For those that are enjoying the math problems..

[StevenSurprenant]

I copied this as is...

Your task, should you decide to do this, is to explain why this works.
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UNBELIEVABLE MATH PROBLEM

Here is a math trick so unbelievable that it will stump you.
Personally I would like to know who came up with this and why that person
is not running the country.
Grab a calculator. (you won't be able to do this one in your Head)

1. Key in the first three digits of your phone number (NOT the Area code...)
2. Multiply by 80
3. Add 1
4. Multiply by 250
5. Add to this the last 4 digits of your phone number
6. Add to this the last 4 digits of your phone number again.
7. Subtract 250
8. Divide number by 2

Do you recognize the answer ??

[Hyfi]

pretty cool

[ForeverAutumn]

Very cool. I tried it with different phone numbers and it worked each time.

[StevenSurprenant]

Very cool. I tried it with different phone numbers and it worked each time.

If you want, figure out why it works. It's fun if you enjoy that kind of thing.

[ForeverAutumn]

I do enjoy that kind of thing. I'm actually taking a college math course right now. It's part of the business/marketing diploma that I'm doing part-time. I barely have time to get my required homework done however, so I'm not sure that I'll have time to figure this out just for fun...at least not right now.

[noddin0ff]

Cute. I was obsessively compelled to break it apart. So...

When you multiply by 80 and then by 250, it's the same as multiplying by

(2 x 2 x 2 x 2 x 5) = 80

(2 x 5 x 5 x 5) = 250

if you regroup these 2 x (2x5) x (2x5) x (2x5) x (2x5) = 80x250
its the same as 2 x 10000.

lets call the first three digits of your phone # "ABC"
2 x 10000 x (ABC) = (2) x A,BC0,000

But, we had added a 1 that also got multiplied by 250. so that 'adds' 250 to this. So the sum at this point is

(2) x A,BC0,000 + 250 = (2) x A,BC0,250

But then all you do is add and subtract for steps 5, 6 and 7 and since it doesn't matter what order you add and subtract stuff, lets do step 7 first and subtract 250.

Now you still have
(2) x A,BC0,000

Let's call your last 4 digits "DEFG"
You add them twice, which is the same as adding (2) x DEFG so we can write this out as

[(2) x A,BC0,000] + [(2) x D,EFG]

equals

(2) x (A,BC0,000 + D,EFG)
(2) x (A,BCD,EFG)

Step 8 divide by 2

and you have A,BCD,EFG

neat :-)

[StevenSurprenant]

Cute. I was obsessively compelled to break it apart. So...

neat :-)

Perfect! Your explanation was very good. There's no need for me to explain, but I wrote it a little different, so I will.

Great job!
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Doing the math using a sample phone number* 932-2563* ...

1. Key in the first three digits of your phone number (NOT*the Area code...)
932

2. Multiply by 80
80(932)

3. Add 1
80(932)+1

4. Multiply by 250
250(80(932)+1) = 20000(932)+250

factoring 10000 out of the 20000(932) we get 2(9320000)* which shows how the first 3 digits were moved to the left to make room for the last 4 digits in step 5 and 6.
so now we have...
2(9320000)+250

5. and 6 adds the last 4 digits (2563) of your phone number twice.
2(9320000)+250 +2(2563)

7. Subtract 250
2(9320000)+250 +2(2563) - 250 = 2(9320000) +2(2563)

8. Divide number by 2
(2(9320000) +2(2563))/2* = 9320000 + 2563 = 9322563 (original phone number)
****************

Basically, he just moved the first 3 digits of the telephone number 4 decimal places to the left and added the last 4 digits.* Plus he added in a lot of unnecessary math to try and confuse.
*
You can see how the numbers from the phone number stays intact during each step and his extra math keeps changing around the phone numbers.