Math Problem

[El Sid]

I think this is the right place to post this. This morning I recieved the following problem via. Email. See if you can work it out.

Choose a number between 1 and 9 (both inclusive)
Multiply with 2
Add 5
Multiply with 50
If you have already had your birthday this year add 1753. If not, add 1752.
Deduct your year of birth. eg 1962

The 3 digit number that remains shows the following:
The first digit is the number you chose and the remaining others indicate your age.

I'm trying but have yet (if at all) to crack it!

[Henrik]

ElSid,

It is actually not very complicated. The figure 1753 or 1752 is what will determine if the year is 2003 or 2002. Then by subtracting your year of birth the result is obviously your age. Add 250 to 1753 and you get 2003.

The first part will always give you a figure of 350 or up by incriments of 100 (350, 450 etc.) This then helps in giving the final answer of the first digit corresponding to the original number chosen.

[El Sid]

Henrik,

Yes, I get that.

Actually it is just a nifty manipulation of numbers, if anything. The same problem won't work, say, for next year. The 1753 or 1752 will increment by 1 to 1754 and 1753.

eg. Go for a choice of n=2 (this is for 2003)

((2 x 2) + 5)*50 = 450
Say y.o.b = 1971 (birth date not reached yet)

Then: 450 + x - 1971 = 231
x = 231 + 1971 - 450
= 1752 (or 1753 if birth date reached)

So for next year 2004, at the corresponding time:
450 + x - 1971 = 232
x = 232 + 1971 - 450
= 1753 (or 1754 if birth date reached)

Go for a choice of n=4 (this is for 2004)

Then: 650 + x - 1971 = 432
x = 432 + 1971 - 650
= 1753 (or 1754 if birth date reached)

Oh, well, did'nt really earn me a beer but what the heck.

[Henrik]

You got that right!

And I'll have me a cold one too...