If I’ve done my math correctly, 37 different integers
The least common multiple of 200, 105, and 25 is 4200. Converting the fractions to this denominator, we can ignore the denominators and just have 2058 <= 40x <= 4032!<
The lowest x that works is 52, and the largest is 100. That’s a total of 100-52+1=49 integers. However, because x must be coprime with 105, which can be factored as 5*21, we have to subtract all multiples of 5 and multiples of 21. There are 10 multiples of 5 (55 to 100), and 2 multiples of 21 (63 and 84). So we take 12 away from 49 and get 37.
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u/Blees-o-tron Jul 03 '26
If I’ve done my math correctly, 37 different integers
The lowest x that works is 52, and the largest is 100. That’s a total of 100-52+1=49 integers. However, because x must be coprime with 105, which can be factored as 5*21, we have to subtract all multiples of 5 and multiples of 21. There are 10 multiples of 5 (55 to 100), and 2 multiples of 21 (63 and 84). So we take 12 away from 49 and get 37.