Question 758470: a box contains 100 tickets 1 to 100 a person picks out 3 such that the product of the number on two on 2 of the tickets yield the number on third ticket . how many tickets can never be picked
options
a.10
b.11
3.25
4.26
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The problem is that the 3 numbers picked must be different, and must be numbers between 1 and 100.
Numbers from 3 to 50 can be picked along with their double, and number 2, as in
2 x 3 = 6, or 2 x 50 = 100.
The number 2 can be picked up along with any other number from 3 to 50 and the double of that other number.
The number 1 cannot be picked up at all, because it cannot be the product, and using it as a factor makes the product equal to the other factor, requiring that the other factor be picked twice, which cannot happen.
Most numbers greater than 50 can be written as the product of two different smaller numbers, but prime numbers and perfect squares of prime numbers could be a problem. (Perfect squares of numbers that are not prime can always be written as products of two different smaller numbers).
Prime numbers that are less than 50, could be one of the factors, as shown above.
Squares of prime numbers turn out not to be a problem. The smallest prime numbers, in order, are:
2, 3, 5, 7, 11, 13, ...,
and their squares are
4, 9, 25, 49, 121, 169, ...
so their squares are not a problem because either they are numbers below 50, or they are larger than 100.
There are 10 prime numbers between 50 and 100:
53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
Each one of those numbers can only be factorized as 1 times itself, so they cannot be picked up.
So, unless I am missing something, ther are exactly  numbers that could not be picked up in the situation described in the problem: the 10 primes from 53 to 97, and number 1.
NOTE:
For problems like this, I would recommend the artofproblemsolving forum. That is a place where you could find aspiring future math olympians. They rise to any challenge, and provide a brief description of their reasoning that is perfectly clear and understandable to them. Here we just try to solve the problem and explain it so that you will understand it, and sometimes we fail.
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