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by goyalsau » Sat Nov 27, 2010 7:19 pm
I have completely no idea How long it will take when i will able to solve problems like these,,,,,,,,,


How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?


165

330

132

440

170

OA - 165
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by Laura GMAT Tutor » Sat Nov 27, 2010 9:07 pm
Where did you find this problem? Are you 100% certain about its wording and the answers?
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by goyalsau » Sat Nov 27, 2010 9:32 pm
Laura GMAT Tutor wrote:Where did you find this problem? Are you 100% certain about its wording and the answers?
I am absolutely sure about the wording and the options, But just don't understand the explanation , If you say i will post the explanation as well,
Please explain it in simple words.....
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by Laura GMAT Tutor » Sat Nov 27, 2010 10:06 pm
I'll read the explanation, but the wording is odd. "How many of the four digit numbers" should be "how many four digit integers" and "have their sum as" should be "have the sum of 12" or "sum to 12."

It's just weird. Please tell me where you got this problem. It may not be worth our time.
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by goyalsau » Sun Nov 28, 2010 12:25 am
Laura GMAT Tutor wrote:I'll read the explanation, but the wording is odd. "How many of the four digit numbers" should be "how many four digit integers" and "have their sum as" should be "have the sum of 12" or "sum to 12."

It's just weird. Please tell me where you got this problem. It may not be worth our time.
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by goyalsau » Sun Nov 28, 2010 12:26 am
Laura GMAT Tutor wrote:I'll read the explanation, but the wording is odd. "How many of the four digit numbers" should be "how many four digit integers" and "have their sum as" should be "have the sum of 12" or "sum to 12."

It's just weird. Please tell me where you got this problem. It may not be worth our time.
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by Laura GMAT Tutor » Sun Nov 28, 2010 8:04 am
I'm sorry, their explanation makes no sense to me. The only way I can think to solve this problem takes a lot longer.

If you'd like to repost this, maybe another person will have a better explanation.

But mostly I'd like to say that in eight years I never saw a question like this. In fact, on my last four real GMATs I never saw a question like this, and I got 780s with 50 subscores in quant.

So.... I wouldn't stress it.
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by goyalsau » Sun Nov 28, 2010 8:12 am
Laura GMAT Tutor wrote:I'm sorry, their explanation makes no sense to me. The only way I can think to solve this problem takes a lot longer.

If you'd like to repost this, maybe another person will have a better explanation.

But mostly I'd like to say that in eight years I never saw a question like this. In fact, on my last four real GMATs I never saw a question like this, and I got 780s with 50 subscores in quant.

So.... I wouldn't stress it.

Thanks a lot, You are like a real tutor..............
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by Rahul@gurome » Sun Nov 28, 2010 10:02 am
goyalsau wrote:How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?
165
330
132
440
170
I do not find the wordings of the problems wrong or confusing. This problem is certainly a difficult one but not uncommon. Finding number of 4 or 5 digit numbers such that the sum of their digits is some constant (9 or 12 or 14 etc) is not a rare problem. The term 'non-zero digits' add the extra constraint in the problem! There is a method which allows us to solve these kind of problems easily and quickly.

The method is the "Inserting Stick" or "Separator" method which is very popular method in solving combinatorics problems. In fact the solution you provided also follow this method but they didn't explained it! This method makes any tough combination problem (this certainly one of them) very easy to approach and solve. Let's see what is this method and how we can apply this method to solve it!

The question asks for the number of four digit numbers such that their sum is 12 and none of the digit is zero. Some possible such numbers are 2415, 5133, 6312 etc.

Let's represent each digit as a combination of '1's and separator between two digit by '|'. Such that 2 is represented as 11, 5 is represented as 11111 etc. Now,
  • 2431 --> 11|1111|1|11111
    5133 --> 11111|1|111|111
    6312 --> 111111|111|1|11
Observe that in this representation the number of '1's is always 12 and number of separators is always 3. Now the problem can be rephrased as: in how many ways three separators can be used to separate twelve 1's such that there is at least one 1 in a group (at least 1 in a group ensures that all the digits are non-zero). There are 11 places between twelve 1's where we can place these three separators.

This can be done in (11P3)/3! = (11!)/(8! * 3!) = (11*10*9)/6 = 11*15 = 165 ways.

The correct answer is A.

Note :
  • (1) Observe that (11P3)/3! is nothing but (11C3) = (12 - 1)C(4 - 1). Now remember there was 12 '1's and 4 '|'s! :)
    (2) This method can also be used to find the number of such numbers without the "non-zero digits" constraint too. In that case the approach is a little different. Think about that! :)
    (3) This method can be used to solve a lot of combinatorics problems. But it takes a lot of practice to understand when you should apply this method and when there is a much easier alternative way.
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by Laura GMAT Tutor » Sun Nov 28, 2010 11:52 pm
I didn't say that the wording was confusing, just that it was odd. "how many of the four digit numbers... have their sum as 12."

And no, I don't think the basic premise is that odd -- but the level of difficulty required is unusual. That's a unique way to solve it, Rahul. I like it. The other way I thought of to solve it would have taken at least 3 minutes. Since I had a migraine coming on, I didn't really feel up to writing it out.
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by gmatter81 » Mon Nov 29, 2010 4:07 am
Hi Rahul,

Request you to post the method for digits with zero as well. This will complete my understanding.

Thanks.

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by beat_gmat_09 » Mon Nov 29, 2010 4:44 am
What a brilliant method Rahul !!
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by rishab1988 » Mon Nov 29, 2010 5:18 am
jeez

Rahul should receive a Fields Medal for this.I think Rahul is much more capable than GMAT.He should work on some kind high level Math research project.

I have no words for the level of knowledge you have demonstrated.

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by Rahul@gurome » Mon Nov 29, 2010 11:25 am
Thanks for the compliments guys!:)
gmatter81 wrote:Hi Rahul,

Request you to post the method for digits with zero as well. This will complete my understanding.

Thanks.
If the same question is asked without the "non-zero digits" constraint, then application of this method doesn't directly yield the answer. In that case we have to proceed in a little different way. (and tricky I must say :) )

Let's take an question for analysis: How many numbers which are less than 10000 are there such that sum of their digits is 10?

The numbers may be a two or three or four digit number. Let's represent the numbers in the same manner, i.e. each digit with '1's and separate each digit with '|'. Some possible numbers and their new representations are,
  • 0082 --> ||11111111|11
    0253 --> |11|11111|111
    4231 --> 1111|11|111|1
    7021 --> 1111111||11|1
Observe that total number of '1's is 10 as the sum should be 10 and total number of separators is 3 as maximum number of digits is 4 (3 separators needed to separate 4 digits). But unlike the previous case there is no constraint in placing the separators between the digits. They can be placed anywhere except a few, which we'll see later. Thus it's just a problem on arranging some items in which some are identical!

The problem can be rephrased as in how many ways ten '1's and three '|' can be arranged?

And the answer is, (10 + 3)!/(10!)*(3!) = 286


Tricky isn't it? But careful! The trick is not over yet! :)
I mentioned there are few limitation in placing the separators!
They can placed anywhere except all of them together at the leftmost or rightmost side! Because doing so will result in |||1111111111 begin_of_the_skype_highlighting              1111111111      end_of_the_skype_highlighting --> 000(10)! That means we have to treat 10 as a digit which is not possible. Same for 1111111111|||.

Thus correct number of numbers = 286 - 2 = 284

Therefore whenever the sum is a two digit number you have to take care of this problem.


Now going back to where we started, if the problem asks for how many four digit numbers are there such that sum of their digits is 10?, the numbers of such numbers will be equal to (Such numbers with 2 or 3 or 4 digits - Such numbers with 2 or 3 digits)! Both of which can be obtained by the above described method!
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by fskilnik@GMATH » Tue Dec 14, 2010 4:36 am
Hi there!

Another BTG post lead me here... beautiful explanation, Rahul, although for a Fields Medal you would have to work a bit harder, as you certainly know... ;)
Rahul@gurome wrote:Therefore whenever the sum is a two digit number you have to take care of this problem.
Perfect. I´d like to add that whenever the sum is a three-, four-, etc digit number, you will have to take "extra care", therefore a fair question would be.... is there another way of dealing with this, in a general context, without this need for "pos-corrections"?

The answer is YES and I leave below a link to some of the ideas/method involved for the interested readers, although we are now really far-beyond GMAT, for sure.

(The problem seems a bit different AT FIRST, but it is easy to figure out how you could adapt my suggestion to the problem-situation presented here.)

Regards,
Fabio.

Link: https://www.beatthegmat.com/a-3-digit-16 ... tml#323769
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