Work this one out for me please

Poststructural Pub
1

This is not one of those waste of time things, it’s fun.

  1. First of all, pick the number of times a week that you would like to go out to eat. (more than once but less than 10)

  2. Multiply this number by 2 (just to be bold)

  3. Add 5

  4. Multiply it by 50 (get the calculator out)

  5. If you have already had your birthday this year add 1759…
    Or if you haven’t, add 1758.

  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 go out to restaurants in a week.)

The next two numbers are

YOUR AGE ! ------ (Oh YES, it is!)

Now how does this work! THIS IS THE ONLY YEAR (2009) IT WILL EVER WORK, SO SPREAD IT AROUND WHILE IT LASTS.

2

Well, next year (2010) use 1760 and 1759!
And in 2011 use 1761 and 1760.
Bingo.
You can figure out the algebra yourself… :party:
(hint: 2009-1759=250=550 and 250=100)

3

It’s just cruelty! Sheer cruelty! Why anyone would want to remind me what age I am… :puppyeyes:

4

bit rude :really:

5

It’s only cruel and rude if you are old. :poke:
It still works for a four year old or a 24 year old like me :laughing: :laughing:

6

okay then…

quite rude :really:

7

It’s so lucky for me that I live far far away in a happy place isn’t it? :laughing: :laughing:

8

Let’s say X is the number you get in step 1.
And Y is your birth year

The equation they give us can be expressed thuswise:
(X*2+5)*50+1759-Y

If we multiply the 50 out using the distributive property we get:
X100+250+1759-Y
Note that 250+1759 = 2009. That’s why it only works this year.
X100+2009-Y

So, X*100 gives you your 1-digit number in the hundreds place
and 2009-Y gives your current age as a 2-digit number.

The whole “multiply by 2 then add five then multiply the result
by 50” thing is to obfuscate the simplicity of the equation
X*100+2009-Y

9

Well I did ask for it!!!

10

Sometimes I like to think I was King Tut. I was born in 1333 BC. It didn’t work out right at all.

11

Yeah, it wouldn’t have worked well for this guy, either.
http://en.wikipedia.org/wiki/Henry_Allingham

12

I had to try it of course and,
I hate to say it, of course you are right

(darn) :tantrum:

13

Didn’t work for a 1983 birth year. Weird.

14

Daaaalllle! They’re doing the math thing again! :laughing:

Redwolf

15

He can’t hear you. He’s busy trying to figure out what happens when 2 trains leave the stations at different times.

16

He marries the blonde, because the down train arrives two minutes before the up train.

17

European trains or North America trains?

18

trains, hmm, that reminds me of this:

One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. The narrow path, no more than a foot or two wide, spiralled around the mountain to a glittering temple at the summit. - The monk ascended the path at varying rates of speed, stopping many times along the way to rest and to eat the dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation, he began his journey back along the same path, starting at sunrise and again walking at variable speeds with many pauses along the way. His average speed descending was, of course, greater than his average climbing speed. - Is there a spot along the path the monk will be on both trips at precisely the same time of day?

19

I know the answer to this, and the proof. But I’ll wait for some other comments before I go into detail.