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

Question

If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?

A. 3980
B. 3840
C. 3780
D. 3760
E. 3700

minustark · Accepted Answer

For 3 digit integers: when 3 is in the hundred's place = 3AB = 1* 10* 10 = 100
3 in the ten's place A3B = 9*1*10 =90
3 in the units place AB3 = 9*10*1 =90
For all 4 digit integer: 3ABC = 10*10*10 = 1000
A3BC =9*10*10= 900
AB3C = 9*10*10 =900
ABC3 = 9*10*10 =900
Total integers= 280+ 2700 +1000 = 3980.
A is the answer

nick1816 · Answer

Number of 3's from 00 to 99 =  \frac{1}{10}(100*2)= 20  {we divided by 10, since number of times each digit got printed is same}

Number of 3's from 0000 to 9999  = \frac{1}{10}(10000*4)= 4000

Number of 3's from 100-10000 = 4000-20 = 3980

sujoykrdatta · Answer

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

GMATinsight · Answer

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

AabhishekGrover · Answer

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

GMATinsight · Answer

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

ankitabth · Answer

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 · Answer

Thank you for your help!

Beko2021 · Answer

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

thr3at · Answer

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

kll08 · Answer

Hi  GMATinsight

I think the double counting is still there.

Eg.

100 - 199 = 1A3 or 13A = 10 + 10 - 1 (For 133 Twice) = 19

So, shouldn't it be 19 instead of 20?

Please clarify

iamjatinagrawal · Answer

Quick and efficient way to solve such questions in under 30 seconds.

Find nearest full 0–999, 0–9999, etc. 
 Each full block contributes  n×10^{n−1} , where n = the number of digits/zeroes after digit 1 (like 4 in case of 9999 or 10,000)

In the given question, we have 10,000 which gives us n = 4 (for 9999)

Hence, the number of times 3 appearing in numbers from 1 to 10000 =  4*10^{4-1} = 4*1000 = 4000

Now, for to find number of 3 appearing in numbers from 1 to 100, n will be 2 which gives us  2*10^{2-1}= 2*10 = 20

Final result for what the question is asking will be = 4000 - 20 = 3980.  Option (A)

pappal · Answer

what if we count the total numbers from 100 through 10000 and subtract the total of all 3 digit and 4 digit numbers without 3 in them.(numbers formed using digits 0 to 9 excluding 3)
I tried but came with an answer quite below i.e.3421
where was i incorrect?

Bunuel · Answer

You undercounted because your method only found how many numbers contain a 3, not how many total 3s appear. Some numbers (like 3333) contribute multiple 3s. That’s why your 3421 is too low; the correct total of 3s printed is 3980.

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

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