Set S consists of numbers 2, 3, 6, 48, and 164. Number K is

Question

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%

A. 10% 
B. 25% 
C. 50% 
D. 90% 
E. 100%

Bunuel · Accepted Answer

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.

s4ur4bh · Answer

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?

milindd · Answer

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)

seekmba · Answer

Bunuel, I agree that 463 is an odd number but 678 is not odd but an even number. What am I missing here...

Bunuel · Answer

It's one number: 678463.

honchos · Answer

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.

shameekv · Answer

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.

FZ · Answer

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%

mathivanan · Answer

z is a multiple of 6 and 678,463 is not a multiple of 6. therefore, the answer is E

EgmatQuantExpert · Answer

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

jimmymat · Answer

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!

Bunuel · Answer

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.

TheNightKing · Answer

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.

HarshavardhanR · Answer

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.

___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

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Set S consists of numbers 2, 3, 6, 48, and 164. Number K is

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