If we listed all numbers from 100 to 10,000, how many times would the

We would consider cases:

# 3 digit numbers:
> Of the form: A B 3 => 9 x 10 = 90
> Of the form: A 3 B => 9 x 10 = 90
> Of the form: 3 A B => 10 x 10 = 100

# 4 digit numbers:
> Of the form: A B C 3 => 9 x 10 x 10 = 900
> Of the form: A B 3 C => 9 x 10 x 10 = 900
> Of the form: A 3 B C => 9 x 10 x 10 = 900
> Of the form: 3 A B C => 10 x 10 x 10 = 1000

Thus, total such numbers: 90 + 90 + 100 + 900 + 900 + 900 + 1000 = 3980

Answer A

Total 3's from 1 to 100 (i.e. 3's at only unit and tens place) = 20
so every 100 numbers have 3's listed 100 times at unit and tens places
Total 100's from 100 to 10000 = 10000/100 - 100/100 = 99
So total 3's at unit and tens places = 99*20 = 1980

Total 3's in every 1000 numbers at hundreds place = 100 (in the series of 300 to 399)
Total series of 1000 from 100 to 10000 = 10
i.e. total 3's at hundreds digit place = 10*100 = 1000

Total 3's at thousands place = 1000 (series from 3000 to 3999)

So total 3's = 1980 + 1000 + 1000 = 3980

Answer: Option A

Wouldn't this lead to double counting? as AB3 might include '333' and 3AB might also include '333'

let's count total 3's from 1-10,000 minus total 3's 1-100

Total 3's 1-100
--> at unit place (03,13,23...) = 10
--> at tenth place (30,31,33..) = 10
Hence total = 10+10=20

Total 3's from 1-10,000

Step 1) unit & tenth digit

Total 3's from 1-100
--> at unit place (03,13,23...) = 10
--> at tenth place (30,31,33..) = 10
Hence total = 10+10=20

Now, 101-200 it will also be 20, 201-300 will also be 20
20 at each segment - till 10,000 = 20x 10 (from 1-1,000) = 200 X 10 (for each 1000 segment i.e 200 3's at unit & tenth place at each 1000 numbers) = 2000

Step 2) hundreds digit

300-399 it is 100 3's . which will repeat in each 1000 segment i.e 1300-1399 (will also have 100 3's )
100x10 (no. of segments till 10,000) = 1000

Step 3) Thousands Digit

3,000-3999 =1000 3's at thousands place

Now, Step 1 + Step 2 + Step 3 = 2000+1000+1000 =4000

4000 3's from 1-10,000

Now subtract no. of 3's from 1-100 (refer steps before step 1)
Hence 4000-20= 3980
Option A

AabhishekGrover
You can check my explanation. We have considered 333 as well

Thank you for your help!

Double counting is present for calculation of number three between numbers 100 to 10000. So we are challenging with repetitions of same numbers beginning by AB3 to ABC3(103-9993).
AB3=90; Here are 90 repetitions.
A3B=90 Here are 90 repetitons minus double counting numbers which are: 133, 233, 333, 433, 533, 633, 733, 833 and 933. So, minus 9 repetitions.
3AB=100; Here are 100 repetitions minus double numbers as: 303, 313, 323, 330, 331, 332, 334, 335, 336, 337, 338, 339, 343, 353, 363, 373, 383 and 393.
So here are minus 18 repetitions of number three.
For numbers from 100 to 1000 we conclude that diference(double counting) is as foĺlow:
AB3=90 repetitions of 'three';
A3B=90 repetitions minus 9 equal to 81 and,
3AB=100 repetitions minus 18 equal to 82 repetitions of number 3!
So, AB3+A3B+3AB=90+81+82=253.
If we continue for double counting numbers on ABC3, AB3C, A3BC and 3ABC of course that sume of double counting is greater.
I think possible answer could be under C) due to raising numbers as repetitors but, need verification!
Regards

Posted from my mobile device

­3AB, A3B, AB3 all this will include 333. 
We want this because 3 is printed 3 times in 333, so we need the overlap.