You have the right answer - we'd write '3' as a units digit exactly 20 times, and as a tens digit exactly 20 times, so we'd write the digit 40 times in total.

'38' is the answer to a different question: how many whole numbers between 1 and 200 contain the digit '3'? Because the answer to the previous question counts "33" and "133" twice, the answer to this question would be 40 - 2 = 38.

So perhaps the source was confused about what question they were asking - where is the question from?

sandipgumtya wrote:If we write numbers from 1 to 200,then how many times will the digit 3 be written?

A-20
B-38
C-40
D-42
E-44

If we write the integers from 0 to 99 as TWO-DIGIT integers (00, 01, 02, 03, ....98, 99), then:
- we have 100 integers, each with 2 digits for a total of 200 digits.
1/10 of those digits are 0's, 1/10 of those digits are 1's, 1/10 of those digits are 2's, AND...
1/10 of those digits are 3's
So, the number of 3's = (1/10)(200) = 20

Next, if we write the integers from 100 to 199, we'll get another 20 3's

So, the number of 3's = 20 + 20 = [spoiler]40 = C[/spoiler]

Cheers,
Brent