**Document ID**: 07_03_03_1

**Date Received**: 2003-07-03 **Date Revised**: 2003-08-27 **Date
Accepted**: 2003-09-04

**Curriculum Topic Benchmarks**: M2.3.11, M8.3.1, M8.3.12

**Grade Level**: Middle School \[6-8\]

**Subject Keywords**: number puzzles, algebra, mathematical tricks

**Rating**: Moderate

**Algebra Magic -- Create Your Own Number Puzzles**

**By**: Ramakrishnan Menon, California State University Los Angeles,
5151 State University Drive, Los Angeles, CA 90032 e-mail:
rmenon@calstatela.edu

**From**: The PUMAS Collection http://pumas.jpl.nasa.gov

©2003, California Institute of Technology. ALL RIGHTS RESERVED. Based on
U.S. Gov\'t sponsored research

Many people think of algebra as dealing with solving equations that have
letters and numbers, and that it is completely unrelated to anything
outside the math classroom. Also, teachers have a difficult time
motivating students to learn algebra. I have found that not only are
number/magic puzzles quite common outside the classroom, but also that
they motivate students, and that teachers can successfully use such
number puzzles as an interesting lead-in to algebra. Here I discuss two
number/magic puzzles that can be demystified and explained by using
algebra.

### Number Puzzle One

An old number trick has directions as follows (the numbers/letters in
parentheses are algebraic equivalents of the steps):

Step 1: Write down the year you were born. \[p\]

Step 2: Write down a year when a memorable event in your life took
place. \[q\]

Step 3: Write down your age as of the last day of this year. \[y -- p\]

Step 4: Write down the number of years that elapsed since the memorable
event (in Step 2) took place. \[y -- q\]

Step 5: Add the 4 numbers obtained from Steps 1 through 4. \[p + q + (y
-- p) + (y -- q)=2y\]

***Prediction***: Everyone who followed all the directions would have
got the number 4006 (if the trick were carried out in the year 2003).

The prediction sounds magical, especially as everyone gets the same
answer, even though all might have started off with different numbers.

So how does this work?

***Explanation***: Look at the variables that were added in Step 5. They
are:

p + q + (y -- p) + (y -- q) = p + q + y -- p + y -- q, by removing
parentheses

= p -- p + q -- q + y + y, by rearranging the terms

= 2 y, as the p's and q's cancel each other out

( = 4006, when y = 2003)

In other words, the number in Step 1 combined with that in Step 3,
*deleted* the number in Step 1, giving the current year, 2003;
similarly, the number in Step 2 combined with that in Step 4, *deleted*
the number in Step 2, giving, once again, the current year, 2003. Hence,
adding all the 4 numbers from Step 1 through Step 4 effectively deleted
every number other than twice the current year, 2003, giving the result
of 2 x 2003 = 4006! (Note that most number puzzles undo what has been
done, so as to "force" a certain result, as can be easily proved by
using deductive logic or by using algebra.)

#### Number Puzzle Two

This number puzzle gives the age of the person, and the number of
siblings in the person's family (sample numbers and algebraic
equivalents are given next to each step/direction).

> Directions Sample #s Algebraic equivalent

1.  Think of your age in years 15 \[10x+y\]

```{=html}
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```
1.  Multiply your age by 2 30 2\[10x+y\]

2.  Add 10 40 2\[10x+y\] + 10

3.  Multiply by 5 200 5{2\[10x+y\]+10}=100x+10y+50

4.  Add \# of siblings you have 203 \[100x+10y+50+z\]

5.  Subtract 50 153 \[100x+10y+50+z\]--50 = 100x+10y+z

***Predication***: You are 15 years old, and you have 3 siblings.

***Explanation***: The first two digits on the left (15) give your age,
and the rightmost digit (3) gives the number of siblings you have.

The algebraic equivalent of the steps result in Step 6, which is 100x +
10y + z, where z\<10. But 100x + 10y + z is the algebraic equivalent of
a three-digit number, with the digits in the hundreds and tens places
giving the age, and the digit in the ones place giving the number of
persons in the house. (Note: the number of siblings has to be less than
10, otherwise this "trick" will not work. Why?)

***Activity***. Now, make up your own number puzzle, justify it
algebraically, and test it out with your friends. Be careful about any
cases where a number is "carried" from one decimal place to another.
Puzzle Two avoided the "carrying" by restricting the number of siblings
to less than 10.