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sum of all

by sanju09 » Thu Apr 02, 2009 4:53 am
Find the sum of all different 5-digit numbers that can be formed by using the digits 2, 3, 4, 5, and 6, where no digit is repeating.
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by adelx » Sat Apr 04, 2009 3:23 am
I think the answer is: 6! * 17 = 720 * 17 = 12240
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by sanju09 » Sat Apr 04, 2009 3:27 am
adelx wrote:I think the answer is: 6! * 17 = 720 * 17 = 12240
please elaborate it a bit :?
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by kapsii » Sat Apr 04, 2009 8:08 am
We have 5 digits and 5 places to fill them, which makes it easier, each digit can occupy either the units place, tens place, hundreds place, thousands place or ten thousands place.
Thus, in all combinations possible, sum of the digit 2 in all the possible places with proper weights would be:
2*10^4 + 2*10^3 + 2*10^2 + 2*10^1 + 2*10^0
Similarly for other digits, hence, together all our digits would contribute to:
(2+3+4+5+6) * 10^4
(2+3+4+5+6) * 10^3
(2+3+4+5+6) * 10^2
(2+3+4+5+6) * 10^1
(2+3+4+5+6) * 10^0

Or 20000 + 2000 + 200 + 20 + 2 = 22222
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by sanju09 » Mon Apr 06, 2009 1:33 am
kapsii! You are near there, see your calculations only... :)
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by sanju09 » Mon Apr 06, 2009 3:54 am
Actually each of the numbers would be in any place 4! times. Hence their contribution when in the ten thousand's place is
(2 + 3 + 4 + 5 + 6) (10000) * 4!
Similarly, when in thousand's place, they contribute
(2 + 3 + 4 + 5 + 6) (1000) * 4!
For 100s, 10s, and 1s places, the contributions are
(2 + 3 + 4 + 5 + 6) (100) * 4!
(2 + 3 + 4 + 5 + 6) (10) * 4!
(2 + 3 + 4 + 5 + 6) (1) * 4!
respectively.
Hence, the total contribution to the sum is

[spoiler](2 + 3 + 4 + 5 + 6) (11111) * 4![/spoiler]
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by ssmiles08 » Wed Apr 08, 2009 3:03 pm
Hi newbie here...I approached the problem the following way... and I got a different answer. Can someone tell me what I might be doing wrong here?

5 numbers can be formed by using 2, 3, 4, 5, 6 = that makes it 120 possible combinations of 5 digit numbers.

Then I took the average of the smallest number and the largest number possible : (65432 + 23456)/2 = 44444


avg = (sum)/(#)

so 44444 = (sum)/(120)

sum = 5333280
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by sanju09 » Thu Apr 09, 2009 12:11 am
ssmiles08 wrote:Hi newbie here...I approached the problem the following way... and I got a different answer. Can someone tell me what I might be doing wrong here?

5 numbers can be formed by using 2, 3, 4, 5, 6 = that makes it 120 possible combinations of 5 digit numbers.

Then I took the average of the smallest number and the largest number possible : (65432 + 23456)/2 = 44444


avg = (sum)/(#)

so 44444 = (sum)/(120)

sum = 5333280
B-) Average (arithmetic mean) of the first and last entries will be the average of the entire data-list only if the data is forming an AP; present is not that case perhaps!
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by gabriel » Thu Apr 09, 2009 9:34 am
ssmiles08 wrote:Hi newbie here...I approached the problem the following way... and I got a different answer. Can someone tell me what I might be doing wrong here?

5 numbers can be formed by using 2, 3, 4, 5, 6 = that makes it 120 possible combinations of 5 digit numbers.

Then I took the average of the smallest number and the largest number possible : (65432 + 23456)/2 = 44444


avg = (sum)/(#)

so 44444 = (sum)/(120)

sum = 5333280
Like Sanju has explained the formula mean*number of elements = sum of all the elements is true only if the numbers are consecutive or in an AP (consecutive numbers are also in an AP). In the question since we are only given the 5 numbers this clearly is not the case.
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by ssmiles08 » Thu Apr 09, 2009 11:27 am
aha gotchya! thanks
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