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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]

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Updated on: 07 Jul 2013, 06:11
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28% (02:27) correct 72% (02:28) wrong based on 1543 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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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  [#permalink]

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23 Aug 2010, 14:51
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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: $$P=1*\frac{9}{10}=\frac{9}{10}$$.

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28 Mar 2008, 05:26
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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.

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Joined: 14 Jul 2004
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28 Mar 2008, 06:53
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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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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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24 Aug 2010, 05:01
2
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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14 Sep 2013, 17:56
1
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: $$P=1*\frac{9}{10}=\frac{9}{10}$$.

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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15 Sep 2013, 00:15
4
2
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: $$P=1*\frac{9}{10}=\frac{9}{10}$$.

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]

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07 Apr 2014, 00:02
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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]

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

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11 Jun 2015, 00:09
6
1
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

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

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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 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: $$P=1*\frac{9}{10}=\frac{9}{10}$$.

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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Posts: 60468
Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is  [#permalink]

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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 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: $$P=1*\frac{9}{10}=\frac{9}{10}$$.

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!

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

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01 Dec 2019, 07:55
Bunuel wrote:
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.

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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Re: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is   [#permalink] 01 Dec 2019, 07:55
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