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Friday, October 14, 2005
Mathematics: 7-Up Trick
Number trick sponsored by 7-Up. (Click on the guy in the lower right corner to move to the next step.) I'll post my explanation in the Comments later. (Hint: most number puzzles I've seen revolve around a special property of 9.)
I noticed that if you add up the digits of your subtraction result (and keep adding digits until you get a 1-digit result), you end up with 9. So when you input 3 digits at the end, it just has to subtract the sum of those digits from 9.
But why does the subtraction result always add up to 9?
Let's say our initial 4-digit number is wxyz. That is (w * 1000) + (x * 100) + (y * 10) + (z * 1).
Shuffle those digits and you get, say, ywzx. That's (y * 1000) + (w * 100) + (z * 10) + (x * 1).
1 comment:
I noticed that if you add up the digits of your subtraction result (and keep adding digits until you get a 1-digit result), you end up with 9. So when you input 3 digits at the end, it just has to subtract the sum of those digits from 9.
But why does the subtraction result always add up to 9?
Let's say our initial 4-digit number is wxyz. That is (w * 1000) + (x * 100) + (y * 10) + (z * 1).
Shuffle those digits and you get, say, ywzx. That's (y * 1000) + (w * 100) + (z * 10) + (x * 1).
Subtract the second from the first and you get:
((w * 1000) + (x * 100) + (y * 10) + (z * 1)) - ((y * 1000) + (w * 100) + (z * 10) + (x * 1))
Which is the same as:
(w * 1000) + (x * 100) + (y * 10) + (z * 1)) - (y * 1000) - (w * 100) - (z * 10) - (x * 1)
Or:
(w * 1000) - (w * 100) + (x * 100) - (x * 1) + (y * 10) - (y * 1000) + (z * 1) - (z * 10)
Or:
(w * 900) + (x * 99) - (y * 990) - (z * 9)
9 is common to all 4 terms:
9 * ( (w * 100) + (x * 11) - (y * 110) - (z * 1) )
And as we know from countless number tricks, when you multiply any number by 9, the sum of the resulting digits will always reduce to 9.
Will 9 always be common to the 4 terms involving w, x, y and z?
I'm pretty sure it will - but I haven't worked out a proof yet.
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