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24 Oct 2010, 11:03
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
94% (00:57) correct
6%
(00:28)
wrong
based on 80
sessions
History
Date
Time
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Not Attempted Yet
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
I will post OA and OE tomorrow.
I will post OA and OE tomorrow.
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Thank you for understanding, and happy exploring!
24 Oct 2010, 11:22
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We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.
When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000
Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200
Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20
Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10
The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555
Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111
Now we can generalize the above
If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)
Consider giving me KUDOS if this is useful.
When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000
Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200
Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20
Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10
The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555
Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111
Now we can generalize the above
If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)
Consider giving me KUDOS if this is useful.
gurpreetsingh
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24 Oct 2010, 11:24
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ankitranjan
Along with the question, do post the options.
if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6
=> 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times.
Similar all the 4 numbers occurs at all the 4 digits - one's,ten's, hundred's, thousand's- 6 times.
=> total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6
= 1111*14*6 = 1111*84 = 93324
