Proving: If n Divisible by 3, Then n^3 Divisible by 3

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SUMMARY

The discussion centers on proving that if an integer n is divisible by 3, then n cubed (n^3) is also divisible by 3. The proof begins with the assumption n = 3m, where m is an integer. By squaring n, it is shown that n^2 = 9m^2, which can be factored to demonstrate that 3 is a factor of n^2. The conclusion is that since n is divisible by 3, it follows that n^3 must also be divisible by 3, confirming the initial statement.

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Homework Statement



Prove the following: If the integer n is divisible by 3 then n^3 is divisible by 3.

Homework Equations



Direct Proof

The Attempt at a Solution



n = 3m

n^2 = 9m^2

n^2 = 3(3m^2)

I think the proof is done at this point because the 3 factors out but I also did this:

n^2 = 9m^2

(n^2)/9 = m^2

(n/3)(n/3) = (m)(m)

which also implies n is divisible by 3 since integer x integer = integer

My professor is kind of harsh on proofs so I am not sure if there are intermediate steps I'm missing. Thanks!
 
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If 3 is a factor of n, it must also be a factor of any natural power of n. Do you mean n^3 or n^2? Your working looks like n^2. Anyway, the first way you did it is correct and satisfactory though if your teacher is really harsh, you may want to go like:

n= 3m for some natural value of m, because n is divisible by 3 (Data).
Squaring both sides, n^2 = 9m^2 = 3 ( 3m^2). Since m is natural, 3m^2 must also be natural, and hence 3 is a factor of n^2 as well.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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