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# how many times will the digit 7 be written when listing the

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Intern
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how many times will the digit 7 be written when listing the [#permalink]

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18 Sep 2007, 08:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

110
111
271
300
304

again any replies i would really appreciate it. this question is giving me this huge headache...lol.
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18 Sep 2007, 09:09
Is it 300? If yes, will explain.
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18 Sep 2007, 09:53
yeah the OA is 300. please explain please.
VP
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18 Sep 2007, 09:59
first take out 7 & 77 & 777 ---> total of six sevens.

next step is to think about x7 or 7x ---> x can be either 1,2,3 ... so for x7 you have 8 ways (excluding 0,7) and for 7x you have 9 ways (excluding 7).

now we have xx7 and x7x ---> every one of them has 72 ways , 8 for the first x and 9 for the second x ---> 8*9 = 72 (excluding 0,7 for the first x and 7 for the second x)

next 7xx ---> 9 ways * 9 ways = 81 (excluding only 7).

and last we have x77 or 77x and 7x7 --> 8 ways, 9 ways, and 9 ways ---> don't forget you get two sevens so multiply by two !!!

total

6+8+9+72+72+81+(9+9+8)*2 = 300

the answer is (D)

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18 Sep 2007, 10:50
thanks so much for the detailed explanation killer. the question is quite a handful, but ur explanation clarifies it very well. i appreciate it man. : )
Director
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Re: digit question [#permalink]

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18 Sep 2007, 11:20
chronolinkz wrote:
how many times will the digit 7 be written when listing the integers from 1 to 1000?

110
111
271
300
304

again any replies i would really appreciate it. this question is giving me this huge headache...lol.

Well I came up with 300 too ...just need to do some mental work.....which is tough to write it down.

[1 7 10 ], [17 , 27 37 …97 100 ], [107 117 127 ..177 ...701..777...997 ..1000]

number of 7s => 1 + 9 + 10 + (1 + 9 + 10 ) 9 + 100
20 + 20*9 + 100
= 300

( will try to write if possible and post later)
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Re: digit question [#permalink]

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18 Sep 2007, 11:24
chronolinkz wrote:
how many times will the digit 7 be written when listing the integers from 1 to 1000?

110
111
271
300
304

again any replies i would really appreciate it. this question is giving me this huge headache...lol.

From 1-100 There are 7,17,27,47,57,67,87,97 So multiply this by 10
=90
From 71-79 There are 10 7's (counting 77) = 9*10=90

We are discounting the 700-799 so far.
So total now is 180.

Now counting 700-799. we have 799-700+1=100 #'s. However this doenst cover all of the 7's just yet.

Now consider: 707,717,727,737,747,757,767,787,797 = 9 additional 7's on the end.

Now from 770-779. we have 770,771,772,773,774,775,776,777,778,779. we have 11 7's b/c we have an additional 7 from 777.

so 180+9+11+100=300!

Ans. D.

I don't really see a quick way to solve this. Just gotta know all the possibilities that 7 can show up.

My original answer was actually 301. So come test day i dunno if i would have picked 304 or 300. Not bad to narrow down to 2 answers, but considering i spent bout 3min on the prob. thats no good.
Director
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19 Sep 2007, 05:01
My approach for such types of questions, which needs count of particular digit in first thousand numbers.

Count the number of digits appearing in 100
7 will appear 20 times in writing first 100 numbers.
In 1000 numbers, total number of digits = 20*100 + 100 (701...800)= 300
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22 Sep 2007, 13:50
22 Sep 2007, 13:50
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# how many times will the digit 7 be written when listing the

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