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Sum of 4 digits

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Sum of 4 digits

by singhsa » Sun Sep 12, 2010 12:18 am
What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

OA - 711040

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by albatross86 » Sun Sep 12, 2010 5:36 am
Alright I'll have a crack at this, Donny ;)

So any four digit number "ABCD" where those are the individual digits, can be represented as follows:

1000A + 100B + 10C + D

First let's figure out how many 4 digit numbers can be formed:

= 4*4*4*4 = 256

Now this next part is going to be a bit tough to explain in words, but I'll try.

None of these 4 digits (1,2,3,4) is going to occur more often than an other since we have no preferences here. So what this means is, if you consider each digit place (A,B,C,D) : Out of the 256 numbers, 64 would have A = 1, 64 would have A = 2 and so on for all the digits and digit-places. So in short, if you add all of them up, you would have the following expression:

64*[ 1000*(1 + 2 + 3 + 4) + 100*(1 + 2 + 3 + 4) + 10*(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4)]

= 64*10*[1111]

= 711040

I hope that made sense, if it didn't let me know and I'll try to explain it again :)

Cheers!
~Abhay

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by singhsa » Sun Sep 12, 2010 6:27 am
albatross86 wrote:Alright I'll have a crack at this, Donny ;)

So any four digit number "ABCD" where those are the individual digits, can be represented as follows:

1000A + 100B + 10C + D

First let's figure out how many 4 digit numbers can be formed:

= 4*4*4*4 = 256

Now this next part is going to be a bit tough to explain in words, but I'll try.

None of these 4 digits (1,2,3,4) is going to occur more often than an other since we have no preferences here. So what this means is, if you consider each digit place (A,B,C,D) : Out of the 256 numbers, 64 would have A = 1, 64 would have A = 2 and so on for all the digits and digit-places. So in short, if you add all of them up, you would have the following expression:

64*[ 1000*(1 + 2 + 3 + 4) + 100*(1 + 2 + 3 + 4) + 10*(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4)]

= 64*10*[1111]

= 711040

I hope that made sense, if it didn't let me know and I'll try to explain it again :)

Cheers!

hmm....tricky...i guess then finding out the total no of 4 digit nos that can be formed was pointless. Also, what do you think will be the level of such type of questions??...cheers!!!
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by albatross86 » Sun Sep 12, 2010 6:30 am
Well it wasn't really pointless because you had to come up with the fact that there would be 64 numbers for each digit place possibility. You could come up with that in other ways too, but I find this total possible numbers divided by number of digits quite easy.

This is definitely a Quant 50 to 51 level question in my opinion.
~Abhay

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by singhsa » Sun Sep 12, 2010 6:42 am
albatross86 wrote:Well it wasn't really pointless because you had to come up with the fact that there would be 64 numbers for each digit place possibility. You could come up with that in other ways too, but I find this total possible numbers divided by number of digits quite easy.

This is definitely a Quant 50 to 51 level question in my opinion.
Oh yea...64...thanks bud !!!
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by The Jock » Sun Sep 12, 2010 7:26 pm
Hey guys,

I know albatross86 has already given an amazing explanation and secondly the explanation given by albatross86 is pure mathematical logic.
Now I have come to know about one formula to solve these type of questions and this formula make these questions very easy.

sum of integers that are formed by the permutations of n digits sum is given by equation:
= (sum of given numbers)*(n-1)!*(111... n times)
if repetition is not allowed.
= (sum of given numbers)*(n)(n-1)*(111... n times)
if repetition is allowed.

Question: What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

for this question, n = 4 and sum of the numbers given is (1+2+3+4) = 10
so according to second formula given above:
= (sum of given numbers)*(n)(n-1)*(111... n times)
= 10*4^3(1111) = 10*64(1111) = 640(1111) = 711040

Let me know if it is of any help.
Thanks and Regards,
Varun
https://mbayogi.wordpress.com/
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by singhsa » Mon Sep 13, 2010 9:47 am
The Jock wrote:Hey guys,

I know albatross86 has already given an amazing explanation and secondly the explanation given by albatross86 is pure mathematical logic.
Now I have come to know about one formula to solve these type of questions and this formula make these questions very easy.

sum of integers that are formed by the permutations of n digits sum is given by equation:
= (sum of given numbers)*(n-1)!*(111... n times)
if repetition is not allowed.
= (sum of given numbers)*(n)(n-1)*(111... n times)
if repetition is allowed.

Question: What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

for this question, n = 4 and sum of the numbers given is (1+2+3+4) = 10
so according to second formula given above:
= (sum of given numbers)*(n)(n-1)*(111... n times)
= 10*4^3(1111) = 10*64(1111) = 640(1111) = 711040

Let me know if it is of any help.
Wow man.....this is great!!!

Is there a formula for similar questions involving product of nos. as well??
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by The Jock » Mon Sep 13, 2010 5:28 pm
Hey singhsa,

Sorry man, not aware right now but If I come to know down the line, I will be very happy to share that.
Thanks and Regards,
Varun
https://mbayogi.wordpress.com/
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by neerajkumar1_1 » Mon Sep 13, 2010 7:32 pm
The Jock wrote:Hey guys,

I know albatross86 has already given an amazing explanation and secondly the explanation given by albatross86 is pure mathematical logic.
Now I have come to know about one formula to solve these type of questions and this formula make these questions very easy.

sum of integers that are formed by the permutations of n digits sum is given by equation:
= (sum of given numbers)*(n-1)!*(111... n times)
if repetition is not allowed.
= (sum of given numbers)*(n)(n-1)*(111... n times)
if repetition is allowed.

Question: What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

for this question, n = 4 and sum of the numbers given is (1+2+3+4) = 10
so according to second formula given above:
= (sum of given numbers)*(n)(n-1)*(111... n times)
= 10*4^3(1111) = 10*64(1111) = 640(1111) = 711040

Let me know if it is of any help.
hey jock ...

according to the formula u have written and the solution u have provided...
just to clarify..

the formula should be
If repetition is allowed:
= (sum of given numbers)*(n)^(n-1)*(111... n times)

and if repetition is not allowed:
= (sum of given numbers)*(n-1)!*(111... n times)

is that correct???
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by The Jock » Tue Sep 14, 2010 2:14 am
Hey Neeraj,

You got it right.
Thanks and Regards,
Varun
https://mbayogi.wordpress.com/
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by alookachaloo » Sun Mar 13, 2011 1:48 am
Thanks Abhay for the wonderful explanation, however how would you calculate that how often a digit will occur at a particular place if the question would have been like :
What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4,5,6 where repetition of digits is not allowed?
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