Lionel has a calculator that displays a maximum of ten digits (for example: 3,478,062,139 and 0.073672932)
• Lionel chose 4 distinct prime numbers p,q,r and s all less than 100 and he calculates p/q – r/s. The result is 0.180451127. Determine p,q,r and s.
• Consider the two base ten numbers 0.728101457 and 0.635149023. One of these numbers is the result displayed by Lionel's calculator of an irreducible fraction u/v with u and v, that are positive integers less than 1000. The decimals of the other number are taken from a table of random numbers.
Determine the terms u and v of the irreducible fraction.
I remembered solving "
Minimizing the Denominator" in the past using continued fractions. That seems like a good way to solve this problem without resorting to brute force computer searches.
I'm not going to rehash how to calculate continued fractions and let Wolfram Alpha do the arithmetic work.
Part 1:
The continued fraction representation of 0.180451127 is [0;5,1,1,5,2,68979,3]. The huge intermediate value of 68979 is an ideal spot to take a truncation.
Then [0;5,1,1,5,2] = 24/133. 133=7*19 is a semiprime which means q and s are in some order 7 and 19. Then its just a bit of fiddling to reconstruct the original difference of 17/19 - 5/7.
Part 2:
The continued fraction representation of 0.728101457 is [0;1,2,1,2,9,1,1,1,1,3,1,15,1,2,1,5,4,1,6] and similarly the continued fraction representation of 0.635149023 is [0;1,1,1,2,1,6,13,1654,1,10,1,2]
The presence of 1654 in the second continued fraction stands out. Then [0;1,1,1,2,1,6,13] = 618/973 = 0.6351490236....
No truncation of the first continued fraction of 0.728101457 matches all ten digits without the numerator and denominator exceeding 1000, so it must be just some random digits.
Continued Fractions Solution
