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sondenso
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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is Updated on: 07 Jul 2013, 06:11
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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29% (02:24) correct 71% (02:26) wrong based on 2033 sessions

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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%
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D
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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 23 Aug 2010, 14:51
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sondenso
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%
Show more

\(S=\{2,3,6,48,164\}\) 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 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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 28 Mar 2008, 05:26
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sondenso
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?
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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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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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 23 Aug 2010, 20:35
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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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 24 Aug 2010, 05:01
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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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It's one number: 678463.
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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 14 Sep 2013, 17:56
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Bunuel
sondenso
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.
Show more

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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 15 Sep 2013, 00:15
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Bunuel
sondenso
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.
Show more

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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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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 10 Jun 2015, 23:35
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sondenso
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%
Show more
z is a multiple of 6 and 678,463 is not a multiple of 6. therefore, the answer is E
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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 11 Jun 2015, 00:09
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matvan
sondenso
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%
z is a multiple of 6 and 678,463 is not a multiple of 6. therefore, the answer is E
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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 non-negative 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 calculation
Probability 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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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is 30 Jun 2017, 21:15
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Bunuel
sondenso
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 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.
Show more


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 non-negative integers. Therefore 6^0 = 1 is not possible and an answer of E 100% results. Thanks!
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Bunuel
sondenso
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 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 non-negative integers. Therefore 6^0 = 1 is not possible and an answer of E 100% results. Thanks!
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Yes, 0 is neither positive nor negative integer but the question talks about first 10 non-negative integers. Non-negative integers are 0 and positive integers, so those which are NOT negative.

Hope it's clear.
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Bunuel
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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...:(

It's one number: 678463.
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I fell for the same thing. :(

Then I realised the grammar part of it. The question says number "is" and not "are". So Yeah. It's one number.
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­Oh wow, what a fun question!

-> We need to find the probability that the number, 678463 is not a multiple of 6^K.
-> K is the product of 2 non-negative integers. Therefore, K is also a non-negative integer.
-> if K = 0, 
then 6^k = 1. 1 is a factor of any number including 678463. In other words, in this case, 678463 would be a multiple of 6^K. 
-> if K > 0 (a positive integer),
then 6^K is even. Then, it is impossible for 678463 to be a multiple of 6^k. 678463 is an odd number (if it contained even one "2", the number would be even, not odd). So, in this case, 678463 would be a multiple of 6^K.

So,
P(678463 is not a multiple of 6^K) = P(K>0) = 1 - P(K=0)

Total Events = total ways of computing K = 5 ways to choose a number from S x 10 ways to choose a number from 1st ten non-negative integers = 5 x 10 = 50.

K = 0 only happens when 0 is the number chosen from the 1st sten non-negative integers (Because Set S has no 0). 

Number of ways of ensuring that K is 0 = 5 x 1 (because 0 is fixed) = 5

P(K=0) = 5/50 = 1/10 = 10%

P(678463 is not a multiple of 6^K) = 1 - P(K=0) = 100 - 10 = 90%. Choice D. 

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