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What is the probability that four whole numbers taken at

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What is the probability that four whole numbers taken at [#permalink] New post 01 Mar 2004, 11:32
What is the probability that four whole numbers taken at random and multiplied together will have 1, 3, 7 or 9 as its unit digit?

16/625
128/625
4/21
:madd
none of these
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Pls include reasoning along with all answer posts.
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 [#permalink] New post 01 Mar 2004, 12:09
Sunny,

Are you competing with me on posting the best questions?
Keep up Sunny!! :twisted:

Answer: A) 16/625.

I'll provide the steps later.

Last edited by kpadma on 01 Mar 2004, 13:44, edited 1 time in total.
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 [#permalink] New post 01 Mar 2004, 12:11
:lol: :beer
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 [#permalink] New post 03 Mar 2004, 09:49
kpadma wrote:
Sunny,

Are you competing with me on posting the best questions?
Keep up Sunny!! :twisted:

Answer: A) 16/625.

I'll provide the steps later.


kpadma,

Please provide your reasoning
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 [#permalink] New post 03 Mar 2004, 11:06
gmatblast wrote:
Please provide your reasoning


Total number of digits = 10
Numbers that yield only 1,3,7 and 9 as unit digit
when two numbers are multiplied are number that have
1,3,7, and 9 in their unit digit.

Potential candidate list = 1,3,7, and 9.
Number of potential candidates = 4

Probabilty to choose a number with a potential candidate = 4/10

Probability to choose all numbers from the potential candidate list
= (4/10)^4 = 16/625
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 [#permalink] New post 03 Mar 2004, 12:18
kpadma wrote:
gmatblast wrote:
Please provide your reasoning


Total number of digits = 10
Numbers that yield only 1,3,7 and 9 as unit digit
when two numbers are multiplied are number that have
1,3,7, and 9 in their unit digit.

Potential candidate list = 1,3,7, and 9.
Number of potential candidates = 4

Probabilty to choose a number with a potential candidate = 4/10

Probability to choose all numbers from the potential candidate list
= (4/10)^4 = 16/625


Well I think the problem talks about "whole numbers" and not about "digits"

In that sense, the probability can not be calculated. Because there are infinite whole numbers and there are infinite numbers ending 1,3,7 and 9.
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 [#permalink] New post 03 Mar 2004, 12:48
Gmatblast, Kpadma is right in his reasoning because all you need to consider is the units digit of those numbers. No matter how large those numbers are, the proportion, through an infinite set of numbers, will be the same.

What I am not certain about is when Kpadma says: "Numbers that yield only 1,3,7 and 9 as unit digit
when two numbers are multiplied are number that have
1,3,7, and 9 in their unit digit."

The original question mentions about four numbers being multiplied together,...
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 [#permalink] New post 03 Mar 2004, 13:57
Paul,

Yeah. I agree now.

Here is how I thought about this problem (considering that it is whole numbers and not digits)

Now the probability = Number of ways in which 4 numbers are selected from the set of numbers that meet our criteria / Number of ways to select 4 numbers from all the possible numbers

For numerator, we have infinite possibilities from which to select 4 numbers (inf C 4)

But I think the key is to understand that there are 4 numbers with 1,3,7,9 in first 10 numbers. Similarly there are 40 numbers with 1,3,7,9 as unit digits in first 100 numbers. and so on...


Quote:
What I am not certain about is when Kpadma says: "Numbers that yield only 1,3,7 and 9 as unit digit
when two numbers are multiplied are number that have
1,3,7, and 9 in their unit digit."


I think what kpadma meant was if you take any four numbers with unit digit = 1,3,7, 9, the product of all the 4 numbers will always have 1,3,7,9 as unit digit. So that is OK. If any of the 4 numbers have a unit digit other that 1,3,7 or 9, the product will also have unit digit other than 1,3,7 or 9
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 [#permalink] New post 03 Mar 2004, 14:19
Agree Gmatblast. So all 4 numbers HAVE to have either 1/3/7/9 as units digit in order to yield a number which is either 1/3/7/9 as units digit.

This is due to the fact that all the above numbers, when put to any exponent, will always give you either 1/3/7 or 9 as unit digit

1^2=1 3^2=3 7^2=9 9^2=1
1^3=1 3^3=9 7^3=3 9^3=9
1^4=1 3^3=7 7^4=1 9^4=1
...and so on

Thus, the answer would be (4/10)^4 = 16/625 because the numbers can be selected with replacement
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  [#permalink] 03 Mar 2004, 14:19
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