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# M01-10

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Math Expert
Joined: 02 Sep 2009
Posts: 52906

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24 Dec 2017, 00:37
Siddhuftr wrote:
ones - x - 1 time
tens - x7 + 7x - 9*1 + 1*10 = 19 times
Hundreds - xx7 + x7x + 7xx = 9*10*1 + 9*1*10 + 1*10*10 = 190
Total = 300

I am unable to understand , how double counting of 77 , 777 is avoided ?

The question asks "How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7. So, for example, for number 777, it should be counted as three 7's not one.
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Intern
Joined: 03 Sep 2017
Posts: 18
Location: Brazil
GMAT 1: 730 Q49 V41

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21 May 2018, 08:25
Bunuel wrote:
Official Solution:

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

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Hello Bunuel

Could you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000?
I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused.

Thank you,
Math Expert
Joined: 02 Sep 2009
Posts: 52906

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21 May 2018, 20:28
vitorcbarbieri wrote:
Bunuel wrote:
Official Solution:

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

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Hello Bunuel

Could you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000?
I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused.

Thank you,

Since neither 0 not 1000 has 7 in it, then excluding/including these values into the range won't change the answer.
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Status: EAT SLEEP GMAT REPEAT!
Joined: 28 Sep 2016
Posts: 166
Location: India

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01 Jun 2018, 07:18
How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Hi Bunuel
Can you please suggest if the below method is correct.

_ _ 7 => 10*10 (since we can use 0 in hundred's place as well in this case)
_ 7 _ => 10*10 (since we can use 0 in hundred's place as well in this case)
7 _ _ => 10*10

Totally = 100+100+100 =300

Option D

Thanks!
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Regards,

Intern
Joined: 01 Aug 2018
Posts: 2
GPA: 3.2

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06 Sep 2018, 04:29
Hello Bunuel

If we look at the approach 1, Doesn't it generalize it for any digit?

Which ever digit you ask for it will come 300 times in 1 to 1000.

Regards,
Amardeep
Intern
Joined: 08 Mar 2018
Posts: 3

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11 Nov 2018, 11:00
approach
For every digit to appear from 1 to 99 -> 20 times any digit will appear
similarly for 100,200,300,400,500,600,800,900 -> total 8*20=160 (Because hundred will remain constant and we know the fact that for 1to99 20 times a digit will appear)
For 700: There will be 100 digits for 7-> 100
For7XX there will be 20 ways ->20
Total (20+160+100+20) there will be 300 digits

I hope it helps a bit. Press Kudos :P
M01-10   [#permalink] 11 Nov 2018, 11:00

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# M01-10

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