sampath 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 678463 is not a multiple of Z?

Ans:[spoiler]9/10[/spoiler]

Hi! This is a truly bizarre question - what's the source?

In any case, it can be solved fairly quickly if you can actually decipher it and understand a few underlying concepts.

First, we note that 6^k will be even for all positive values of k. Since 678463 is odd, if k is positive then 678463 will NOT be a multiple of 6^k.

Second, we note that "non-negative" does NOT mean "positive"; "non-negative" includes 0. So, the first 10 non-negative integers are:

{0, 1, 2, 3, ... , 9}.

Third, we note that any positive number to the exponent 0 equals 1, which is a factor of 678463.

Putting it all together:

there's a 1/10 chance that we're multiplying a number from set S by 0; so, there's a 1/10 chance that k=0. Consequently, there's a 9/10 chance that k is positive. Accordingly, the answer is 9/10.

The question is interesting, but unlike anything I've seen on the GMAT or from a reliable source - for those who read it and shook your heads, I wouldn't worry about it too much.

Hi,
Z= 6^k means Z is an even number unless k = 0(Z^0=1)
The number 678463 is odd. So it cant be a multiple of an even number. This number can be a multiple of Z only if k = 0
For k to be zero, the number selected from the first 10 non -ve integers must be '0'
Probability of picking 0 from {0,1,2,3..,9} is 1/10
So, the probability that 678463 is not a multiple of Z is 1-(1/10) = 9/10.

Cheers!

sampath 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 678463 is not a multiple of Z?

Ans:[spoiler]9/10[/spoiler]

k can be formed in 10C1*5C1 ways; (10C1 indicates no. of ways of selecting 1 number from 10 non-negative integers, and 5C1 indicates no. of ways of selecting 1 number form set S(2,3,6,48,164);

now here; 678463 will be the multiple of Z=6^k if, k=0;
now k will be zero if we select 0 from first 10 non negative integers and multiply it with any of 5 integers of set S; hence; total no. of ways of having k=0 will be 1*5C1;

therefore required probability is ; 1- 5/50=1-1/10=9/10