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May 7, 2011


Filed under: Uncategorized — aspensmonster @ 5:10 am

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Just stumbled across this little gem. Some sort of “favorite movie test” game:


And now for the mathematics behind it. Consider your number to be ‘Z’

Z = n

And now let us multiply by 3

Z =  3n

and then add 3

Z = 3n + 3 = 3\left(n+1\right)

and then multiply by three again

Z = 9\left(n+1\right)

So no we have an expression for whatever your number is, given

1 \le n \le 9

Now the test wants us to add the digits of this number to get a resultant which is supposed to be our favorite movie of all time (OF ALL TIME!). Let us consider the digits of Z to be a and b, wherein a is the tens and b is the ones. Consequently

Z = 10a + b

is a true statement, given the nature of the decimal system. The question becomes, under what conditions will equality hold? That is, when will

Z = 10a + b = 9\left(n+1\right)

all equal each other? One possibility is if the digits behave as so: a=n and b=9-n . In this case, we have

Z = 10a + b = 10n + \left(9-n\right) = 9n + 9 = 9\left(n+1\right)

Given this constraint on the nature of the digits, we can conclude that the sum of these digits would always be

a + b = n + 9 - n = 9

So, if 1 \le n \le 9, and if a=n, and if b=9-n, then the sum of the digits will always be nine, and you will be forever doomed to having The Joy of Anal Sex With a Goat be your favorite movie of all time, OF ALL TIME!

I’m sure there’s a way to use modular arithmetic to better “prove” this problem without resorting to my cheap shot of “well, if a and b act like this, then POOF!”, but I don’t know modular arithmetic. Maybe I’ll update this later.

UPDATE: Yeppers. Theorem stating that any number divisible by 9 must also have a digit sum that is divisible by 9. If digit sum is divisible by nine, then the number is divisible by nine (iff statement. goes both ways. same goes for 3). I’ll update later with a better exposition.

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