# Fun with MATH...

### #1

Posted 26 May 2004 - 03:24 PM

1. Grab a calculator. (you won't be able to do this one in your head)

2. Key in the first three digits of your phone (NOT the area code)

3. Multiply by 80

4. Add 1

5. Multiply by 250

6. Add the last 4 digits of your phone ~~~~~~

7. Add the last 4 digits of your phone ~~~~~~ again.

8. Subtract 250

9. Divide by 2

Recognize the number?

### #2

Posted 26 May 2004 - 04:11 PM

### #3

Posted 26 May 2004 - 10:20 PM

so we have FUN in Mathematics too?

**Take my hand and we'll go riding through the sunshine from above**

### #4

Posted 27 May 2004 - 03:59 AM

Second, it doesn't work for brittish numbers we only have six digits in our land lines without the area code or 11 digits in our mobile phones.

The point which I should first wish to understand is whether the pious or holy is beloved by the gods because it is holy, or holy because it is beloved of the gods.

Sonnet XCIVBut if that flower with base infection meet,The basest weed outbraves his dignity:For sweetest things turn sourest by their deeds;Lilies that fester smell far worse than weeds### #5

Posted 27 May 2004 - 04:57 AM

Chocolate Math

1. First of all, pick the number of times a week that you >> would like to have chocolate. (try for more than once but >> less than 10) >> >> >> >> >>

2.. Multiply this number by 2 (Just to be bold) >> >> >> >>

3. Add 5. (for Sunday) >>

4. Multiply it by 50 I'll wait while you get the calculator................. >> >> >> >> >>

5. If you have already had your birthday this year >> add 1754.... If you haven't, add 1753 ......

6. Now subtract the four digit year that you were born.

>> You should have a three digit number. >> >> >> >> >> The first digit of this was your original number (i.e., how many >>times >> you want to have chocolate each week). >> >> >> The next two numbers are ........ >> >> YOUR AGE! ( Oh YES, it IS!!!!! ) >> >> THIS IS THE ONLY YEAR IT WILL EVER WORK, SO >> SPREAD IT AROUND WHILE IT LASTS. ;D ;D

### #6

Posted 27 May 2004 - 08:13 AM

The choclate thing didn;t work for me. Maybe I don't want enough chocolate

### #7

Posted 27 May 2004 - 01:51 PM

Ok, a buddy sent this too me and it's just way too fun [smiley=cwm35.gif].

1. Grab a calculator. (you won't be able to do this one in your head)

2. Key in the first three digits of your phone (NOT the area code)

3. Multiply by 80

4. Add 1

5. Multiply by 250

6. Add the last 4 digits of your phone ~~~~~~

7. Add the last 4 digits of your phone ~~~~~~ again.

8. Subtract 250

9. Divide by 2

Recognize the number?

The minute I saw this, I knew I wouldn't be able to sleep until I proved it true. So here it is, in case anyone's interested:

a) Let the telephone number be XYZABCD; breaking it into its component parts, XYZABCD=1,000,000X + 100,000Y + 10,000Z + 1,000A + 100B + 10C + 1D

We now break this number into two parts: the first three numbers and the last four numbers:

XYZ = 100X + 10Y + Z

c) ABCD = 1000A + 100B +10C + 1D

So following the steps provided:

3) 8000X + 800Y + 80Z

4) 8000X + 800Y +80Z + 1

5) 2,000,000X + 200,000Y + 20,000Z + 250

d) Addition is associative (or Commutative? I always confuse the two). This means that E + F + G = G + E + F, and so forth. Since steps 6, 7, and 8 are all addition, we may do step 8 first; this will prove easier.

8) 2,000,000X + 200,000Y + 20,000Z

e) Adding a number twice is the same as adding 2 times that number one. We therefore combine steps 6 and 7.

6/7) 2,000,000X + 200,000Y + 20,000Z + 2(1,000A + 100B + 10C +D)

9) 1,000,000X + 100,000Y + 10,000Z + 1000A + 100B + 10C +D = XYZABCD

Q.E.D.

You were right; that was fun. I haven't been able to prove the chocolate one yet, though.

Edited: Yes, I have

Let x be and arbitrary number from 1-10, exclusive

Let yr be the year you were born

Let age be your age

We will demonstrate that this is true if you haven't had your birthday yet; proving it true if you have follows the same formula

1) x

2) 2x

3) 2x + 5

4) 100x + 250

5) 100x + 250 + 1753

It is demonstratably true that if you haven't had your birthday yet, 2004 - yr = age + 1

thus age = 2003 - yr

6) 100x + 250 + 1753 - yr

100x + 2003 - yr

100x + age

Thus x will be in the hundreds place, and since

adding a number to a whole number between 100 and 900 that is evenly divisible by hundred merely involved replacing the last two zeroes with the number your adding, the last two digits will be age. I don't think this will work if you're over 100, though.

### #8

Posted 27 May 2004 - 04:21 PM

### #9

Posted 27 May 2004 - 07:03 PM

### #10

Posted 27 May 2004 - 07:46 PM

*lol* are you serious the most fun you had all day :'( j/kAre you kidding? Proving those was the most fun I had all day! Very basic, but I always get a rush out of that sort of thing.

### #11

Posted 28 May 2004 - 08:08 AM

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users