Recent content by vidyarth

True, I missed one. Corrected though

The meaning of an element of a ring being irreducible is, that it cannot be expressed as a product of two distinct elements in the ring upto units barring itself and unity, upto equivalence, i.e., its associates. Thus, since $2$ is irreducible in the ring $\mathbb{Z}+\mathbb{Z}\sqrt{5}$...

The question seems interesting. The answer is $24$. That is, about half of the odd 2-didit numbers cannot be expressed as a sum of two primes. The proof is simple. The sum if two primes must contain $2$, for otherwise the number cannot be odd. Thus, we have just check the primality of $n-2$ ...

What you did upto the last but one step is perfect, but your showing the contradiction is not that correct, as $\cos (\frac{\pi}{2})=0$ is true only for one value, i.e. at $\pi/2$. The correct contradiction comes from the fact that $2\sin x \cos x\le0 \ \forall x\in[\frac{\pi}{2},\pi]$ from the...

If $a^2b^2+b^2c^2+c^2a^2-69abc=2016$, then, what can be said about the least value of $\min(a, b ,c)$? This problem is unyielding to the major inequalities like AM-GM, Cauchy-Schwarz, etc. I also tried relating it to $x^3+y^3+z^3-3xyz=(x+y+z)(\sum_{cyc}x^2+\sum_{cyc}xy)$, but of no use. Any...

yes. But can you explain how you get it, thoroughly?

The answer found [here][1]is $5989\overbrace{\ldots}^{\text {217 9s}}989888$ which is much shorter than your example. [1]:number theory - Property of 2016 - Mathematics Stack Exchange

This is because the least number of digits required to get a digit sum of $2016$ is $\frac{2016}{9}=224$. But since a string consisting of only $9$s is not divisible by $2016$, therefore it can be made up by using one extra digit. And I think the number should end in $8$ which is the second...

What is the least multiple of 2016 such that the sum of its digits is 2016. I think the answer must be a 225 digit long number ending in 8 but do not know the exact value nor how to prove it. Any ideas. Thanks beforehand.