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# Find the sum of 4 digit numbers

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Senior Manager
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Find the sum of 4 digit numbers [#permalink]

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24 Aug 2012, 02:31
Find the sum of all the 4 digit numbers which are formed by the digits 1,2,5,6.

a) 933510
b) 93324
c) 65120
d) 8400
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Re: Find the sum of 4 digit numbers [#permalink]

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24 Aug 2012, 02:52
Find the sum of all the 4 digit numbers which are formed by the digits 1,2,5,6.

a) 933510
b) 93324
c) 65120
d) 8400

A 4 digit number abcd is written as 1000*a + 100*b + 10*c + d

Possible 4digit numbers starting with 1 in thousdands digit are
1256
1265
1526
1562
1625
1652
As, you will notice the pattern in the hundred's ten's and unit's digit then 2,5 and 6 each occur twice in hundred's ten's and unit's digit
So Sum of all the numbers in which 1 is in the thousand's digit is given by
1000*6*1 + 100*2*(2+5+6) + 10*2*(2+5+6) + 1*2*(2+5+6)
= 6000 + (2+5+6)*2*(100+10+1)
= 6000 + 13*2*111
= 8886

Similarly when 2 is in the thousand's digit then the sum of all the numbers will be
1000*6*2 + 100*2*(1+5+6) + 10*2*(1+5+6) + 1*2*(1+5+6)
= 12,000 + 12*2*111
=> Sum = 14664

Similarly when 5 is in the thousand's digit then the sum of all the numbers will be
1000*6*5 + 100*2*(1+2+6) + 10*2*(1+2+6) + 1*2*(1+2+6)
= 30,000 + 111*2*9
=> Sum =31,998

Similarly when 6 is in the thousand's digit then the sum of all the numbers will be
1000*6*6 + 100*2*(1+2+5) + 10*2*(1+2+5) + 1*2*(1+2+5)
= 36,000 + 111*2*8
=> Sum = 37,776

Total Sum = 8886 + 14664 + 31,998 + 37,776 = 93,324

ONe MOre way of doing this is taking all the sums together then we have
1000*6*(1+2+5+6) + 100*2*3*(1+2+5+6) + 10*2*3*(1+2+5+6) + 1*2*3*(1+2+5+6)
= (1+2+5+6) * (6000+600+60+6)
= 14 * 6666
= 93,324

So, Answer is B

Hope it helps!
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Re: Find the sum of 4 digit numbers [#permalink]

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26 Aug 2012, 07:26
1
This post was
BOOKMARKED
Here is the short cut :

4 digit number is in the form of abcd
total number formed by different 4 digit numbers = 4! = 24
This 24 numbers are formed in which one of the four digits, say a, take all four positions.
So, a takes unit position > 6 nos.
a takes tens position > 6 nos.
a takes hundreds position > 6 nos.
a takes thousands position > 6 nos.

This mean number all the four numbers takes unit position 6 numbers, tens - 6 numbers, hundreds - 6 numbers and thousands - 6 numbers.

so, sum of thousands position = ((1+2+5+6)*6)* 1000 = 84000
sum of hundreds position = ((1+2+5+6)*6)* 100 = 8400
sum of tens position = ((1+2+5+6)*6)* 10 = 840
sum of unit position = ((1+2+5+6)*6) = 84
-----------------------------------------------------------------------------------
Total = 93324
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Re: Find the sum of 4 digit numbers [#permalink]

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26 Aug 2012, 08:06
premnath wrote:
Here is the short cut :

4 digit number is in the form of abcd
total number formed by different 4 digit numbers = 4! = 24
This 24 numbers are formed in which one of the four digits, say a, take all four positions.
So, a takes unit position > 6 nos.
a takes tens position > 6 nos.
a takes hundreds position > 6 nos.
a takes thousands position > 6 nos.

This mean number all the four numbers takes unit position 6 numbers, tens - 6 numbers, hundreds - 6 numbers and thousands - 6 numbers.

so, sum of thousands position = ((1+2+5+6)*6)* 1000 = 84000
sum of hundreds position = ((1+2+5+6)*6)* 100 = 8400
sum of tens position = ((1+2+5+6)*6)* 10 = 840
sum of unit position = ((1+2+5+6)*6) = 84
-----------------------------------------------------------------------------------
Total = 93324

Yes This is a simpler way to solve this.. nice job
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan
Kudos drives a person to better himself every single time. So Pls give it generously
Wont give up till i hit a 700+

Kudos [?]: 113 [0], given: 282

Re: Find the sum of 4 digit numbers   [#permalink] 26 Aug 2012, 08:06
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# Find the sum of 4 digit numbers

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