Brain Play

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!

A Magic Trick - Solution

→ Print-friendly version After fiddling with some expressions on paper, Harriet called Gus back. “I have the solution to your mystery,” she told him. Gus was excited. “I’ve been stuck on this for days! How did you solve it?” “First,” Harriet told him, “I wrote the two-digit number as $10a + b$, where $a$ and $b$ are variables representing digits between 0 and 9.” “Okay,” said Gus. “That’s what I did, too.

A Phony Proof - Solution

→ Print-friendly version “The problem,” Detective Harriet Hesterton told Gus, “happens when you divide both sides of the equation by $x - y$. You started by saying that $x = y$, but that means $x - y = 0$. When you divide both sides by $x - y$, you’re really dividing by zero.” “Why is that a problem?” Gus asked, not giving up yet. “Say you start with the equation $0 = 0$,” Harriet explained.

A Recurrence Riddle - Solution

→ Print-friendly version Let’s start by computing the first few terms of the sequence: $$ \begin{align} a_1 &= 1 \\ a_2 &= 1 + 2(2) - 1 = 4 \\ a_3 &= 4 + 2(3) - 1 = 9 \\ a_4 &= 9 + 2(4) - 1 = 16 \\ a_5 &= 16 + 2(5) - 1 = 25 \\ a_6 &= 25 + 2(6) - 1 = 36 \end{align} $$

A Triangle of Prime Opportunity - Solution

→ Print-friendly version Let’s call the side lengths $a$, $b$, and $c$, and the altitudes $x$, $y$, and $z$. (By the way, a fun fact: the altitudes of any triangle intersect in a single point called the orthocenter.) What next? Well, we have altitudes; maybe that’s a clue. So what’s special about altitudes? One thing that altitudes let us do is very easily calculate the area of the triangle. And since we have all three side lengths and all three altitudes, we can calculate the same area in three different ways:

An Interplanetary Currency - Solution

Let’s start with the 6-blyp case. The first thing we notice is that the ATM will never give Alex an 8-blyp coin for this problem, since 6 < 8. So it’s enough for us to solve this problem just with 1, 2, and 4 blyp coins. After playing around with this for a bit, we might notice that some ways of making 6 blyps involve a 4-blyp coin while others don’t.

Bridging the Gap - Solution

→ Print-friendly version a. After a bit of thinking, the first thing we notice about this problem is that the bridges are the only things that matter: the shape and size of the islands are irrelevant. So we can simplify each terrestrial region down to a single point – let’s call the points A, B, C, and D – and let each bridge correspond to a line between two points:

City Council - Solution

→ Print-friendly version Let’s say that there are $x$ knights on the city council. If we can figure out what the value of $x$ is, we’ll be practically done. We can start our exploration with a couple of extreme cases: If $x=0$, then all of the councilmembers are liars. But then the first councilmember’s statement is true, and since liars never tell the truth, that means the first person is a knight.

Counterfeit Detective - Solution

→ Print-friendly version Harriet pulled out 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third bag, and so on, up through 10 coins from the tenth bag. “If all of these were real,” she thought to herself, “they would weigh $(1 + 2 + \ldots + 10) \cdot 5 \textrm{ grams} = 275 \textrm{ grams}$. But some of them are fake, so the total will be less than that – and the difference will tell me which bag holds the fakes.

Cut Corners - Solution

→ Print-friendly version Let’s take a closer look at that grid. Maybe you’ve played with this for a while without coming up with a valid domino tiling. It’s definitely not as easy as it looks! To see why, let’s color this grid like a checkerboard: (Give this problem another try before reading more. I’ll wait.) … … … … … Okay. The key to finishing here is to notice that every time we place a domino on this checkerboard, we cover up one white square and one green square.

Duo Dominoes - Solution

→ Print-friendly version Let’s take another look at that 2xn region. The hard thing about this problem is the n. A good strategy here is to start with small numbers and build up and generalize. So let’s look at n=1. This case is pretty simple: there’s only room for one domino on this grid, and there’s only one way to place it. Now let’s look at n=2. This case is also pretty simple: there’s room for 2 dominoes side by side, and we could either place them both horizontally or both vertically.