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29/20 answers
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7h
answered $n$ such that $9$ divides $(n+3)(n-3)(n+1)(n-1)(n-100)$
9h
answered Congruence problem $12x\equiv3\pmod{45}$
9h
answered Integrate $\int \frac{x^7}{(1-x^4)^2}dx$
10h
answered Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals
10h
comment $17x+11y \equiv 7 \pmod {29}$ and $13x+10y \equiv 8 \pmod {29}$. What are x and y?
@MartinSleziak, Updated
10h
revised $17x+11y \equiv 7 \pmod {29}$ and $13x+10y \equiv 8 \pmod {29}$. What are x and y?
deleted 5 characters in body
20h
comment series $\sum_{n=0}^{\infty}9^{n}z^{2n}$
@lucad93, See en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
1d
comment How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
@AayushAgrawal, See also: math.stackexchange.com/questions/189621/…
1d
comment How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
@AayushAgrawal, What is $\tan(A+B+C)$ in terms of $\cot$ ?
1d
revised How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
added 4 characters in body
1d
comment How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
Probably the most natural way.
1d
revised How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
added 5 characters in body
1d
comment How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
Multiply out by $$\tan x\tan2x\tan3x$$ and use $$3x+(-2x)+(-x)=0\cdot\pi$$ and math.stackexchange.com/questions/477364/… OR math.stackexchange.com/questions/477364/…
1d
answered How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
1d
comment If $\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r} $
@Iaamuseruser, Trying figure out the nature of $a,b,c; p,q,r$
1d
answered series $\sum_{n=0}^{\infty}9^{n}z^{2n}$
1d
comment If $\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r} $
@Iaamuseruser, What is the source of the problem?
1d
comment If $\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r} $
Why $$cp+bq+ar > 0?$$
1d
comment limit of the following telescoping sequence
$$\dfrac4{a_ma_{m-1}}=\dfrac{a_m-a_{m-1}}{a_ma_{m-1}}=?$$
1d
comment If $\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r} $
@vrugtehagel, I've established that we need another condition for the proposition to be true.