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sum of numbers

by gmatrix » Mon Sep 06, 2010 8:22 pm
find the sum of all the four digit numbers which are formed by the digits 1,2,5,6
[spoiler]OA:93324[/spoiler]
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by Rahul@gurome » Mon Sep 06, 2010 9:08 pm
Solution:
I think you need to mention that there is no repetition of digits.
When 1 is in the thousands place, we can have 3! or 6 such numbers by arranging the remaining three digits (2, 5 and 6).

Similarly when 2, 5 and 6 are in the thousands respectively, places there will be 3! or 6 numbers each.

So the sum of all the digits in the thousands place will be 1*6 + 2*6 + 5*6 + 6*6 =14*6 = 84.

Similarly the sum of all digits in hundreds, tens and units place is also 84 each.

So the sum of all numbers formed is 84*1000+84*100+84*10+84 = 93324.
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by sanju09 » Mon Sep 06, 2010 9:17 pm
Rahul@gurome wrote:Solution:
I think you need to mention that there is no repetition of digits.
When 1 is in the thousands place, we can have 3! or 6 such numbers by arranging the remaining three digits (2, 5 and 6).

Similarly when 2, 5 and 6 are in the thousands respectively, places there will be 3! or 6 numbers each.

So the sum of all the digits in the thousands place will be 1*6 + 2*6 + 5*6 + 6*6 =14*6 = 84.

Similarly the sum of all digits in hundreds, tens and units place is also 84 each.

So the sum of all numbers formed is 84*1000+84*100+84*10+84 = 93324.
Bang on target so marvelously, Rahul@99

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by sanju09 » Mon Sep 06, 2010 9:27 pm
gmatrix wrote:find the sum of all the four digit numbers which are formed by the digits 1,2,5,6
[spoiler]OA:93324[/spoiler]
Hi gmatrix,

Rahul@gurome has already explained and answered your query so brilliantly that I really have to say nothing about it. Just two things to submit beyond your query:

(1) I liked your profile picture very much.

(2) I didn't find your signature lines reading "[spoiler]Life is all about ass; you're either covering it, laughing it off, kicking it, kissing it, busting it, trying to get a piece of it, or behaving like one[/spoiler]." as fit for the forum.

Regards
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by scorpionz » Mon Sep 06, 2010 9:51 pm
Rahul@gurome wrote:Solution:
I think you need to mention that there is no repetition of digits.
When 1 is in the thousands place, we can have 3! or 6 such numbers by arranging the remaining three digits (2, 5 and 6).

Similarly when 2, 5 and 6 are in the thousands respectively, places there will be 3! or 6 numbers each.

So the sum of all the digits in the thousands place will be 1*6 + 2*6 + 5*6 + 6*6 =14*6 = 84.

Similarly the sum of all digits in hundreds, tens and units place is also 84 each.

So the sum of all numbers formed is 84*1000+84*100+84*10+84 = 93324.
Awesome!!!!

Never thought that the solution would be so simple..

Thanks a ton!
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