A man and a woman standing side by side begin walking so that their right feet hit the ground at the same time. The woman takes 3 steps for each 2 steps of the man. How many steps does the man take before their left feet simultaneously reach the ground?
I gave this problem to three gifted friends of mine and they all failed. I did not, as I solved the problem musically. I am very curious and the fact I solved it was not enough. I was interested how the problem would look like whatever number of steps the man and woman take. What I found is following:
1. We have to decide who is faster, i.e. who takes more steps per unit of time. If no one is faster then it will take them only 1 step to simultaneously reach the ground with their left feet. If the woman (or the man) is faster, we have to solve the problem as stated by James L. Adams. Let us follow the assumption that the number of steps the woman takes is always greater than the number of steps the man takes (as on average men are taller than women).
2. In a unit of time a woman takes y steps and the man takes x steps. Let us explore the various ratios x/y. There are 2 types of problems then, each type depending on the ratio:
2.1. Never-problems, where the people will never reach the ground simultaneously with their left feet. Such is the problem stated by James L. Adams;
2.2. Sometimes-problems, where the people will simultaneously reach the ground (at a certain time/distance from the start). The relevant question here is when/where.
3. My solution:
3.1. If a problem is a sometimes-problem, then x/y is (or could be reduced to) a proper fraction where the numerator and the denominator are odd numbers. People will simultaneously reach the ground with their left feet on the y-th step of the woman (if x/y can be reduced) or on the xy-th step of the woman (if x/y is a fraction that cannot be reduced further).
3.2. All other problems, such as 2/3 problem above (2 is not odd), are never-problems.
Can you prove me wrong?