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Rebaz
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unicornilove
Aren't there 19 and not 20 numbers from 1-99?
­No. Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). 
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Numbers from 1 to 1000 could be divided in 3 groups
Single digits 1~9, 2 digit numbers 10~99 and 3 digits numbers 100 to 999

Case 1—-~~ 1 digit number only 7

Case 2 ~~ 2 digits number
We have to fill the gap _ _

case a ) the first number is 7 then the second number can be 1 number from 0 to 9 (1C10)
1*10=10
Case b) 1 number is any number from 1 to 9 with 0 excluded (to be considered as a 2 digit number)
9*1=9


Case 3 ) ~~~ 3 digits numbers

_ _ _
Case a) 1 *10*10 =100
Case b) 9*1*10 =90
Case c) 9*10*1= 90

Total outcome 100+90+90+19+1 =300

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I got it doing it manually like you but with the counting method I do not understand why it does not work.

Spaces method
All - Not including 7
1000 - (9)(9)(9)
1000 - 729 = 271
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isaisrz
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

I got it doing it manually like you but with the counting method I do not understand why it does not work.

Spaces method
All - Not including 7
1000 - (9)(9)(9)
1000 - 729 = 271

The question does not ask about the number of integers containing the digit 7; rather, it asks about the number of times the digit 7 will be written when listing the integers from 1 to 1000. For example, the number 77 will count the digit 7 twice, and the number 777 will count it three times.
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I solved it with a lengthy approach

  • 1 Digit Number = 1 Seven

  • 2 Digit Numbers We have


* 7 X = 9 (as X can't be 7)
* X7 = 8 (as X can't be 7 and 0)
* 77 = 2 Seven


Total We got 19 Seven

  • 3 Digit Number We have


XX7 = 8*9 Number= 72 Seven
X7X = 8*9 Number = 72 Seven
7XX = 9*9 Number = 81 Seven
X77 = 8 Numbers = 8*2 = 16 Seven
77X = 9 Numbers = 9*2 = 18 Seven
7X7 = 9 Numbers = 9*2 = 18 Seven
777 = 1 Number = 1*3 = 3 Seven

Total We got 280 Seven


hence Total = 300 Seven we have

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Hey all,
My approach was, there are 998 numbers between 1-1000 (as it does not say it is inclusive)
So, I subtract the times when 7 does not appear

For X
- 7 times (2,3,4,5,6,8,9)

For XX
- 8x9 = 72
- Here for the tens (1,2,3,4,5,6,8,9)
- For the units (0,1,2,3,4,5,6,8,9)

For XXX
- 8x9x9 = 648

So, 998 - 7 - 72 - 648 = 271

Where did I go wrong?

Bunuel
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mbaaccount1234
Hey all,
My approach was, there are 998 numbers between 1-1000 (as it does not say it is inclusive)
So, I subtract the times when 7 does not appear

For X
- 7 times (2,3,4,5,6,8,9)

For XX
- 8x9 = 72
- Here for the tens (1,2,3,4,5,6,8,9)
- For the units (0,1,2,3,4,5,6,8,9)

For XXX
- 8x9x9 = 648

So, 998 - 7 - 72 - 648 = 271

Where did I go wrong?

Bunuel

First of all, "when listing the integers from 1 to 1000" implies that both 1 and 1000 are included.

However, this is not the main issue. The question does not ask for the number of integers that contain the digit 7; rather, it asks for the number of times the digit 7 is written when listing the integers from 1 to 1000.

For example, the number 77 contains the digit 7 twice, and the number 777 contains it three times. In your method, you're only counting how many numbers include the digit 7 at least once, but some of those numbers contain it more than once. To answer the actual question, you'd need to account for repeated occurrences of 7 within a single number and add those accordingly.
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Bunuel Bunuel,

Please let me know if my analysis/conclusion below is correct:

- When the question asks.. How many times the digit (say 7) appears from 1 to 1000, the instance (number) 777, is counted as 3 times.
-Whereas, when the question asks In how many integers does digit 7 appears (1-1000), the instance 777 is counted as one (integer).

This is the reason why number of times in this example is 300 and and number of integers in which digit 7 appears is 271.

Am I correct?

Juan C. Avellan
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avellanjc
Bunuel Bunuel,

Please let me know if my analysis/conclusion below are correct:

- When the question asks.. How many times the digit (say 7) appears from 1 to 1000, the instance (number) 777, is counted as 3 times.
-Whereas, when the question asks In how many integers does digit 7 appears (1-1000), the instance 777 is counted as one (integer).

This is the reason why number of times in this example is 300 and and number of integers in which digit 7 appears is 271.

Am I correct?

Juan C. Avellan

Yes, you're absolutely correct.

  • When the question asks how many times the digit 7 is written, every appearance counts separately. So 777 is counted three times (one for each 7).
  • When the question asks in how many integers does the digit 7 appear, each number is only counted once, no matter how many 7s it contains. So 777 is counted once.

That’s exactly why the answer to the first question is 300, and to the second one would be 271. Well understood.
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