Question number 26, using the whole numbers from 1 to 22 inclusive, Horatio wants to form 11 fractions by choosing one number as the numerator and one number as the denominator. Every number will be used exactly once. What is the maximum number of Horatio's fractions that could have an integer value? So, first of all, let's focus in on the prime numbers less than 22 because they will cause us the most trouble in fractions that simplify to an integer value. So, let's say that from the primes less than 22, that is the numbers 3, 5, 7, 11, 13, 17, and 19, we can form the fractions 22 divided by 11. And the reason we do want to choose this fraction is that 11 here is the largest number that divides 22. And just one of the following, 13 divided by 1, 17 divided by 1, and 19 divided by 1. So, let's go ahead and keep this one and keep 13 divided by 1. And so, we will have the fraction either 17 divided by 19 or 19 divided by 17 will not work. And so, we have at most 10 integers in this construction. Of the 22 numbers that we use, 2 have to drop out, 2 primes that will remain. And then, after that, it's a guess and check process. So, with 22 over 11 and 13 over 1, I came up with the following 10 fractions. And then, we have, of course, one more, the fraction made by using 17 and 19. But here is an example, the maximum example that we can come up with. And so, the answer here is 10.

fractions

integer values

prime numbers

Horatio

number pairing