little math trick for classroom setting

Ok This is a good one for the classroom. Write on the board 1011010 and tell the students/audience that you need to somehow get nine-fifty out of those numbers and you are allowed to moved one of the numbers, only once, but you can move them anywhere (turn them sideways and upside down if you want) Ok, now once nobody had gotten it which is most likely; take the second 1 and turn it sideways to make the letter TO so it will now be 10t010 (in other words- nine-fifty is the same as ten to ten)

Applying a Fibonacci series

Here's one that uses a sort of Fibonacci series to determine the result of a sum rather quickly. The setup and explanation seems quite long, but it is easy to accomplish. I didn't want to spare any detail. Try it.

1. On a worksheet (blackboard) put a column of numbers from 1 to 10 with a dash after each number. This has a dual purpose. It makes sure we have exactly **10 numbers to add**, and that we can pick out the **7th number** at a glance.

2. Have someone choose a number between 5 and 15 (actually, you could have them choose any number at all, any size and the trick will work). Jot that number after **1-**

3. Have another person choose another number between 10 and 20 (Here again, you could have them choose any number they want). Jot that number after **2-**

4. Have either person add these two numbers together to get the **3-** number and so on, adding each sum to the previous sum to arrive at the next sum just like the **Fibonacci sum**. Stop at the **10-** number.

5. Immediately produce the sum and say **"Ta da!"**

**How it's done**

Each term in a Fibonacci series containing 10 numbers is a fraction of the total. It happens that the 7th number is always precisely **1/11** of the final total. No other number in the series provides a consistent result. So you just **multiply the 7th number by 11.**

**How do you multiply a number by 11 mentally?**

In this example, easy. But if you use larger numbers to start with, maybe a little tougher.

If you use the restriction placed upon choosing the initial 2 numbers, then the 7th number will be a 3-digit number. Here' s how you multiply a 3-digit number by 11 mentally. I know most of your probably already know how, but indulge me. If you don't know how, you might look here: Math shortcut trick 11 times tables or study these examples.

Suppose the 7th number was **227**.

1) First, bring down the **7** as the ones digit

2) Second, **add the 2nd and 3rd digit** to get **9** as your **tens digit**.

3) Third, **add the 1st and 2nd digit** to get **4** as your **hundreds digit.**

4) Fourth, bring down the 1st digit **2** to finish.

5) Result: **2,497**

Sometimes you have to carry digits as you would in normal addition. For example. Suppost the 7th number was **168.**

1) First, bring down the **8** as the ones digit

2) Second, **add the 2nd and 3rd digit** to get **14,** but just place the **4** as your **tens digit** and carry the **1**.

3) Third, **add the 1st and 2nd digit** to get **7**, and then add the **1** that you are carrying to get **8** as your **hundreds digit**.

4) Fourth, bring down the 1st digit **1** to finish.

5) Result: **1,848**

__Here's a complete setup and finish:__

1- 15 <===== first number chosen

**2-** 10 <===== second number chosen

**3-** 25 <===== Fibonacci sums continue to 10 numbers

**4-** 35

**5-** 60

**6-** 95

**7-** 155 <===== this is the number you multiply by 11 to get final total

**8-** 250

**9-** 405

**10- **655

---------

**1,705** <=====**155 X 11**

Bring down **5**

5+5=10, jot down **0** and carry 1

1+5=6 and carry 1 to get **7**

Bring down **1**