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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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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.
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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.

Answer: E.

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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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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New post 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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New post 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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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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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.

Answer: E.

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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New post 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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New post 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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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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New post 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.

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New post 30 Jun 2017, 09:11
What is the sum of all possible three digit numbers that can be constructed using the digits 3,4,5 if each digit can only be used once in a number?

The explanation:

6 ways to arrange the digits: 345, 354, 453, 435, 543, 534.
Each digit appears twice in the hundreds column, twice in the tens column, and twice in the ones column.
Because each digit appears twice in the hundreds column, you have 3+3+4+4+5+5 = 24 in the hundreds column. *** this is where I lose it. why are we adding the the digits in the first place and why twice?
ans: 100(24) + 10(24) + 24 = 2664

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New post 30 Jun 2017, 10:48
aadnub1 wrote:
What is the sum of all possible three digit numbers that can be constructed using the digits 3,4,5 if each digit can only be used once in a number?

The explanation:

6 ways to arrange the digits: 345, 354, 453, 435, 543, 534.
Each digit appears twice in the hundreds column, twice in the tens column, and twice in the ones column.
Because each digit appears twice in the hundreds column, you have 3+3+4+4+5+5 = 24 in the hundreds column. *** this is where I lose it. why are we adding the the digits in the first place and why twice?
ans: 100(24) + 10(24) + 24 = 2664


The possible combination of three digit numbers are listed below. I have also broken down the numbers into hundreds, tens and units.

345 = 300 + 40 + 5
354 = 300 + 50 + 4
435 = 400 + 30 + 5
453 = 400 + 50 + 3
534 = 500 + 30 + 4
543 = 500 + 40 + 3

So if you want to add the 6 numbers above, you can go about adding the hundreds, tens and then units.

In each you will notice, two 3's, two 4's and two 5's.

So what we have is (3+3+4+4+5+5)*(100+10+1) = 24*(100+10+1) = 24*111 = 2664

Hope this helps.

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

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New post 30 Jun 2017, 11:07
aadnub1 wrote:
What is the sum of all possible three digit numbers that can be constructed using the digits 3,4,5 if each digit can only be used once in a number?

The explanation:

6 ways to arrange the digits: 345, 354, 453, 435, 543, 534.
Each digit appears twice in the hundreds column, twice in the tens column, and twice in the ones column.
Because each digit appears twice in the hundreds column, you have 3+3+4+4+5+5 = 24 in the hundreds column. *** this is where I lose it. why are we adding the the digits in the first place and why twice?
ans: 100(24) + 10(24) + 24 = 2664


Merging similar topics. Please refer to the solutions above.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Thank you.
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Re: What is the sum of all possible 3-digit numbers that can be   [#permalink] 30 Jun 2017, 11:07
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