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Pls explain

by Nycgrl » Mon Aug 25, 2008 12:41 pm
What is the sum of all possible 3 digit numbersthat can be constructed using 3,4,5.If each digit can be used only once in the number.

There are no options.

A = 2664

Can somebody please explain how to solve above question.
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by anju » Mon Aug 25, 2008 1:01 pm
two ways 2 solve this :
a lil' length way is to write all possible numbers and add it up like
345
354
435
453
534
543
addition gives 2664 or the other way is
if you observed, for each place the number is repeated 2 times so add 3+4+5 = 12 * 2 = 24
now each place sums up to 24 so unit's place is 4 then for ten's place is 4+2 = 6 and hunderd's place is 26 so
number is 2664

I am sure there are some more easy methods.... can someone provide a btr way
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by Suyog » Mon Aug 25, 2008 1:03 pm
345+354+435+453+534+543=2664
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by Nycgrl » Mon Aug 25, 2008 1:04 pm
now each place sums up to 24 so unit's place is 4 then for ten's place is 4+2 = 6 and hunderd's place is 26 so
number is 2664
I could not understand this part. Can you please explain how did u get 26 for hundred's place?

I know adddition is better method for this question but in case there are more complex numbers involved.
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by anju » Mon Aug 25, 2008 1:14 pm
Nycgrl wrote:now each place sums up to 24 so unit's place is 4 then for ten's place is 4+2 = 6 and hunderd's place is 26 so
number is 2664
I could not understand this part. Can you please explain how did u get 26 for hundred's place?

I know adddition is better method for this question but in case there are more complex numbers involved.
let's look at this :
345
354
435
453
534
543

add all the digits in unit's place so 5+4+5+3+4+3 = (5+4+3)*2 = 12*2 = 24
so we write 4 and carry forward 2
ten's digit also has the same number so now 24+ 2 (which we carried fwd) = 26 so write 6
similarly hundred's place, 24+2 = 26 where 6 is hunderdth digit and 2 is thousandth digit
so number is 2664.

hope it is clear now
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by Nycgrl » Mon Aug 25, 2008 1:17 pm
Thnaks
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by mehravikas » Tue Aug 26, 2008 10:45 pm
I think its a good idea to do permutations on these sort of problems. It gives you the total number of options you can have, and then you can list and add them.
Nycgrl wrote:Thnaks
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Re: Pls explain

by sudhir3127 » Tue Aug 26, 2008 11:05 pm
Nycgrl wrote:What is the sum of all possible 3 digit numbersthat can be constructed using 3,4,5.If each digit can be used only once in the number.

There are no options.

A = 2664

Can somebody please explain how to solve above question.
yeah the answer shud be 2664

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

(3+4+5) *(3-1)!*(111)

12*2*111= 2664

hope that helps..
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Re: Pls explain

by mehravikas » Tue Aug 26, 2008 11:39 pm
Sudhir, just a quick question. You mentioned 111...n times..does it mean that the value of n increases with each number?

As in, if the numbers are 3, 4, 5, 6 -

(3 + 4 + 5 + 6) (4 - 1)! * (1111) ??

sudhir3127 wrote:
Nycgrl wrote:What is the sum of all possible 3 digit numbersthat can be constructed using 3,4,5.If each digit can be used only once in the number.

There are no options.

A = 2664

Can somebody please explain how to solve above question.
yeah the answer shud be 2664

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

(3+4+5) *(3-1)!*(111)

12*2*111= 2664

hope that helps..
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Re: Pls explain

by sudhir3127 » Wed Aug 27, 2008 12:09 am
mehravikas wrote:Sudhir, just a quick question. You mentioned 111...n times..does it mean that the value of n increases with each number?

As in, if the numbers are 3, 4, 5, 6 -

(3 + 4 + 5 + 6) (4 - 1)! * (1111) ??

sudhir3127 wrote:
Nycgrl wrote:What is the sum of all possible 3 digit numbersthat can be constructed using 3,4,5.If each digit can be used only once in the number.

There are no options.

A = 2664

Can somebody please explain how to solve above question.
yeah the answer shud be 2664

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

(3+4+5) *(3-1)!*(111)

12*2*111= 2664

hope that helps..
Exactly vikas !!..hope that formula helped
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Re: Pls explain

by mehravikas » Wed Aug 27, 2008 10:50 am
Oh absolutely, it makes life easier for these sort of questions.

Thanks heaps :-)
sudhir3127 wrote:
mehravikas wrote:Sudhir, just a quick question. You mentioned 111...n times..does it mean that the value of n increases with each number?

As in, if the numbers are 3, 4, 5, 6 -

(3 + 4 + 5 + 6) (4 - 1)! * (1111) ??

sudhir3127 wrote:
Nycgrl wrote:What is the sum of all possible 3 digit numbersthat can be constructed using 3,4,5.If each digit can be used only once in the number.

There are no options.

A = 2664

Can somebody please explain how to solve above question.
yeah the answer shud be 2664

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

(3+4+5) *(3-1)!*(111)

12*2*111= 2664

hope that helps..
Exactly vikas !!..hope that formula helped
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by pre-gmat » Wed Aug 27, 2008 11:01 am
Wow this is a cool solution.
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by pre-gmat » Wed Aug 27, 2008 7:51 pm
Can this be any set of numbers or just the consecutive numbers?
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by parallel_chase » Wed Aug 27, 2008 7:52 pm
pre-gmat wrote:Can this be any set of numbers or just the consecutive numbers?
Any set of integers will do.
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by pre-gmat » Wed Aug 27, 2008 7:53 pm
Thanks
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