# Puzzles from Undergraduate Newsletters

**Puzzle from Fall 2012 Newsletter**

**Puzzle from Spring 2012 Newsletter**

**Puzzle from Fall 2011 Newsletter**

7638

**Puzzle from Spring 2011 Newsletter**

**Riddle #1 from Fall 2010 Newsletter**

What number comes next in the sequence: 61, 691, 163, 487, 4201, ?

Solution:

9631. The sequence consists of the prime numbers which, when their digits are reversed, are perfect squares.

**Riddle #1 from Spring 2010 Newsletter**

Bob and John form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years. How old are Bob and John?

Solution:

Let us call the current age of Bob B and of John J.

The first fact is that Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages.

When Bob was half as old as the sum of their current ages, he had reached the age of (B + J) / 2 years. This is now B – (B + J) / 2 = B/2 – J/2 years ago. At that moment, John was J – (B/2 – J/2) = 3/2 J – B/2 years old. If Bob is twice as old as John at that moment, then Bob is (3/2 J – B/2) * 2 = 3 J – B years old. That moment is 3 J – 2 B years from now. Then John will be J + 3 J – 2 B = 4 J – 2 B years old. And that is exactly the age of Bob, so B = 4 J – 2 B which gives that 3 B = 4 J.

The second fact is that John is as old as Bob was when John was half as old as he will become over ten years.

When John had half the age as he will have over ten years, he was (J+10)/2 years old. This is now J – (J + 10) / 2 = J/2 – 5 years ago. At that moment Bob was B – (J/2 – 5) years old. According to the second fact, John is now as old as Bob was at that moment, so J = B – (J/2 – 5). It now follows that 3 J / 2 = B + 5.

Summarizing, we end up with two equations.

3 B = 4 J

B + 5 = 3 J / 2

Multiplying the bottom equation with -3 gives after adding the top equation -15 = (4 – 9/2) J

Solving this equation gives J = 30. And now it easily follows that B = 40.

So John is 30 years old and Bob 40.

**Riddle #1 from Fall 2009 Newsletter**

What is the integral of “one over cabin” with respect to “cabin”? http://www.onlinemathlearning.com/calculus-riddles.html

Answer:

Natural log cabin + c = houseboat.

Sudoku Solution (pdf)

**Riddle #1 from Spring 2009 Newsletter**

A man was found murdered on Sunday morning. His wife immediately called the police. The police questioned the wife and staff and got these alibis:

The Wife said she was sleeping.

The Cook was cooking breakfast.

The Gardener was picking vegetables.

The Maid was getting the mail.

The Butler was cleaning the closet.

The police instantly arrested the murdered. Who did it and how did they know?

Answer:

It was the Maid. She said she was getting the mail. There is no mail on

Sunday! (e-mail does not count)

**Riddle #1 from Spring 2008 Newsletter**

Driving along, Terry notices that the last four digits on the odometer are palindromic. A mile later, the last five digits are palindromic. Two miles later, all six are palindromic. What was the odometer reading when Terry first looked at it?

Answer:

1-9-8, 8-8-8. That’s what he first saw when he looked.

One mile later it was 1-9-8, 8-8-9, the last five are palindromic.

One mile later, 1-9-8, 8-9-0, the middle four are palindromic. You

drive one more mile and it’s 1-9-8-, 8-9-1, all six are palindromic.

**Riddle #1 from Fall 2007 Newsletter**

A boat has a ladder that has six rungs, each rung is one foot apart. The bottom rung is one foot from the water. The tide rises at 12 inches every 15 minutes. High tide peaks in one hour. When the tide is at it’s highest, how many rungs are under water?

Answer: No rungs will be under water; the boat floats.

**Riddle # 2 from Fall 2007 Newsletter**

There are 4 men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man.

Man 1: 1 minute to cross

Man 2: 2 minutes to cross

Man 3: 5 minutes to cross

Man 4: 10 minutes to cross

Answer:

Man 1 and Man 2 cross 2 min

Man 2 comes back 2min

Man 3 and Man 4 cross 10 min

Man 1 comes back 1 min

Man 1 and Man 2 cross 2 min

Total 17 min