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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
28 Mar 2008, 01:32

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Difficulty:

95% (hard)

Question Stats:

29% (02:00) correct
71% (01:53) wrong based on 1453 sessions

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?

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.

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)

Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
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 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: \(P=1*\frac{9}{10}=\frac{9}{10}\).

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. _________________

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%

Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
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%

Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
13 May 2014, 05:10

Bahh..mistook non negative for non zero integers.. Quite easy...I checked zero too..but it was not in my set anyways..not a 700 I think _________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
20 May 2015, 18:36

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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
10 Jun 2015, 22:35

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 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

Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is [#permalink]
10 Jun 2015, 23:09

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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 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

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 = \(1 * \frac{9}{10} = 90\)%.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...