Question 1: What is the least possible value of integer K such that K^3 is divisible by 240 ?

Posting options for the 1st question could have helped solve the problem in less than 30 seconds. Nonetheless, here it goes

240 = (2^4)*5*3.So the value of k^3 should be divisible by 2^4,5,3.
Let us say k^3 = (2^4)*5*3, but powers of the prime factors of cube of a number should be in multiples of 3. Nearest multiple of 3 greater than 4(2^4) is 6, greater than 1(5^1) is 3, greater than 1(3^1) is 3. So, the value of k^3 = (2^6)*(5^3)*(3^3). Implies k = 60

Question 2: Each of the first 100 positive integers is written on a different blank slip of paper. The 100 slips are then placed in an empty hat. If a slip of paper is drawn from the hat at random, what is the probability that the number written on the slip is the product of three distinct prime numbers?

30,42,66,70 and 78 are the only numbers below 100 which can be expressed as a product of three distinct prime numbers
IMO : [spoiler]1/20[/spoiler]

Will update the complete solution in a while !

confused..
can you explain them more clearly plz

shikh wrote:confused.. can you explain them more clearly plz

First approach

240 = 2^4 * 3 * 5. For k^3 to be a multiple of 240, k^3 should be divisible by 2*2*2*2*3*5 (240).
Now that we know k^3 contains just 2's,3's and 5's, let us assume that k =2*3*5(Least number).Then the value of k^3 = 2^3 * 3^3 * 5^3 = 2*2*2*3*3*3*5*5*5.k^3 should be divisible by 2*2*2*2*3*5 but it is not. k is short of one 2. Let us multiply the value of k by 2 making it k = 2^2 * 3 * 4 .Now, k^3 = 2^6 * 3^3 * 4^3, which satisfies the minimum criteria of four 2s, one 3 and one 5.

k = 60

Another approach :

240 = 2^4*3*5, so k should be a multiple of 2*3*5 = 30.
Let k =30, then k^3 = 30*30*30 => (k^3)/240 = 900/8 = Not an Integer
Let k =60(next multiple of 30), then k^3 = 60*60*60 => (k^3)/240 = 900 = Integer, So 60 is the least value of k

got it:)
thanks a lottt..
wat abt the second one:s

chaitanya.bhansali wrote: Question 2: Each of the first 100 positive integers is written on a different blank slip of paper. The 100 slips are then placed in an empty hat. If a slip of paper is drawn from the hat at random, what is the probability that the number written on the slip is the product of three distinct prime numbers?

The first three prime number are 2,3,5.
The least possible number below 100 which can be formed by multiplying three distinct prime number is 2*3*5 = 30.
The largest possible number below 100 which can be formed by multiplying three distinct prime numbers is 2*3*13 = 78
So, all numbers less than 30 and greater than 78 cannot be written as a product of three distinct prime numbers. 2*3*7,2*3*11,2*5*7 are the rest

you need not consider prime numbers beyond 13 as 2*3*17 > 100. In all the combinations formed using 2,3,5,7,11,13 the value of 2*3*5, 2*3*7, 2*3*11, 2*3*13, 2*5*7 (30,42,66,70,78)is less than 100.

Total possibilities= 100
Conditional possibilities = 5
Probability = 5/100 = 1/20

Correct me if I am wrong.