one too many sevens, in which case the answer might be E.

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Manager
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Re: PS: P/C Ascent 4/02 [#permalink]

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29 Oct 2005, 20:42
GMATT73 wrote:
How many number of times will the digit â€˜7' be written when listing the integers from 1 to 1000?

(1) 271
(2) 300
(3) 252
(4) 304
(5) None of the above

Let us take 3 digit numbers first -

Total number of combinations 9*10*10 = 900.
Combinations with NO 7 in it. 8*9*9 = 648.
Combinations with 7 = 900-648 = 252.

Now take 2 digit numbers that is easy 10+9 = 19.

252+19 = 271.
A.

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Director
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29 Oct 2005, 20:58
B) 300

7XX - 100
X7X - 100 (10 times in every 100 - 70-79;170-179...)
XX7 - 100 (10 times in every 100 - 07,17,27...107,117...)

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Current Student
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29 Oct 2005, 21:20

--------------------------------------------------------------------------------
Solution:

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 <= x, y, z <= 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300

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Manager
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29 Oct 2005, 22:45
Javin,

Quote:
Now take 2 digit numbers that is easy 10+9 = 19.

Look at my post. I have listed 2 digit numbers and the amount of sevens in them. It is not 19 as you can see - I think you are forgetting 77, and for that matter other numbers where 7 is written more than once.[/quote]

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29 Oct 2005, 22:45
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