Brain Play is a collection of mathematical puzzles meant to stretch your logical and numerical reasoning. Playing with these puzzles is a great way to make friends with math!
Puzzles are posted periodically here, each followed by a solution. Print-friendly versions of all problems and solutions to date can be found here.
Happy puzzling!
→ Print-friendly version If you’ve played with this problem for a while, you might have noticed that getting to (1, 1) is quite troublesome. Every time the frogs seem to get close, they just miss the mark.
There’s a reason for that!
To see why this is so hard, let’s take a closer look at what a leap looks like.
To get from one frog’s location to another frog’s location, you can always take some number of steps to the right and some number of steps up.
→ Print-friendly version Imagine that each light bulb is numbered: Bulb 1, Bulb 2, Bulb 3, etc. When Rachel pulls the chain of every nth bulb, only the bulb numbers divisible by n are affected. To see how the procedure affects an individual bulb, then, we just need to look at its divisors.
Take the case of Bulb 12. The divisors of 12 are 1, 2, 3, 4, 6, and 12, as we can see when we multiply:
→ Print-friendly version The first thing we notice is that the choices for the basketballs, volleyballs, baseballs, and golf balls are all independent: how Leslie assigns one type of ball to the different trucks doesn’t restrict how she might assign a different type of ball to the different trucks. So we can start with just the basketballs and hope that we can re-use our technique on the other kinds of balls.
→ Print-friendly version When we roll the outer penny around the central cluster, we’re rolling it from bump to bump 6 times. If we can figure out what happens just going from one bump to the next, we can just repeat that 6 times to get our overall result. Let’s zoom in on just the top part and see what it looks like geometrically.
Notice how we’ve connected the centers of the pennies together.
→ Print-friendly version At every decision point, the five pirates are assessing what would happen if they accepted or rejected the proposal. If we start with the first decision, we’re going to need to handle some pretty complex decision criteria. So instead, we’ll start with a much simpler case: let’s imagine what would happen if most of the pirates were thrown overboard, and only 2 pirates – Desdemona and Ed – were left.
→ Print-friendly version We don’t know the size of the salt and pepper shakers. Is this a problem?
It turns out not to be, as long as we keep our eye on what happens with the teaspoon.
Step 1 in the procedure is to move a teaspoon of pepper to the salt shaker. This is pretty straightforward: one teaspoon of pepper is transferred.
Step 2 is to move a teaspoon of salt-and-pepper mix back to the pepper shaker.
→ Print-friendly version To find the probability that Harvey picks a valid order for his socks and boots, we want to count the number of valid orders for his socks and boots, and then divide that by the total number of ways Harvey might order his socks and boots.
Let’s start with the total number of orderings, since that’s the simpler part of this puzzle. Harvey has 16 pieces of footwear to choose from (8 socks plus 8 boots), so he has 16 choices for the first piece of footwear he puts on; after that, he has 15 choices for his second piece of footwear, and then 14 choices for his third, and so on, for a total of
→ Print-friendly version At first glance, we might think of the star as consisting of many pieces that add together to form the overall shape. But there are a lot of pieces to handle if we think about it this way!
Let’s instead look at the area that’s not covered by the star:
That looks a lot simpler! Because of the star’s symmetry, all of the green triangles are congruent– they have the same shape and the same size.
→ Print-friendly version It’s simplest to think about the entire flock taking off by working one swan at a time.
After the first swan takes off, we’re at $k=1$. The probability that a second swan takes off is $1 - \frac{1}{1+3} = 1 - \frac{1}{4} = \frac{3}{4}$.
After the second swan takes off, we’re at $k=2$. The probability that a third swan takes off is $1 - \frac{1}{2+3} = 1 - \frac{1}{5} = \frac{4}{5}$.
→ Print-friendly version “Ohhh, I see,” Janelle said, after a long pause staring at the rectangle. “It doesn’t line up all the way!”
“What do you mean?” Terrence asked his big sister.
“There’s a really thin slit on the diagonal that’s not covered with paper,” Janelle explained, and she pulled apart the pieces of paper just slightly so the slit was more visible.
“Look at the slope of the diagonal cuts,” she continued.
