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pfftt... this is retarded. division properties of natual numbers, anyone?
1st. the subtraction of two natural numbers that are composed of the same digits but in different order ALWAYS yields a number that is aliquot to 9.
2nd. a natural number is divisible by 9 if and only if the sum of its digits is divisible by 9 (that is, it's a multiple of 9)
[for example, 10347428372867454356401527 is divisible by 9 because 1+0+3+4+7+4+2+8+3+7+2+8+6+7+4+5+4+3+5+6+4+0+1+5+2+7=108 is divisible by 9 as well]
so if, at the end, u have a 3-digit number, and u give away 2 digits, it doesn't take a whole genius to figure out what the third digit is 
example`
6387 <-- random number
jumble up --> 8763
8763-6387=2376 <-- subtract
circle any digit --> 3 for example
we have 2376=abcd <-- a+b+c+d=9p
u give away 2,7 and 6 in any order (u circled out 3, remember?)
therefore --> 6+2+7+b=9p <=> 15+b=9p
now, p is a natural number, b is a digit, ie b < 10, therefore in order for (15+b)/9 to be a whole number, b HAS to be 3
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Last edited by Noisician on Nov-11-2003 at 04:35
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Want to let me read your mind
