# Number of turns for two wheels to bring edge-points together

The puzzle is as follows:

We present you below this riddle. A certain mechanism opens a gate of a maximum security lab. It just happens that a glass lets you see the mechanism and you know the password to enter is related with the number of turns given by the small wheel. The condition is that the password is the minimum number of turns that the wheel with the smallest radius must make, so that points $$A$$ and $$B$$ (where the opening latches are located) are in contact for the second time. Using this information, find the password.

The alternatives given are:

1. 11.25
2. 11
3. 13.5
4. 6.25

Is there a gentle way to solve this without much fuss? (note that the word takes inspiration from this source)

My issue is that I was only able to figure out that the number of turns the smaller wheel makes will be less than the larger.

Note that although not specifically mentioned in the problem, I'm assuming that the points' positions are with respect to the center.

The labels are as follows:

$$b$$: Turns given by the bigger wheel

$$s$$: turns given by smaller wheel.

Then:

$$s=\frac{18}{10}b$$

Thus the smaller wheel will get more turns than the bigger one. This makes sense. But I don't know how to relate this with the given positions in the graph.

I can't say that both will cover the same angle as that's not the case. But they will attain the same length.

But here I'm stuck. Is my logic correct? It would help me a lot if answers could explain step-by-step the necessary details so I will not be confused about matching points.

I have already attempted a few methods but could not find a logical answer.

This puzzle is from my Logic and Challenges book from 2000's and from the looks of it seems to be an adaptation from a reprinted copy of Martin Gardner's 1950's book on Puzzles.

I'll use your notation of $$b$$ for the number of turns of the big wheel, and $$s$$ for the number of turns of the small wheel.

You already deduced that $$s=\frac{18}{10}b$$.

From the picture we can also deduce that:

After a quarter turn of the small wheel, point B is in contact with the large wheel as required. This happens again after any number of additional full rotations. Therefore $$s=n+\frac14$$ for some whole number $$n$$.
The same reasoning goes for point A on the big wheel, so we also have $$b=k+\frac14$$ for some whole number $$k$$. Of course this will not the same number of rotations as the small wheel, because the large wheel turns more slowly than the small one.

Now you can just substitute and see where it gets us:

$$s=\frac{18}{10}b\\n+\frac14 = \frac{18}{10}\left(k+\frac14\right)\\n = \frac{9k+1}{5}$$
But $$n$$ is a whole number, so we need $$9k+1$$ to be divisible by $$5$$.
This first happens when $$k=1$$, $$n=2$$. The second time it happens, which is what the question asks for, is when $$k$$ is increased by a further $$5$$ rotations, i.e. when $$k=6$$, and $$n=11$$.

The total number of rotations of the small wheel is therefore:

$$s=n+\frac14 =11+\frac14$$ which is multiple choice option 1.

• Your answers are always the best! But I wonder, how did you concluded that for the second time that both wheels coincide is when $k$ is increased by further $5$?. Was this due trial and error on $n=\frac{9k+1}{5}$?. I mean after replacing $2$,$3$,$4$,$5$ and finally getting to $6$?. Or did you obtained it from a straight conclusion that because the denominator is $5$, then I must add $5$?. Which of these was the reason. Can you attend this doubt please? Feb 27, 2021 at 2:18
• @ChrisSteinbeckBell Both those methods work. You can do it by trial and error, but it is quicker when you realise that when you add something to $k$, the numerator increases by $9$ times that much, i.e. a multiple of $9$. We need that numerator increase to be a multiple of $5$, and the first multiple of $9$ that is also a multiple of $5$ is $9\cdot5$, so $k$ must increase by $5$. Feb 27, 2021 at 9:29