But my point is that all these "unbelievable" math "coincidences" usually involve some rather simple math functions with a little bit of juggling to hide what's happening. In this case, a couple of the steps involve adding 1, multiplying by 250, and subtracting 250. Others involve multilying by 2 and then dividing by 2. It's a cute trick, but hardly "unbelievable."
Here's a slightly more complex one:
Pick any number between 10 and 10,000.
Calculate the sum of the digits. (For example: 453 -- 4+5+3 = 12)
Subtract that number from the original number.
Calculate the sum of the digits of your result.
Is the number greater than 10? Then calculate the sum of the digits of THAT number.
Repeat the last step until the result is less than 10.
The result is 9. (No matter what your starting number was. In fact, this works for ANY integer greater than 10.)
You could have just said the sum of the digits of any number minus that number is always divisible by nine, you know. :)
Why is that, anyway?
Let's say we have some number of indeterminate length, but its last five digits are edcba. (That is, the whole number is ...edcba.) That really equals 10,000e +1000d +100c +10b +a. If we subtract the sum of the digits (e+d+c+b+a), we get 9999e +999d +99c +9b (+0), which is clearly divisible by nine. It should be apparent that no matter how large the number is, the pattern will hold, as each term will be divisible by nine.
You may have learned as a child that if the sum of the digits of a number is divisible by nine, then the number is divisible by nine. The paragraph above explains that rule. A number minus the sum of its digits is always divisible by nine, so if the sum of the digits is also divisible by nine, then the whole number must be, as well. (Since the number minus the sum of its digits is also divisible by three -- every number divisible by nine is divisible by three -- then, if the sum of the digits is divisible by three, the number is divisible by three, something else you may have learned in grade school.)
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krys
Holy cow!
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Sudo
As my kids would say..
Kewl!!
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learning
To be free and to be funny my only response is:
JESUS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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GeorgeStGeorge
Well, gee... let's see:
1. is wrong. I did this in my head
3 and 5 together multiply your phone exchange by 20,000 (doubling it and adding four zeros)
5 and 8 together negate 4
6 and 7 together add double your last four phone number digits, so now we have twice your phone number
9 then gives you your phone number
George
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krys
There is no magic in math. I couldn't find the factors like George did, but I was surprised that I got the result in a relatively few moves.
You have mathematically inclined way of thinking George - - you see patterns in things easily. I admire that.
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GeorgeStGeorge
Thanks, Krys.
But my point is that all these "unbelievable" math "coincidences" usually involve some rather simple math functions with a little bit of juggling to hide what's happening. In this case, a couple of the steps involve adding 1, multiplying by 250, and subtracting 250. Others involve multilying by 2 and then dividing by 2. It's a cute trick, but hardly "unbelievable."
Here's a slightly more complex one:
Pick any number between 10 and 10,000.
Calculate the sum of the digits. (For example: 453 -- 4+5+3 = 12)
Subtract that number from the original number.
Calculate the sum of the digits of your result.
Is the number greater than 10? Then calculate the sum of the digits of THAT number.
Repeat the last step until the result is less than 10.
The result is 9. (No matter what your starting number was. In fact, this works for ANY integer greater than 10.)
George
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Raf
You could have just said the sum of the digits of any number minus that number is always divisible by nine, you know. :)
Why is that, anyway?
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topoftheworld
Can I get you guys to do my taxes?
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dmiller
Dang -- It fit's like a hand in a glove!!
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GeorgeStGeorge
Let's say we have some number of indeterminate length, but its last five digits are edcba. (That is, the whole number is ...edcba.) That really equals 10,000e +1000d +100c +10b +a. If we subtract the sum of the digits (e+d+c+b+a), we get 9999e +999d +99c +9b (+0), which is clearly divisible by nine. It should be apparent that no matter how large the number is, the pattern will hold, as each term will be divisible by nine.
You may have learned as a child that if the sum of the digits of a number is divisible by nine, then the number is divisible by nine. The paragraph above explains that rule. A number minus the sum of its digits is always divisible by nine, so if the sum of the digits is also divisible by nine, then the whole number must be, as well. (Since the number minus the sum of its digits is also divisible by three -- every number divisible by nine is divisible by three -- then, if the sum of the digits is divisible by three, the number is divisible by three, something else you may have learned in grade school.)
George
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dmiller
Every time I log on here, I realize how ignorant I am!
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Raf
Here are the tricks I know:
1: Every whole number greater than zero (is that redundant?) is divisible by 1.
2: Look at the last digit. If it's even, the number is divisible by 2.
3: Add the digits. If the sum is divisible by 3, the number is divisible by 3.
4: Look at the last TWO digits. If that number is divisible by 4, the entire number is divisible by 4.
5: Look at the last digit. If it's 5 or 0, the number is divisible by 5.
6: Apply the rules for both 2 and 3.
7: There's no rule for 7. Get busy dividing.
8: Look at the last THREE digits. If that number is divisible by 8, the entire number is divisible by 8 (I never found that one very helpful. :))
9: Add all the digits. If the sum is divisible by 9, the number is divisible by 9.
10: If it ends in 0, it's divisible by 9.
20: If it ends in 0 and the number before it is even, it's divisible by 20.
Now this is a question:
30: If the last number is 0 and the rule for 3 applies, then it's divisible by 30. True? False? I just made that up, but it seems rational.
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GeorgeStGeorge
True. Likewise, if the rule for 3 applies and the number ends in 5, it's divisible by 15.
George
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2fortheroad
OK - I've done the math 4 times, and I don't get my phone number.
I did it with my office phone number and it came out just fine, but every time I do my home number, it doesn't work.
Are there some number combinations that just don't compute??
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templelady
It was still Kewl-even if itis a math trick
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Belle
:unsure:
I can tie a cherry stem into a knot with my tongue.
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templelady
Belle,
I am AWESTRUCK!!
All that happens when I try it is that my tongue sticks the darn stem in the gap between my molars :(
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rhino
That is so much more entertaining than those silly math games :)
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