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
OA is B.
But I am getting C as answer.
Experts Pl help.
Sandip Gumtya
How many 3s
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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?
'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?
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I think the answer should be 40. If we write number from 1 to 100 then each digit appears 20 times. So from 1 to 200 '3' should appear 40 times
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If we write the integers from 0 to 99 as TWO-DIGIT integers (00, 01, 02, 03, ....98, 99), then: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
- 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
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If I am correct you counted 30 to 39 as 10 that includes 33 also, so it should be total 9 and same for 133 as well
-Satya Achanta