The given question is GMAT 700 level problem solving question combining number properties and probability concepts. The focus is on understanding what kind of numbers have exactly 3 factors and elementary counting methods.

Question 4: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors ?

- \\frac{\text{4}}{\text{25 × 99}})
- \\frac{\text{2}}{\text{25 × 99}})
- \\frac{\text{8}}{\text{25 × 99}})
- \\frac{\text{16}}{\text{25 × 99}})
- \\frac{\text{32}}{\text{25 × 99}})

@ INR

Any positive integer will have '1' and the number itself as factors. That makes it a minimum of 2 factors (except '1' which has only one factor). If the positive integer has only one more factor, then in addition to 1 and the number, the square root of the number should be the only other factor.

There are two key points in the above finding. The number has to be a perfect square. And the only factor other than 1 and the number itself should be its square root.

Therefore, if a positive integer has only 3 factors, then it should be a perfect square and it should be the square of a prime number.

Let us look at an example. 4 has the following factors: 1, 2, and 4 (exactly 3 factors). It is the square of '2' which a prime number.

Squares of numbers that are not prime numbers will have more than 3 factors. For instance, 36 is a perfect square. But it has 9 factors.

Number of squares of prime numbers from 1 to 100 that have exactly 3 factors are 4, 9, 25, and 49. i.e., 4 numbers

Number of ways of selecting two distinct integers from the set of first 100 positive integers = 100C_{2} ways.

i.e., 100C_{2} = \\frac{\text{100 × 99}}{\text{2}})

The product of two distinct numbers 'a' and 'b' will be 4 when one of the numbers is 1 and the other is 4. There is only one set that will result in this product.

The same holds good for the other 3 numbers as well. Product of two distinct numbers 'a' and 'b' will be 9 when one of the numbers is 1 and the other is 9 and so on.

Therefore, there are 4 outcomes in which the product of the two numbers will result in a number that has exactly 3 factors.

Probability(Event occuring) = \\frac{\text{Number of favourable outcomes}}{\text{Total number of possibilities}})

i.e., Required Probability = \\frac{4}{\frac{\text{100 × 99}}{\text{2}}}) = \\frac{\text{2}}{\text{25 × 99}})

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