Dreamtime Posted April 21, 2008 Share Posted April 21, 2008 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 3. Add 5 4. Multiply it by 50 5. If you have already had your birthday this year add 1758 If you haven't, add 1757. 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!) THIS IS THE ONLY YEAR (2008) IT WILL EVER WORK Link to comment Share on other sites More sharing options...

gretsch Posted April 21, 2008 Share Posted April 21, 2008 That is spooky. Link to comment Share on other sites More sharing options...

vipernut Posted April 21, 2008 Share Posted April 21, 2008 Thats pretty good,always like things like that. Stephen Link to comment Share on other sites More sharing options...

jewgaffer Posted April 22, 2008 Share Posted April 22, 2008 (edited) That's very interesting Marina 31. Works out correct applying pre calculated common denominators to 2008. Edit. To find those common denominators, allowing for the variables in birth years for ages between 1 year olds and 100 year olds plus eat out variables of more than once (2 x) and less than 10 (9 x) add to the variables and require complicated mathematical equations and formulas. Equations are needed to precalculate hidden denominators to find the two successive numbers common to the year 2008. Clever maths. Cheers jewgaffer Edited April 22, 2008 by jewgaffer Link to comment Share on other sites More sharing options...

Josh88 Posted April 22, 2008 Share Posted April 22, 2008 That's very interesting Marina 31. Works out correct applying pre calculated common denominators to 2008. Edit. To find those common denominators, allowing for the variables in birth years for ages between 1 year olds and 100 year olds plus eat out variables of more than once (2 x) and less than 10 (9 x) add to the variables and require complicated mathematical equations and formulas. Equations are needed to precalculate hidden denominators to find the two successive numbers common to the year 2008. Clever maths. Cheers jewgaffer ......what he said Pretty amazing these things work like that Link to comment Share on other sites More sharing options...

jewgaffer Posted April 23, 2008 Share Posted April 23, 2008 1. First of all, pick the number of times a week that you would like to go out to eat........... ...........The next two numbers are YOUR AGE ! ------ (Oh YES, it is!) THIS IS THE ONLY YEAR (2008) IT WILL EVER WORK Hi Mariner31. Looking at the merits and the complications to present his work as a puzzle, finding the hidden numbers used to make the calculation for the successive numbers given by the mathematician alone needs much more than a calculator. A formula has been applied by the mathematician to allow for birthdays in both parts of the year 2008 so that there won't be a discrepency of one year in the answer, the person's age. So the only numbers that can be used must be successive and are based on hidden pre calculated numbers using a formula to establish them and naturally those "hidden numbers" are not revealed to us. They are saying that it will only work for the year 2008 yet mathematics is an exact science. Cheers jewgaffer Link to comment Share on other sites More sharing options...

Dreamtime Posted April 25, 2008 Author Share Posted April 25, 2008 So How Jewgaffer, Have you got the answer or is it all gobbltygook? Link to comment Share on other sites More sharing options...

jewgaffer Posted April 25, 2008 Share Posted April 25, 2008 Hi Mariner31. It is not all gobbltygook as you say, the topic is an interesting example of a complex maths problem when you look beyond the surface. The problem has been set up perfectly. It's not a riddle or a play on words. The answer works out correct for the persons age. I was interested in how it was presented as a puzzle and how they pre-calculated and arrived at the figures which they required interested readers to use, and why it could only apply to 2008 as they stated. That's quite correct. What they mean is that you can't use those figures they gave for any other year except for 2008. I always treat complicated mathematics in an inquisite manner and I was surprised by the maths work needed to arrive at the figures i.e. the 1757 and the 1758 which were set as the two datum points for the puzzle, know what I mean. Obviously they didn't just pull the figures 1757 and 1758 out of a revolving drum with the thousands of other combinations in it. Cheers jewgaffer Link to comment Share on other sites More sharing options...

cupster Posted April 25, 2008 Share Posted April 25, 2008 (edited) THIS IS THE ONLY YEAR (2008) IT WILL EVER WORK it will work in 2009 as well,all you have to do is change 1758-1757 to 1759-1758 {and so on }...maths is a constant,but your age isn't.. or did jewgaffer already explain that??. Edited April 25, 2008 by hooked4life Link to comment Share on other sites More sharing options...

Hotzy Posted April 25, 2008 Share Posted April 25, 2008 Now I have a headache jewgaffer. I think I need a beer Link to comment Share on other sites More sharing options...

Dreamtime Posted April 25, 2008 Author Share Posted April 25, 2008 it will work in 2009 as well,all you have to do is change 1758-1757 to 1759-1758 {and so on }...maths is a constant,but your age isn't.. or did jewgaffer already explain that??. Mate, I know what year I was born in but still get it wrong sometimes. I have every respect for people that have this high level of intelligence. It's still got me buggered though? Will it work every year just by changing the figures as you have quoted? Link to comment Share on other sites More sharing options...

cupster Posted April 26, 2008 Share Posted April 26, 2008 (edited) Will it work every year just by changing the figures as you have quoted? so long as the end sum remains 3 digits long it will,but at some point as you keep adding to the figure the end total will be at least 4 digits long..ie 1000 or more.since you can't start with a number higher than 9 then the first two digits won't match the original number you thought of..[won't matter,we'll be long gone by then].but the last two numbers should [i think!!] still corrispond with whatever the last two digits of your age happen to be in that year...so yeah,it should be good for at least a few more 100 years. edit...wait,i'm starting to have second thought's on this.. .,no,i think i'm right....getting a headache...time to bail!!! Edited April 27, 2008 by hooked4life Link to comment Share on other sites More sharing options...

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