The following was posted on an online horse forum I regularly read. Then someone asked, "How does this really work?" I've seen this sort of trick before, and have never taken the time to work through the math. Here's the problem, followed my a simple explanation of how it works. Let me know if it doesn't make sense.
YOUR AGE BY RIDING
Don't tell me your age; you probably would tell a falsehood anyway... but, your instructor may know!!
YOUR AGE BY RIDING MATH
DON'T CHEAT BY SCROLLING DOWN FIRST!
It takes less than a minute. Work this out as you read. Be sure you don't read the bottom until you've worked it out!
This is not one of those waste of time things, it's fun.
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1. First of all, pick the number of times a week that you would like to ride your horse.
(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
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 ride in a week.)
The next two numbers are...
YOUR AGE ! ------ (Oh YES, it is!)
THIS IS THE ONLY YEAR (2008.) IT WILL EVER WORK, SO SPREAD IT AROUND WHILE IT LASTS.
Okay. Now that you've done it, and are thoroughly impressed by this little "magical" trick, here's how it works. (And also why, with a slight
modfication, it will work perfectly well in 2009, and the year after that, and the year after that.)
First, let's establish a few variables to make the problem easier to look at:
Current Year: 2008
Let your age = x
Let the year of your birth = y
Naturally, from these first two statements we can already see that if you have already had your birthday this year, 2008 - y will equal your age.
If you have not had your birthday so far this year, 2008 - y - 1 will equal your age.
Now, let the number you want to ride in a week (a digit at least 1 but less than 10) = n.
Here's what essentially happens during the problem:
Steps 1-3
2n +5
Step 4: Multiply by 50
50 (2n + 5)
Step 5: Add 1758 (For the purposes of the explanation, we'll assume you've had your birthday. The other case, i.e., haven't had birthday yet,
is fairly trivial once the first case is understood).
50 (2n + 5) + 1758
Step 6: Subtract the year you were born
50 (2n + 5) + 1758 - y
Now, lets rearrange a few of those terms
50 (2n + 5) = 100n + 250. So the statement now becomes:
100n + 250 + 1758 - y
Add the second and third terms together:
100n + 2008 - y
Earlier, we said that 2008 - y should equal your age. So, the statement can further be re-written as:
100n + x
Conclusions
If you take a number between 1 and 9 and multiply it by 100 (essentially, 100n) the the hundreds digit of the number will be the original number. Or,
as in the problem, the first digit is still your original number. And, as we can see from the statement 100n + x, the second two digits of the number
will be your age. So, if you pick the number 8, and are 16, you'll end up with 100n + x = 100*8 + 16 = 816. If you are 8 and pick the number 3, you'll
end up with 100n + x = 100*3 + 8 = 308.
The number you pick and your age essentially function as 2 independent components of the same number, your picked number forms the hundreds place,
and your age the tens and ones place.
Interestingly, the original problem states that your number must be 1 or greater, but less than 10. This is not true.
If we consider the statement 100n + x, we see that this is not true. The problem will work perfectly fine for zero or any positive number.
However, it won't work if you are more than 99 years of age. (And I do know a man who was still riding his Arabians well into his 100s!) However, you
must pick at least 0, the problem doesn't work for negative numbers. See below for examples of some of these cases. Most likely, the problem claims
that you must pick a number greater than 1 because if you pick the number zero, it's a lot easier to see how the trick works. Also, the game likely
tells you to pick a number less than 10 merely for the convience that the problem becomes a lot more fun if you're not trying to multiply and add
huge numbers.
Also, this will work perfectly fine next year and the year after. However, next year, you'll need to use 1759 and 1758 instead (and just keep adding 1
for each additional year). This is because the "magic number" of 1758 is really just 250 + 1758 = 2008. Add 1 to both sides of the equation, and you get
2009 and 1759.
Case 1: Number picked is zero. Problem still works. (For each of these cases, I'll use my age, 23, and my year of birth, 1985. I've also already
had my birthday this year.)
Step 1: 0
Step 2: 0*2 = 0
Step 3: 0*2 + 5 = 5
Step 4: 50 (0*2 + 5) = 250
Step 5: 50 (0*2 + 5) + 1758 = 2008
Step 6: 50 (0*2 + 5) + 1758 - 1985 = 23
Result: 023
As you can see, the first digit is the original number I picked, and the second and third are my age.
Case 2: Number picked is greater than 10. Let's pick 72. The problem will still work.
Step 1: 72
Step 2: 72*2 = 144
Step 3: 72*2 + 5 = 149
Step 4: 50 (72*2 + 5) = 7450
Step 5: 50 (72*2 + 5) + 1758 = 9208
Step 6: 50 (72*2 + 5) + 1758 - 1985 = 7223
Result: 7223
As you can see, the first two digits are the original number I picked, and the second and third are my age. Brilliant! It still works.
Case 3: Number picked is negative. Let's pick -5. The problem will not work.
At the end of step 6, you'll end up with 100n + x. So:
-5 * 100 + 23 = -500 + 23 = 477.
If n is negative, the problem won't work.
Back to my blog!