Tessellate S.T.E.M.S. (2019) - Computer Science - College - Set 4 - Objective Problem 5

How many rational numbers in between 0 and 1 are such that, when expressed as a fraction in the lowest form, the product of their numerator and denominator is 20 ! 20! ?


This problem is a part of Tessellate S.T.E.M.S (2019)

32 256 144 128

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1 solution

Avner Strum
Dec 24, 2018

To construct a coprime pair whose product is 20! we take the prime factors of 20! (ignoring their exponents) and split them into two distinct subsets from which we build the nominator and denominator (with their original exponents). The set of primes generating 20! is S={2,3,5,7,11,13,17,19} So let's say we start with the nominator - for each element of S we decide whether to take it or not to build the nominator, and the complelment subset determines the value of the denominator (preserving for each factor its original exponent). That makes a total of 256 options. Finally, it's clear that half of the options we constructed are greater than 1 so all in all we've got 128 options.

That sounds like a correct approach. Good job!

Tessellate STEMS Computer Science - 2 years, 5 months ago

How are you sure that only half values are less than 1

V i S i o N . - 2 years, 4 months ago

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For each fraction x x that we constructed, we constructed also 1 x \frac{1}{x} . Since x 1 x\neq 1 , for each of these couples of reciprocal numbers, one is bigger than 1 1 and the other is between 0 0 and 1 1

Marco Brezzi - 2 years, 4 months ago

I got stumped

Spriha Basir - 2 years, 3 months ago

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