In 1 to 100 total "7" appearance is 20 times..

so 20 x 10 = 200
+ 100 times appearance of 7 from 700 to 799

So, 200 + 100 = 300

rakeshd347 wrote: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

OA is D

Please explain your approach.
Source Jeff sackman's book

From 1 to 100 7 as last digit 9*10=90

Then 7 as 2nd digit from 1 to 100 9*10=90

Total 90+0=180

Then 700 to 799 100 times

So 90+90+100=300 times

I hope I am correct.

More Algebraic approach:

We have options 0 to 9(including 7 ) to fill two places and "7" to fill 3rd spot.

So, 3 ( 10C1 x 10C1 x 1C1) = 3 ( 10 x 10 x 1 ) = 300

vinay1983 wrote:
From 1 to 100 7 as last digit 9*10=90

Then 7 as 2nd digit from 1 to 100 9*10=90

Total 90+0=180

Then 700 to 799 100 times

So 90+90+100=300 times

I hope I am correct.

Vinay, the appearance of "7" at unit place from 1 to 100 will be 10 times and not 9 times.. so the count in your first step should be 10*10 = 100
similarly, in second step it should be 10*10 = 100
so, 200 + 100 = 300

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

(a) 271
(b) 300
(c) 252
(d) 304
(e) 512

If we use 0 as a placeholder -- so that 007 represents 7, 072 represents 72, etc. -- then we need to count the number of times that 7 will appear among the 3-digit integers from 000 to 999, inclusive.

Total number of 3-digit integers from 000 to 999, inclusive = biggest - smallest + 1 = 999-000+1 = 1000.

Each of these 1000 integers includes 3 digits.
Thus, the total number of digit appearances = 3*1000 = 3000.

Among these 3000 digit appearances, each of the 10 digits will appear the same number of times.
Thus, the number of times that 7 will appear = 3000/10 = 300.

The correct answer is B.

theCodeToGMAT wrote:

vinay1983 wrote:
From 1 to 100 7 as last digit 9*10=90

Then 7 as 2nd digit from 1 to 100 9*10=90

Total 90+90=180

Then 700 to 799 100 times

So 90+90+100=300 times

I hope I am correct.

Vinay, the appearance of "7" at unit place from 1 to 100 will be 10 times and not 9 times.. so the count in your first step should be 10*10 = 100
similarly, in second step it should be 10*10 = 100
so, 200 + 100 = 300

Yes it so.Thanks for pointing out the mistake.

rakeshd347 wrote: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

Use the slot method to find the number of appearances of 7:

1 - 9 --> only 7 --> 1

10 - 99 --> _ _ --> _ 7 & 7 _
First case: tens digit can take 1 - 9 = 9 Values
Second case: units digit can take 0 - 9 = 10 Values

100 - 999 --> _ _ _ --> _ _ 7 & _ 7 _ & 7 _ _
First case:
hundreds digit can take 1 - 9 = 9 Values
tens digit can take 0 - 9 = 10 Values
total ways = 90
Second case:
hundreds digit can take 1 - 9 = 9 Values
units digit can take 0 - 9 = 10 Values
total ways = 90
Third case:
tens digit can take 0 - 9 = 10 Values
units digit can take 0 - 9 = 10 Values
total ways = 100

Total = 1 + 9 + 10 + 90 + 90 + 100 = [spoiler]300; Answer D[/spoiler]