two-digit multiples of 6 -> 6,12,18 or simply (99-n)/6, {here we can use 6=2*3 must be even with digits' sum divisible by 3} 96/6 and we have 16 numbers which are multiples of 6
two-digit prime numbers -> 11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 ~ total 21 numbers
random selection of one multiple of 6 out of 16 and adding this to randomly selected 2-digit prime number out of 21, which is followed by reducing the result by half should result in how many real numbers and integer numbers?
first is even and the second is odd always (prime), the result is divided by 2 and will result in not integer number. Hence the probability is 0.
p.s. we actually don't need to find all primes and multiples of 6. This is concept question testing even-odd
sampath wrote:Q: Operation '#" is defined as adding a randomly selected two-digit multiple of 6 to a randomly selected
two-digit prime number and reducing the result by half. If operation '#' is repeated 10 times, what is the
probability that it will yield at least two integers?
