How many times 3 appear in 1 to 1000 ?

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How many times 3 appear in 1 to 1000 ?

by crackitpal » Tue Aug 14, 2007 2:44 pm
Hello Everybody,

How can I find out how many times number 3 appears in 1 to 1000 ? The answer is 300. Please let me know the method to calculate it.

Thank you

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Re: How many times 3 appear in 1 to 1000 ?

by veekay » Tue Aug 14, 2007 4:00 pm
crackitpal wrote:Hello Everybody,

How can I find out how many times number 3 appears in 1 to 1000 ? The answer is 300. Please let me know the method to calculate it.

Thank you
This is P&C question


Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

Case 1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits

You have 1*9*9 = 81 such numbers. However, 3 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 3 will appear only once.

Case 2. The numbers in which 3 will appear twice. In these numbers, one of the digits is not 3 and it can be any of the 9 digits.
There will be 9 such numbers. However, this digit which is not 3 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 3 is written twice. Therefore, 3 is written 54 times.

Case 3. The number in which 3 appears thrice - 333 - 1 number. 3 is written thrice in it.

Therefore, the total number of times the digit 3 is written between 1 and 999 is 243 + 54 + 3 = 300

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by crackitpal » Thu Aug 16, 2007 9:03 am
Thank you ! I took time to think and solve it in my own words. I understood this now. Just a question, in GMAT, what is the difficulty level of this level ? Easy, medium difficult or difficult ?

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by aim-wsc » Thu Aug 16, 2007 10:13 am
I think we already have discussed this problem earlier....
:?

Please use search function effectively :) before starting a new topic in GMAT subjects section.
https://www.beatthegmat.com/viewtopic.ph ... ears++1000

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by jairajcool » Thu Nov 20, 2014 6:10 pm
A simple method to solve this problem would be

First we need to know How many 3's are there in 1 to 100

3, 13,...33,...93 = ten 3s at the unit's place
30,31...39= ten 3s at the ten's place.
So a total of 20 threes make an appearance here

so for 1 to 1000, 3's in tens and units place would be 20X10="200" ---(1)
Here 10= is for every 100 digits till 1000

and for from 300 to 399 there are "100" 3's in hundred's place ---(2)
so, therefor considering 1+2
100+200= 300..

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by Brent@GMATPrepNow » Thu Nov 20, 2014 7:00 pm
How can I find out how many times number 3 appears in 1 to 1000 ?


Here's another way to look at it.
Write all of the numbers as 3-digit numbers.
That is, 000, 001, 002, 003, .... 998, 999

NOTE: Yes, I have started at 000 and ended at 999, even though though the question asks us to look at the numbers from 1 to 1000. HOWEVER, notice that 000 and 1000 do not have any 3's so the outcome will be the same.

First, there are 1000 integers from 000 to 999
There are 3 digits in each integer.
So, there is a TOTAL of 3000 individual digit. (since 1000 x 3 = 3000)

Each of the 10 digits is equally represented, so the 3 will account for 1/10 of all digits.

1/10 of 3000 = 300

So, there are 300 0's, 300 1's, 300 2's, 300 3's, . . ., and 300 9's in the integers from 000 to 999

Cheers,
Brent
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by [email protected] » Thu Nov 20, 2014 7:13 pm
Hi All,

Since the two "recent" posts are from people offering solutions to the original question (from over 7 years ago), I won't rehash any of that here. To anyone who might be coming across this post for the first time, it's worth noting that on Test Day there would be 5 answer choices to this question. The answer choices sometimes provide big hints as to what the correct answer is, or even how to get to the correct answer. The way this question was originally posted forces us to deal with the prompt as if it were a "math" question, which is rather limiting. Take advantage of all the information that prompts give you and learn to master more than one approach to answering questions.

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by Brent@GMATPrepNow » Thu Nov 20, 2014 7:22 pm
How can I find out how many times number 3 appears in 1 to 1000 ?
To be "nitpicky," I should point out that the number 3, appears only once in the integers from 1 to 1000.
We have 1, 2, 3, 4, 5, ....

To be SUPER "nitpicky," the digit 3, appears INFINITELY MANY TIMES in 1 to 1000.
Since the question doesn't say that we're dealing with INTEGERS, we can even say that the digit 3 appears INFINITELY MANY TIMES in the (rational) numbers from 1 to 2.
There's 1.33333333...., 1.2333333333... etc.

A true GMAT question would look something like this:
How many times does the digit 3 appear in the integers from 1 to 1000 inclusive?
A) 100
B) 250
C) 300
D) 333
E)334

OA: C
Cheers,
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by Mathsbuddy » Fri Nov 21, 2014 7:34 am
Brent@GMATPrepNow wrote:
How can I find out how many times number 3 appears in 1 to 1000 ?
To be "nitpicky," I should point out that the number 3, appears only once in the integers from 1 to 1000.
We have 1, 2, 3, 4, 5, ....

To be SUPER "nitpicky," the digit 3, appears INFINITELY MANY TIMES in 1 to 1000.
Since the question doesn't say that we're dealing with INTEGERS, we can even say that the digit 3 appears INFINITELY MANY TIMES in the (rational) numbers from 1 to 2.
There's 1.33333333...., 1.2333333333... etc.

A true GMAT question would look something like this:
How many times does the digit 3 appear in the integers from 1 to 1000 inclusive?
A) 100
B) 250
C) 300
D) 333
E)334

OA: C
Cheers,
Brent
To be even more nitpicky: the digit 3 appears INFINITELY MANY TIMES in the (rational AND IRRATIONAL) numbers from 1 to 2.

Just having a laugh!

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by Brent@GMATPrepNow » Fri Nov 21, 2014 8:44 am
Mathsbuddy wrote:
Brent@GMATPrepNow wrote:
How can I find out how many times number 3 appears in 1 to 1000 ?
To be "nitpicky," I should point out that the number 3, appears only once in the integers from 1 to 1000.
We have 1, 2, 3, 4, 5, ....

To be SUPER "nitpicky," the digit 3, appears INFINITELY MANY TIMES in 1 to 1000.
Since the question doesn't say that we're dealing with INTEGERS, we can even say that the digit 3 appears INFINITELY MANY TIMES in the (rational) numbers from 1 to 2.
There's 1.33333333...., 1.2333333333... etc.

A true GMAT question would look something like this:
How many times does the digit 3 appear in the integers from 1 to 1000 inclusive?
A) 100
B) 250
C) 300
D) 333
E)334

OA: C
Cheers,
Brent
To be even more nitpicky: the digit 3 appears INFINITELY MANY TIMES in the (rational AND IRRATIONAL) numbers from 1 to 2.

Just having a laugh!
So, maybe we should say that 3 appears (2)(infinity) times in the rational AND IRRATIONAL numbers from 1 to 2 :-)

Cheers,
Brent
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by Mathsbuddy » Fri Nov 21, 2014 9:38 am
How can I find out how many times number 3 appears in 1 to 1000 ?

Assuming "number" means digit,
in binary it doesn't appear at all!

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by Brent@GMATPrepNow » Tue Apr 11, 2017 6:55 am
Mathsbuddy wrote:How can I find out how many times number 3 appears in 1 to 1000 ?

Assuming "number" means digit,
in binary it doesn't appear at all!
I just noticed this response. So, here's a belated "ha!" :-)
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by moshe.k » Sun Jun 25, 2017 1:13 am
Sorry, but i think all above are wrong!! since they calculate some numbers twice!!
between 1 and 100 we have 3,13,23,43,53,63,73,83,93 = 9 times
and we have the numbers 30-39 = 10 times which gives us total of 19 nunbers!! (not 20)
So between 0 and 1000 this metod will repeat itself 9 times for, 0-99, 100-199, 200-299, 400...
and that will five us 9*19=171 and we have 100 nunbers 300-399
so the correct answer is 271!! not 300

the mistake of "veekay" was that we only have 27 combinations in which 3 appears twice in the number and not 54
so the calculation is 243 (for 1 digit only) + 27 (for 2 digits only) + 1 (for 333)=271!!

Sorry guys, i was wrong since i forgot that between 30-39 we have 11 times the digit 3, and between 300-399 we have 120 times the digit 3 so it comes up to 300 times in total.
Last edited by moshe.k on Sun Jun 25, 2017 2:48 am, edited 1 time in total.

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by GMATGuruNY » Sun Jun 25, 2017 1:59 am
moshe.k wrote:Sorry, but i think all above are wrong!! since they calculate some numbers twice!!
between 1 and 100 we have 3,13,23,43,53,63,73,83,93 = 9 times
and we have the numbers 30-39 = 10 times which gives us total of 19 nunbers!! (not 20)
So between 0 and 1000 this metod will repeat itself 9 times for, 0-99, 100-199, 200-299, 400...
and that will five us 9*19=171 and we have 100 nunbers 300-399
so the correct answer is 271!! not 300
The portion in red is incorrect.
Between 30 and 39, inclusive, the digit 3 appears 11 times:
30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
Because 3 appears TWICE in 33, the total number of appearances is not 10 but 11.
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by rsarashi » Sun Jun 25, 2017 9:46 am

Here's another way to look at it.
Write all of the numbers as 3-digit numbers.
That is, 000, 001, 002, 003, .... 998, 999

NOTE: Yes, I have started at 000 and ended at 999, even though though the question asks us to look at the numbers from 1 to 1000. HOWEVER, notice that 000 and 1000 do not have any 3's so the outcome will be the same.

First, there are 1000 integers from 000 to 999
There are 3 digits in each integer.
So, there is a TOTAL of 3000 individual digit. (since 1000 x 3 = 3000)

Each of the 10 digits is equally represented, so the 3 will account for 1/10 of all digits.

1/10 of 3000 = 300

So, there are 300 0's, 300 1's, 300 2's, 300 3's, . . ., and 300 9's in the integers from 000 to 999

Cheers,
Brent
[/quote]

Hi Brent ,

Just a quick question. From 1 to 1000, we will have 33, 333. So these will be counted one 3 or more than one 3?

Please explain.