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

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)

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.

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.

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%

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

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.

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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