It is clear that any integer can be expressed in one of the three ways, 3n, 3n+1, 3n+2, where n is an integer.
Using the fact above to show that if k is an integer and (k^2) is divisible by 3, then k is divisible by 3.
Maths question
2013-02-28 4:13 am
回答 (1)
2013-02-28 4:49 am
✔ 最佳答案
Let k = 3p.Assuming k = 3n + 2 ,k² = (3n + 2)² = 9n² + 12n + 4 = 3(3n² + 4n + 1) + 1 ≠ 3p
since L.H.S. is not divisible by 3.
(Contradiction)Assuming k = 3n + 1 ,
k² = (3n + 1)² = 9n² + 6n + 1 = 3(3n² + 2n) + 1 ≠ 3p.
(Contradiction)Hence k must be 3n ,then k is divisible by 3 ,
where k² = (3n)² = 9n² = 3p ,
p = 3n²
2013-02-27 20:56:33 補充：
Let k² = 3p
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