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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:

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Question Stats:

27% (03:12) correct
73% (01:53) wrong based on 400 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 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^kIS 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}.

Set S consists of numbers 2, 3, 6, 48 and 164. Number K [#permalink]
13 Aug 2013, 21:21

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 = 6k, what is the probability that 678463 is not a multiple of Z?

Re: Set S consists of numbers 2, 3, 6, 48 and 164. Number K [#permalink]
14 Aug 2013, 00:07

Expert's post

Countdown 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 = 6k, what is the probability that 678463 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?

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^kIS 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^kIS 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

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%

gmatclubot

Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is
[#permalink]
06 Apr 2014, 23:02