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# What is the sum of all possible 3-digit numbers that can be

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What is the sum of all possible 3-digit numbers that can be [#permalink]

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26 Apr 2012, 00:59
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Difficulty:

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Question Stats:

87% (01:49) correct 13% (00:42) wrong based on 187 sessions

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What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?

A. 2660
B. 2661
C. 2662
D. 2663
E. 2664
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Apr 2012, 01:21, edited 1 time in total.
Edited the question
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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26 Apr 2012, 01:20
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sugu86 wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?

A) 2660 B) 2661 C) 2662 D) 2663 E) 2664

Thanks,

Suganth

What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A. 2660
B. 2661
C. 2662
D. 2663
E. 2664

Any 3-digit number can be written as: 100a+10b+c.

# of three digit numbers with digits {3, 4, 5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664.

Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:

(n-1)!*(sum of the digits)*(111…..n times)

In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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20 Jan 2013, 17:33
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Another way to approach this problem is to recognize that the way the sequence increases from the min (345) is symmetrical to the way it decreases from the max (543). Therefore if you find the average of the min and max and multiply it by the number of possibilities (3! or 6) then you'll have your answer.

$$\frac{345+543}{2} = 444$$

$$444*3! = 444*6 = 2664$$
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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02 Oct 2012, 08:59
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sugu86 wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?

A. 2660
B. 2661
C. 2662
D. 2663
E. 2664

the unit digits of all possible 3-digit numbers are supposed to have a sum of 3 +3 +4+4+5+5=24, so the sum of numbers should have 4 as a unit digit - 2664 is the only possible option.
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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02 Oct 2012, 09:36
I think Bunuel got the basic way to solve this kind of question. if the number is 4 digit or the answer has 4 number with last number is 4 then U should follow Bunuel
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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06 Oct 2012, 19:53
thaihoang305 wrote:
I think Bunuel got the basic way to solve this kind of question. if the number is 4 digit or the answer has 4 number with last number is 4 then U should follow Bunuel

Many thanks to Bunuel for his very clear explanations, I am going through all problems with his explanations in forum's PS part. For this very problem I just wanted to find out the fastest way to solve as far as you need to take time into account as well.
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Unable to comprehend this question [#permalink]

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07 Nov 2012, 11:26
I encountered the below question on MGAT 5th Edition FDP guide, and the last part (underlined) does not make sense to me.

What is the sum of all the possible three-digit numbers that can be constructed using the digits 3, 4, and 5 if each digit can be used only once in each number

The answer provided by the guide is below: (and I know how to solve the problem that results in the below answer, but that underlined part of the question threw me away):

[Reveal] Spoiler: Answer to the above question
2664

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Re: Unable to comprehend this question [#permalink]

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07 Nov 2012, 15:00
Expert's post
megafan wrote:
I encountered the below question on MGAT 5th Edition FDP guide, and the last part (underlined) does not make sense to me.

What is the sum of all the possible three-digit numbers that can be constructed using the digits 3, 4, and 5 if each digit can be used only once in each number

The answer provided by the guide is below: (and I know how to solve the problem that results in the below answer, but that underlined part of the question threw me away):

[Reveal] Spoiler: Answer to the above question
2664

Merging similar topics. Please refer to the solutions above.

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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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21 Jan 2013, 03:08
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Bunuel wrote:
sugu86 wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?

A) 2660 B) 2661 C) 2662 D) 2663 E) 2664

Thanks,

Suganth

What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A. 2660
B. 2661
C. 2662
D. 2663
E. 2664

Any 3-digit number can be written as: 100a+10b+c.

# of three digit numbers with digits {3, 4, 5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664.

Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:

(n-1)!*(sum of the digits)*(111…..n times)

In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.

wow on this type of a question i was only able to come up with the computation of possible number of ways of arranging 3 digits,but the rest part gave me problems

truly speaking @bunuel i am complete lost on this part(
These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664.)...but i guess the formula would make it easier..you should add it in the topic of number theory in the gmat math book.. Rgrds

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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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27 Aug 2013, 23:42
I like that approach because it has precise formula, but could you please clarify this part:

Bunuel wrote:

(n-1)!*(sum of the digits)*(111…..n times)[/b]

So when to use 111 and when to use different number? You stated 111.....n times, isn't it should be 3 times in our case?
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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27 Aug 2013, 23:42
I like that approach because it has precise formula, but could you please clarify this part:

Bunuel wrote:

(n-1)!*(sum of the digits)*(111…..n times)[/b]

So when to use 111 and when to use different number? You stated 111.....n times, isn't it should be 3 times in our case?
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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28 Aug 2013, 09:26
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ziko wrote:
I like that approach because it has precise formula, but could you please clarify this part:

Bunuel wrote:

(n-1)!*(sum of the digits)*(111…..n times)[/b]

So when to use 111 and when to use different number? You stated 111.....n times, isn't it should be 3 times in our case?

111... n times mean that if we have 2 digits it should be 11, if 3 digits 111, if 4 digits it should be 1,111.

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Hope it helps.
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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06 Sep 2014, 18:27
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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26 Sep 2015, 20:01
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Re: What is the sum of all possible 3-digit numbers that can be [#permalink]

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17 Apr 2016, 09:29
there are 6 ways to answer to arrange the 3 digits: 345, 354, 435, 453, 534, 543, with repetitions of 3+3+4+4+5+5 = 24 each time you go down. If you multiply the 24*100 for the hundred, 24*10 for the tens and 24 *1 for the single digits, then you can put it together like this: 24*100 + 24*10 + 24 = 2,400 + 240 + 24 = 2,664. this is the answer.
Re: What is the sum of all possible 3-digit numbers that can be   [#permalink] 17 Apr 2016, 09:29
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