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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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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 ?

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.

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.

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.

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

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.

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):

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):

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.

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

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.

gmatclubot

Re: What is the sum of all possible 3-digit numbers that can be
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17 Apr 2016, 09:29

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