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M01-10

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Re: M01-10  [#permalink]

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New post 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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Re: M01-10  [#permalink]

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New post 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.


Answer: D


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,
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Re: M01-10  [#permalink]

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New post 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.


Answer: D


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M01-10  [#permalink]

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New post 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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Re: M01-10  [#permalink]

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New post 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
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M01-10  [#permalink]

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New post 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. :) :thumbup: Press Kudos :P
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M01-10 &nbs [#permalink] 11 Nov 2018, 11:00

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