## Saturday, August 26, 2017

### Replacing Marbles

We place 15 black marbles and 15 white marbles in an urn. We have 30 additional black marbles in a bag. Then we follow these rules.

1. Remove two marbles from the urn.

2. If they are different colors, put the white marble back in the urn and the black marble in the bag.

3. If they are the same color, put both marbles in the bag, then put one black marble from the bag into the urn.

Continue following these rules until only one marble is left in the urn. What color is that marble? Click below for the answer.

Labels:
logic puzzles,
math

## Saturday, August 19, 2017

### Factor Sums

Not counting itself, the number 6 has the factors 1, 2, and 3, which add to 6. The number 28 has the same property (its factors are 1, 2, 4, 7, and 14). Can you come up with a three-digit number that has this property? What about a four-digit number? Click below for the answers.

If you knew that a number that is the sum of its own proper divisors is called a Perfect number, this puzzle was pretty easy. You could just search for that name and find the solutions are

The formulations states that $q(q + 1) / 2$ is a perfect number whenever $q$ is a prime of the form $2^p - 1$ for prime $p$ (now known as a Mersenne prime). So, if we know the first few Mersenne primes, we can calculate the first few perfect numbers.

$3(3 + 1) / 2 = 6$

$7(7 + 1) / 2 = 28$

$31(31 + 1) / 2 = 496$

$127(127 + 1) / 2 = 8,128$

$8,191(8,191 + 1) / 2 = 33,550,336$

The ancient Greek mathematicians would not have known that 8,191 was a prime, so Euclid would only have known the first four Perfect numbers. Now you can say you know something that Euclid didn't!

**496**and**8,128**. Perfect numbers have been known at least as far back as Euclid (323–283 BCE), who included a formulation for then in his book of Elements.The formulations states that $q(q + 1) / 2$ is a perfect number whenever $q$ is a prime of the form $2^p - 1$ for prime $p$ (now known as a Mersenne prime). So, if we know the first few Mersenne primes, we can calculate the first few perfect numbers.

$3(3 + 1) / 2 = 6$

$7(7 + 1) / 2 = 28$

$31(31 + 1) / 2 = 496$

$127(127 + 1) / 2 = 8,128$

$8,191(8,191 + 1) / 2 = 33,550,336$

The ancient Greek mathematicians would not have known that 8,191 was a prime, so Euclid would only have known the first four Perfect numbers. Now you can say you know something that Euclid didn't!

## Saturday, August 12, 2017

### Minimum Percentage

75% of men from a certain group are tall, 75% have brown hair, and 75% have brown eyes. What is the minimum percentage that are tall, have brown hair,

*and*have brown eyes? Click below to see the answer.

Instead of thinking in percentages to solve this problem, it's helpful to think back to the Pigeonhole Principle. Think of a group of 100 men, then 75 are tall, 75 have brown hair, and 75 have brown eyes. That's 225 individual attributes to assign to 100 men, so at least 25 of them (or

**25%**) must have each of the three attributes.## Saturday, August 5, 2017

### A Two-Digit Number

Find a two-digit number that's equal to two times the result of multiplying its digits. Click below to see the answer.

My first attempt at solving this puzzle was to set it up as an equation and try to solve it algebraically. Let's say the two digits are $x$ and $y$. Then the equation would be:

$10x + y = 2xy$

The left-hand side is the two-digit number ($x$ in the tens place, $y$ in the ones place) and the right-hand side is two times the result of multiplying its digits. If you try to isolate either $x$ or $y$, you'll see that it's not very easy to come up with a clean solution. That's because the equation above describes a hyperbola.

That's not exactly a dead end, but it isn't the kind of easy-to-understand (once you see it) solution I like in a logic puzzle. Luckily, there's an easier way. There aren't that many possibilities (we're only dealing with two digits), and we can eliminate a lot of them.

For example, we know that neither digit is 0. Also, we know that $2xy$ is an even number, so $y$ must be even (because the result of adding it to an even number is even). We also know that the product of the digits must be less than 50, otherwise $2xy$ would have three digits. That gets us down to only 32 possibilities to test.

Any other shortcuts that I can think of would only eliminate a few possibilities, but it's easy to just test the remaining ones (I went through them manually, but you could write a short script or use a spreadsheet), and find that the solution is

$36 = 2 * 3 * 6$

$10x + y = 2xy$

The left-hand side is the two-digit number ($x$ in the tens place, $y$ in the ones place) and the right-hand side is two times the result of multiplying its digits. If you try to isolate either $x$ or $y$, you'll see that it's not very easy to come up with a clean solution. That's because the equation above describes a hyperbola.

That's not exactly a dead end, but it isn't the kind of easy-to-understand (once you see it) solution I like in a logic puzzle. Luckily, there's an easier way. There aren't that many possibilities (we're only dealing with two digits), and we can eliminate a lot of them.

For example, we know that neither digit is 0. Also, we know that $2xy$ is an even number, so $y$ must be even (because the result of adding it to an even number is even). We also know that the product of the digits must be less than 50, otherwise $2xy$ would have three digits. That gets us down to only 32 possibilities to test.

Any other shortcuts that I can think of would only eliminate a few possibilities, but it's easy to just test the remaining ones (I went through them manually, but you could write a short script or use a spreadsheet), and find that the solution is

$36 = 2 * 3 * 6$

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