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Director
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If one has 4 digits {1,2,8,9} what are the last two digits [#permalink]
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29 May 2005, 15:30
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If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?



SVP
Joined: 05 Apr 2005
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Re: PS III [#permalink]
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29 May 2005, 16:27
Quote: {1,2,8,9}
i got the followings:
8+2 (or 9+1) =10
9+2 (or 8+1+2) = 11
9+1+2 = 12
9+8 = 17
9+1+8= 18
9+2+8= 19
9+8+2+1= 20



Director
Joined: 18 Apr 2005
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The answer is a two digit number
sum = 1289 + 1298 + 2189 + ... etc



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Re: PS III [#permalink]
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30 May 2005, 07:04
sparky wrote: If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?
80.
As we would have a total of 256 different no. with a equal representation of each of the 4 digits in all the places, we would have the sum of all units digits as (1+2+8+9)*64 = 1280. So the units digits is 0.
The tens digit is 1280 + 128 ( carry ) = 8 with 140 carried to the hundreth place.Hence 80.
HMTG.



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Re: PS III [#permalink]
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30 May 2005, 15:53
sparky wrote: If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?
Total no of 4 digits nos = 4! = 24.
Each digit comes an equal no of times in units and tens place.
sum of units = (1+2+8+9) * 4 = 80
sum of tens= (1+2+8+9) * 4 = 80
therefore sum of all units and tens = 880
last 2 digits = 80
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Director
Joined: 18 Apr 2005
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Re: PS III [#permalink]
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01 Jun 2005, 10:37
sparky wrote: If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?
My solution
Here we go,
Sum of units digits is 3!*(1+2+8+9) = 6*20 = 120 => put 0 carry over 12
Sum of tens digits is 3!*(1+2+8+9) + 12 = 120 + 12 = 132 => put 2 carry over 13
tens and units digits are, therefore, 20
The entire sum is 133,320, if I am not mistaken



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Re: PS III [#permalink]
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01 Jun 2005, 21:40
sparky wrote: sparky wrote: If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits? My solution Here we go, Sum of units digits is 3!*(1+2+8+9) = 6*20 = 120 => put 0 carry over 12 Sum of tens digits is 3!*(1+2+8+9) + 12 = 120 + 12 = 132 => put 2 carry over 13 tens and units digits are, therefore, 20 The entire sum is 133,320, if I am not mistaken
sparky, I think that you are assuming that the numbers cannot be repeated.
there can be a total of 4^4 numbers, yours just give 4*3!(24)
HMTG.



Director
Joined: 18 Apr 2005
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hehe, when I wrote the problem I meant no repeats, and I said in the problem itself that 'all possible four digit numbers that can be constructed from those 4 digits'
if you did it for 4^4, it's cool as long as you understand the concept it all that matters, the rest is details.



Director
Joined: 18 Apr 2005
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That's what I got. I don't have an answer from a book for this one.



Senior Manager
Joined: 17 May 2005
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sparky wrote: That's what I got. I don't have an answer from a book for this one.
well thats what i got too...so hopefully its right! hehe










