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Growing fast (Posted on 2015-03-23) Difficulty: 3 of 5
Given a set of 4 random integers a,b,c,d.
Let's create another set:
a1=a-b, b1=b-c, c1=c-d, d1=d-a;
from the above set we can create in similar fashion a2, b2,c2, and d2,
then a3, b3,c3, and d3, etc ...

Prove that if none of the numbers a100, b100, c100, and d100, is more than 1012 then a=b=c=d=0.

No Solution Yet Submitted by Ady TZIDON    
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re: correction- ............... see before solving | Comment 3 of 4 |
(In reply to correction- ............... see before solving by Ady TZIDON)

Given that change to the problem, then the item to be proved should be changed to say that one or more of the numbers a100..., d100, will be more than 10^12, as the conditions are already set that the numbers are distinct, and therefore not equal.
  Posted by Charlie on 2015-03-24 07:43:06

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