You're exactly right, the answer is $11$ times every $12$ hours, or $1$ time every $\frac{12}{11}$ of an hour, which works out to about once every $1$ hour, $5$ minutes, and $27.27$ seconds. This isn't an average either; the amount of time that passes between each meeting of the two hands is constant.

Just think about the time 11:59, and count through $12$ hours. You'll notice that there is a meeting for each hour $12,1,2,3,\ldots,10$, but no meeting for the $11$th hour because you stop at 11:59. This makes $11$ meetings in $12$ hours.

It's not too hard to see why the time passing between two meetings is constant. Just think about the rotational symmetry.

Another (and more difficult way) to look at the problem is to calculate the angular speed of each hand. Using convenient units, we have the minute hand traveling at a rate of $60$ units per hour, and the hour hand travels at a rate of $5$ units per hour. Thus, we would like to solve the following equation:

$$60t=5t\operatorname{mod}60$$

$0$ clearly solves the equation, so we look for the smallest positive value of $t$ solving it. Thus we want to solve:

$$55t=60$$

which gives us $t=\frac{12}{11}$ hours as before.

The overlap (possibility that both children are of the specific kind mentioned) is the issue, I think. Here’s a picture that helped me understand. Consider a random point $(x,y)$ in the unit square.

If you know that one of the two coordinates is bigger than $1/2$ (in set $A$), you have narrowed down the possibilities to the solid red region in the left picture. The chance that the other coordinate is also bigger than $1/2$ (you’re in $A\times A$) is about $1/3$ (blue-spotted part of the red region). It’s significantly less than the size of $A$ ($1/2$), because $A\times A$ is a significant part of $A\times I\cup I\times A$, the red region you know you’re in.

However, if you know that one coordinate is not only in $A$, but also in $B$ (say between $0.67$ and $0.71$), you are in the cross-shaped region on the right ($B\times I\cup I\times B$). The blue part of that region ($A\times B\cup B\times A$) is pretty close to $1/2$ of the cross, because the overlap is not so significant.

## Best Answer

Since Javes has

more than$40$ books, he hasat least$41$ books. Thus, Aslam has at least $5\cdot 41=205$ books. Since his number of books is a multiple of $4$ and $5$, it is a multiple of $20$. Thus, he has at least $220$ books.Say that $a$ is the number of books Aslam has. We know that $4|a$ ($4$ divides $a$) and that $5|a$. In general, if $k|x$ and $l|x$, it follows that $\text{lcm}(k,l)|x$, where $\text{lcm}(k,l)$ is the least common multiple of $k$ and $l$. Since the smallest number that is a multiple of both $4$ and $5$ is $20$, it follows that $20|a$.