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M01-10

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Re: M01-10  [#permalink]

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New post 24 Dec 2017, 01:37
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, 09:25
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, 21:28
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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Re: M01-10  [#permalink]

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New post 03 May 2019, 22:15
Bunuel chetan2u

The answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000).

Let me know if my understanding is correct?
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Re: M01-10  [#permalink]

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New post 03 May 2019, 23:38
1
SpiritualYoda wrote:
Bunuel chetan2u

The answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000).

Let me know if my understanding is correct?


Hi..
if we write all as 000 to 999, we will have 3000 digits as 3(number of digits in each number )*1000(total numbers)=3000..
In these 3000, all 10 digits will be 3000/10=300..

But we do not have 0s in single digit and two digits..
so 000,as we do not have 0, so 3 0s missing
00a.. from 1 to 9, they are not written as 001,002...009, so 2*9=18 x 0s missing
0ab.....from 10 to 99, they are not written as 010,011...099, so 1*90=90 x 0s missing
Total 3+18+90=111 number of 0s are less..

Thus, from 1 to 999, we have 300 of all 9 digits - 1 to 9, while we have 300-111=189 of 0s.
Add 1000,
digit 0 is used 189+3=192 times
digit 1 is used 300+1=301 times
digits 2 to 9, 300 times each

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Re: M01-10  [#permalink]

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New post 29 May 2019, 16:25
Hey Bunuel, what is meant by "In 10 hundreds 7" in approach 3? I'm having a hard time understanding this.
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Re: M01-10   [#permalink] 29 May 2019, 16:25

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