[Shahbaz Khan]

I believe it should be a removable discontinuity at pi/2. Feel free to text me on Skype (shahbazakhan410@gmail.com) if you want to discuss. I do not know LaTeX, so discussing here is going to be painful for me. Cheers.

[Shahbaz Khan]

Use Cauchy-Schwarz on a/b b/c c/a and a,b,c. Should follow easily from this.

[Shahbaz Khan]

Uh, I'm not sure how I'm supposed to type math here, so I'll just type in the concept of how I would prove this question.

So, any integer can be written as the sum of various terms of the form a*(10^k) where k is a whole number, a is an integer belonging to [0,9]

Now, we have to note that 10 is congruent to 1 modulo 3. Keeping this in mind, we can assert that in the above terms, the remainder when dividing each by 3 will be the constant, that is, the "a" term, and hence, the entire number is congruent to the sum of the constant terms modulo 3.

Now, the initial number and the sum of the constants is the same modulo 3. That is, when the number is divisible by 3, the sum will be divisible too.

This proves the given statement.

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