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jzh200
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Post subject: FDP Pg 23 Chapter 1 Question 2  Posted: Tue May 12, 2009 12:15 pm |
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Question 2:
What is the sum of all the possible 3-digit numbers that can be constructed using the digits 3,4, and 5, if each digit can used only once in each number?
MGMAT Answer:
2664: There are 6 ways to arrange these digits: 345, 354, 435, 453, 534, and 543. Notice that each digit appears twice in the hundreds column, twice in the tens column, and twice in the ones column.
100(24)+10(24)+(24) = 2400+240+24= 2664
My Comments:
I fully understand that there are 6 ways of arranging these digits (3!). My problem is understanding this equation:
100(24)+10(24)+(24) = 2664
Where is the 24 coming from? I mean the 100, 10, and 1 stand for the digit values but I am failing to understand the 24?
I would really appreciate your help!
Last edited by jzh200 on Wed May 13, 2009 10:12 am, edited 1 time in total.
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Ricardo_lamour
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Tue May 12, 2009 10:55 pm |
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those are my thoughts exactly, i was stumped by the answer. the 24 came out of no where.
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espritology
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Thu May 14, 2009 2:02 am |
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As mentioned by you earlier, each digit appears twice in hundreds, tens, and units place value. Hence, adding all these "number-combinations" would result into 100*(5 + 5 + 4 + 4 + 3 + 3) + 10*(5 + 5 + 4 + 4 + 3 + 3) + 1*(5 + 5 + 4 + 4 + 3 + 3) => 100*24 + 10*24 + 1*24 = 2664
I hope it will help.
[EDIT]
As a shortcut, you can always use the following for such question types (digits repeating in all possible place values) -
Two Digit Sum = 11 * (a + b)
Three Digit Sum = 222 * (a + b + c)
Four Digit Sum = 6666 * (a + b + c + d)
[/EDIT]
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jzh200
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Thu May 14, 2009 9:55 am |
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espritology wrote: As mentioned by you earlier, each digit appears twice in hundreds, tens, and units place value. Hence, adding all these "number-combinations" would result into 100*(5 + 5 + 4 + 4 + 3 + 3) + 10*(5 + 5 + 4 + 4 + 3 + 3) + 1*(5 + 5 + 4 + 4 + 3 + 3) => 100*24 + 10*24 + 1*24 = 2664
I hope it will help.
[EDIT]
As a shortcut, you can always use the following for such question types (digits repeating in all possible place values) -
Two Digit Sum = 11 * (a + b)
Three Digit Sum = 222 * (a + b + c)
Four Digit Sum = 6666 * (a + b + c + d)
[/EDIT] Excellent! It makes perfect sense to see it broken down. So essentially we are multiplying place value with the sum of numbers (repeated twice). I appreciate your help.
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StaceyKoprince
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Fri May 15, 2009 2:43 pm |
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Thanks, espritology, for explaining. We could / should have written out the explanation more fully.
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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jzh200
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Sat May 16, 2009 7:36 pm |
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StaceyKoprince wrote: Thanks, espritology, for explaining. We could / should have written out the explanation more fully. Thanks Stacey, for the follow up.
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StaceyKoprince
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Thu May 28, 2009 1:51 pm |
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No problem! We always try to weigh in even if another student has already provided the right answer or explanation - if only to confirm that the student has, in fact, answered correctly. :)
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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sanyu4
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Fri Feb 25, 2011 6:53 pm |
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sanyu4
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Fri Feb 25, 2011 7:10 pm |
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As a shortcut, you can always use the following for such question types (digits repeating in all possible place values) -
Two Digit Sum = 11 * (a + b)
Three Digit Sum = 222 * (a + b + c)
Four Digit Sum = 6666 * (a + b + c + d)
I understand the answer to the question in the text book however I do not understand the general scenario given above. For the three digit Sum example, is that 2*2*2 (2+2+2)? So if we were to use the same concept in the problem given:
3*4*5 (3+4+5) = 720 (not 2664, which is the correct answer) Please explain. thank you
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jnelson0612
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Sun Feb 27, 2011 7:30 pm |
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sanyu4 wrote: As a shortcut, you can always use the following for such question types (digits repeating in all possible place values) -
Two Digit Sum = 11 * (a + b)
Three Digit Sum = 222 * (a + b + c)
Four Digit Sum = 6666 * (a + b + c + d)
I understand the answer to the question in the text book however I do not understand the general scenario given above. For the three digit Sum example, is that 2*2*2 (2+2+2)? So if we were to use the same concept in the problem given:
3*4*5 (3+4+5) = 720 (not 2664, which is the correct answer) Please explain. thank you It's just 222 * the sum of the three numbers that are used. In this case, it is 222(3+4+5), which is 222(12), or 2664.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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kitsgq
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Sun May 15, 2011 5:59 pm |
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jnelson0612
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Mon May 16, 2011 1:16 am |
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kitsgq wrote: 222 * ( 3+4+6) = 2664 Did you mistype above? It should be (3+4+5).
_________________
Jamie Nelson
ManhattanGMAT Instructor
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shoumik
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Sat Jun 04, 2011 9:46 pm |
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Hey guys,
I am still having a little trouble understanding the answer of this question.
I am not sure how it is taking account of numbers like: 334, 335, 336, 337,338,339 - in which there are two 3's which is restricted as per question.
So why would we multiply 3 by 100 when we should be multiplying 3 by 80? Since 330-339 has repeating 3's just like 303, 313,323, 333, 343, 353, 363,373,383,393.
Please explain this,Thanks!!
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jnelson0612
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Post subject: Re: FDP Pg 23 Chapter 1 Question 2 Posted: Wed Jun 08, 2011 8:06 pm |
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shoumik wrote: Hey guys,
I am still having a little trouble understanding the answer of this question.
I am not sure how it is taking account of numbers like: 334, 335, 336, 337,338,339 - in which there are two 3's which is restricted as per question.
So why would we multiply 3 by 100 when we should be multiplying 3 by 80? Since 330-339 has repeating 3's just like 303, 313,323, 333, 343, 353, 363,373,383,393.
Please explain this,Thanks!! Check out what the question says: "What is the sum of all the possible 3-digit numbers that can be constructed using the digits 3,4, and 5, if each digit can used only once in each number?" Since we can only use each number once we cannot have numbers such as 334, 335, etc. Hope this helps!
_________________
Jamie Nelson
ManhattanGMAT Instructor
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