Reason out the logic behind this chocolate Math!!!
It tells your Age. :D

1) Pick the number of times you would like to have chocolate in a week(It should be more than 1 and less than 10).
2) Multiply the number by 2(Just to be bold)
3) Add 5 to it.
4) Multiply the result by 50 (I will wait... as you launch that calculator)
6) If you have already had your bithday this year, Add 1755 and if not Add 1754.
7) Now subtract the 4 digit year that you were born.
.....You should be having a three digit number.
The first digit was your orignal number(how many times you would like chocolate in a week.)
The Next Two Numbers are Your AGE!!!(Oh YES, It is!!!)

Now Reason out.

yeah nice mathematical equations!thanks that's cool...

it's all about math! and it's fun doing it!

Probably the best way to see why this works is to avoid committing to
a specific number and see what happens. Instead of picking a number,
I'm going to use 'n' to represent some choice I haven't yet made.

Let's see what happens as I follow the steps:

1) n

2) 2n

3) 2n + 5

4) 100n + 250

5) 100n + (2001 or 2002)

6) 100n + (2001 or 2002) - (year of birth)

Now, let's look at what I ended up with. If we write it this way,

100n + (2001 or 2002) - (year of birth)

= 100n + [(year of my last birthday) - (year of my birth)]

= 100n + [my age]

This might be a good time to note that if you start with a number
greater than 9, or if you're more than 99 years old, the trick won't
work.

So basically, the trick is saying that

50(2n + 5) + 1750 + i - (year of birth)

where i (for 'increment') is the number of years past 2000 for the
year when you had your last birthday, equal to

100n + (your age)

which is, of course, true (for 0 < n < 10, and age < 100):

50(2n + 5) + 1750 + i - (yob)

= (100n + 250) + 1750 + i - (yob)

= 100n + (250 + 1750 + i) - yob

= 100n + (2000 + i) - (yob)

= 100n + (your age)

It _is_ true that with these specific numbers, the trick will work
only this year. However, if you want it to work next year, you can
just change i to '2 or 3' instead of '1 or 2'.

Some other variations on the problem would be

25(4n + 10) + 1750 + i - (year of birth)

20(5n + 10) + 1800 + i - (year of birth)

4(25n + 127) + 1492 + i - (year of birth)

4(25n + 224) + 1776 + i - (year of birth)

In general, you have

a(bn + c) + d + i - (year of birth)

and the constraints you have to meet are

1) ab = 100

2) ac + d = 2000

Let's say I want to make d = 1908, the last year the Chicago Cubs won
the World Series. Then

ac + 1908 = 2000

ac = 2000 - 1908

ac = 92

= 2 * 2 * 23

The possible values of a and c are:

a = 1, c = 92
a = 2, c = 46
a = 4, c = 23
a = 23, c = 4
a = 96, c = 1

But 2 and 4 are the only values of a that divide 100 evenly, so I can
have either

2(50n + 46) + 1908 + i - (year of birth)

or

4(25n + 23) + 1908 + i - (year of birth)


Got it from here (http://mathforum.org/library/drmath/view/61702.html)