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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is

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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink] New post 28 Mar 2008, 01:32
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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?

A. 10%
B. 25%
C. 50%
D. 90%
E. 100%
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Jul 2013, 05:11, edited 1 time in total.
Edited the question and added the OA.
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Re: Probability [#permalink] New post 28 Mar 2008, 04:26
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sondenso wrote:
Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?


I first looked at 678,463. The number is not a multiple of 2,3,7 or 9.

Then I looked at Z.
Z = 6*6*6* ...*6 (k times).

If 678,463 has to be a multiple of Z, it has to be a multiple of 6.

Another case is that the integer we pick is 0. Probability of picking 0 as integer is 1/10. If integer is 0, Z becomes 1 and 678,463 becomes a multiple of Z.

Hence answer is D 90%.

What is the answer?
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Re: Probability [#permalink] New post 28 Mar 2008, 05:53
I think the answer is 90%, because the probablity of choosing 0 from 0-9 is 10%.
If 0 is chosen than only we have 678463 is multiple of 6^0 (= 1).
If any other number is chosen, then 678463 is not multiple of 6 (because 6^k)
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Re: Probability [#permalink] New post 04 Aug 2010, 10:20
Are the exact numbers in the set irrelevant to finding the answer? Would any positive integers have worked in lieu of 2, 3, 6, 48, and 164?
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Re: Probability [#permalink] New post 23 Aug 2010, 12:34
I was not able to understand the solution here.

Bunuel, do you mind explaining the approach to such problems?
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Re: Probability [#permalink] New post 23 Aug 2010, 13:51
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sondenso wrote:
Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?

a. 10%
b. 25%
c. 50%
d. 90%
e. 100%


S=\{2,3,6,48,164\} and set of first 10 non-negative integers, say T=\{0,1,2,3,4,5,6,7,8,9\}.

K=s*t, where s and t are random numbers from respective sets.

678,463 is an odd number.

The only case when 6^k IS a factor of 678,463 is when k equals to 0 (in this case 6^k=6^0=1 and 1 is a factor of every integer). Because if k>0, then 6^k=even and even number can not be a factor of odd number 678,463.

Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}.

Answer: D.
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Re: Probability [#permalink] New post 23 Aug 2010, 19:35
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Bunuel, I agree that 463 is an odd number but 678 is not odd but an even number. What am I missing here...:(
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Re: Probability [#permalink] New post 24 Aug 2010, 04:01
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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink] New post 07 Jul 2013, 05:13
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Re: Probability [#permalink] New post 14 Sep 2013, 16:56
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Bunuel wrote:
sondenso wrote:
Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?

a. 10%
b. 25%
c. 50%
d. 90%
e. 100%


S=\{2,3,6,48,164\} and set of first 10 non-negative integers, say T=\{0,1,2,3,4,5,6,7,8,9\}.

K=s*t, where s and t are random numbers from respective sets.

678,463 is an odd number.

The only case when 6^k IS a factor of 678,463 is when k equals to 0 (in this case 6^k=6^0=1 and 1 is a factor of every integer). Because if k>0, then 6^k=even and even number can not be a factor of odd number 678,463.

Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}.

Answer: D.


Couldn't understand this-
Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}.




I simply calculated probability like this-

45/50

45- when 6^k IS EVEN, 50 total number of outcomes.
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Re: Probability [#permalink] New post 14 Sep 2013, 23:15
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honchos wrote:
Bunuel wrote:
sondenso wrote:
Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?

a. 10%
b. 25%
c. 50%
d. 90%
e. 100%


S=\{2,3,6,48,164\} and set of first 10 non-negative integers, say T=\{0,1,2,3,4,5,6,7,8,9\}.

K=s*t, where s and t are random numbers from respective sets.

678,463 is an odd number.

The only case when 6^k IS a factor of 678,463 is when k equals to 0 (in this case 6^k=6^0=1 and 1 is a factor of every integer). Because if k>0, then 6^k=even and even number can not be a factor of odd number 678,463.

Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}.

Answer: D.


Couldn't understand this-
Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}.




I simply calculated probability like this-

45/50

45- when 6^k IS EVEN, 50 total number of outcomes.


First of all the total number of outcomes will be 10 * 6 = 60 (10 from 0 to 9 and 6 from Set S)
6^k will be even for all the numbers of K but 0.
Therefore number of cases when 6^k will be even will be 9*6 = 54 i.e. (9 from 1 to 9 excluding 0 and 6 from Set S). Since K can take any value from 1 to any multiple of 1.

Therefore 54/60 is the probability i.e. 9/10 = 90%.

Regarding what Bunuel has posted "Hence 6^k NOT to be a factor of 678,463 we should pick any number from S and pick any number but 0 from T: P=1*\frac{9}{10}=\frac{9}{10}."

He means Probability to pick any number from S will be 6/6 i.e. 1 and probability to pick any number from T but 0 will be 9/10. Since K is multiplication of these probabilities it will be 1*9/10 = 90%

Hope it helps.

Consider Kudos if it helped. :)
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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink] New post 06 Apr 2014, 23:02
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Z=6^K, so Z is even or Z = 1 (K=0)
if K is not equal to zero than Z is even and 678,463 is not a multiple of Z
if K is equal to zero than z is equal to 1 and 678,463 is a multiple of Z (Z=1)
the propability that K is equal to zero is 1/10 =10% (K=a*b where a is one random number from set S whose numbers are all not equal to zero, and b is one of the first 10 non-negative integers)
So the propability that 678,463 is not a multiple of Z is 100% - 10% = 90%
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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink] New post 13 May 2014, 05:10
Bahh..mistook non negative for non zero integers.. :evil:
Quite easy...I checked zero too..but it was not in my set anyways..not a 700 I think
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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is   [#permalink] 13 May 2014, 05:10
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