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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
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Updated on: 07 Jul 2013, 06:11
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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 nonnegative 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%
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Originally posted by sondenso on 28 Mar 2008, 02:32.
Last edited by Bunuel on 07 Jul 2013, 06:11, edited 1 time in total.
Edited the question and added the OA.




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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
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23 Aug 2010, 14:51
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 nonnegative 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 nonnegative 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 cannot 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
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28 Mar 2008, 05:26
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 nonnegative 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
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28 Mar 2008, 06:53
I think the answer is 90%, because the probablity of choosing 0 from 09 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
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23 Aug 2010, 20:35
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
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24 Aug 2010, 05:01
seekmba wrote: Bunuel, I agree that 463 is an odd number but 678 is not odd but an even number. What am I missing here... It's one number: 678463.
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Re: Probability
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14 Sep 2013, 17:56
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 nonnegative 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 nonnegative 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
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15 Sep 2013, 00:15
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 nonnegative 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 nonnegative 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
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07 Apr 2014, 00:02
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 nonnegative 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
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10 Jun 2015, 23:35
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 nonnegative 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% z is a multiple of 6 and 678,463 is not a multiple of 6. therefore, the answer is E



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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
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11 Jun 2015, 00:09
matvan 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 nonnegative 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% z is a multiple of 6 and 678,463 is not a multiple of 6. therefore, the answer is E Hi matvan, The answer is D i.e. 90%. The question is asking the probability of \(\frac{678463}{6^k}\) not being an integer. For a number to be divisible by any positive multiple of \(6\), it should at least be divisible by both \(2\) and \(3\). Since \(678463\) is not an even number, it is not divisible by \(2\). So for every positive multiple of \(6\), \(\frac{678463}{6^k}\) is not an integer. However the question talks of \(k\) as one of the first ten nonnegative numbers which also includes 0. If \(k = 0\) , then \(6^k = 6^ 0 = 1\). In that case \(678463\) will be a multiple of \(6^0\) i.e.\(1\). Hence the probability of \(678463\) not being a multiple of \(6^k\) is only possible when \(k = 0\) AND any random number being picked from set S. Probability calculationProbability of any random number being picked from set S = \(1\) Probability of \(k\) not being \(0\) = \(\frac{9}{10}\) ( as there are total of \(10\) ways to pick up \(k\) and \(9\) ways for \(k\) not being \(0\)) Since it's an AND event , we will multiply the probabilities of both the events. Hence total probability = \(1 * \frac{9}{10} = 90\)%. Hope it's clear Regards Harsh
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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
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30 Jun 2017, 21:15
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 nonnegative 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 nonnegative 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 cannot 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. My understanding was that 0 is a number that is neither negative nor positive and therefore should not form part of the set of the first ten nonnegative integers. Therefore 6^0 = 1 is not possible and an answer of E 100% results. Thanks!



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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
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30 Jun 2017, 23:39
jimmymat 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 nonnegative 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 nonnegative 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 cannot 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. My understanding was that 0 is a number that is neither negative nor positive and therefore should not form part of the set of the first ten nonnegative integers. Therefore 6^0 = 1 is not possible and an answer of E 100% results. Thanks! Yes, 0 is neither positive nor negative integer but the question talks about first 10 nonnegative integers. Nonnegative integers are 0 and positive integers, so those which are NOT negative. Hope it's clear.
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