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40m
comment Proof of $p_n<n^2$ by Elementary Means
@user170039 Chebyshev already proved something of the strength $p_n \le (1.2 + o(1)) n \log n$; this was nearly 50 years before PNT was established. The proof that karvens refers to is probably the simpler one due to Erdős.
1h
comment Last 3 digits of $7^{12341}$
@rsadhvika $7 \equiv -1 \pmod 8$ so it is easy to see that any even exponent satisfies the $2$-adic part of the congruence. For the $5$-adic part, we easily observe that $7^2 = 49 \equiv -1 \pmod{25}$, so then $7^4 \equiv 1 \pmod{25}$, and an easy application of binomial theorem shows that $(7^4)^5 \equiv 1 \pmod{125}$.
3h
comment Convergence/Divergence of the series $\sum\limits_{n=1}^{\infty}\tan(1/n)$
"for small nonnegative $x$"
7h
answered On the difference between consecutive primes
18h
comment A general solution for a (x+y)^n = x^n + y^n
If both $(x,y)$ are not zero then divide through by $x^n$ and let $z = y/x$. Can you see why equality can't hold for $z>0$? Can you see why it's false for $-2\le z<0$? (assuming $n$ is even) Can you see why it's false for $z < -2$?
18h
comment Pigeon hole principle based puzzle question
The question is worded poorly: "12 pairs each of 3 different types of gloves" is $36$ pairs. "12 pairs of gloves, of 3 different types" is $12$ pairs.
18h
comment Pigeon hole principle based puzzle question
"Any other scenario involves extracting less gloves, so 13 is the $minimum to be sure$." This makes it sound like there are no other scenarios where you need $13$ gloves, which isn't true. You don't need the first $12$ gloves to all have the same handedness, so I think this part deserves a little more justification.
2d
comment How to prove that $\lim_{u\to \infty} (1+\frac 1u)^u = e$
@user3835165 That is not a definition. Many numbers that are not equal to $e$ match the same description. The limit in your question can't possibly be equal to all of them.
Oct
17
comment What are the odds of hitting exactly 100 rolling a fair die
Expectation by itself is not enough for the general case. Suppose $S = \{2,4,6\}$. Then the probability of hitting exactly $101$ is much lower than $2/7$ :).
Oct
16
answered These symbols are referring to? BOOK: Introduction to Genetic Algorithms by S.N.Sivanandam · S.N.Deepa
Oct
13
answered Complexity of finding $\alpha(G) + \omega(G)$
Oct
13
comment Matrix exponential converse. Baker-Campbell-Hausdorff
Do you mean $e^X e^Y = e^{X+Y}$ in the second-last equation, and $e^{Xt}e^{Yt} = e^{(X+Y)t}$ in the last?
Oct
11
comment A continuous nonconstant function has uncountable range
Sure, can you prove that a countable subset of $\mathbb R$ having at least two points is disconnected?
Oct
9
comment Find a possible n such that $(2^n +3^n)/{113}$ is integer
Actually this is a pretty simple method, it just completely fails to solve the question :). Also, two increasing functions can easily have more than one point of intersection: $x + \sin x$ and $x$, for instance.
Oct
8
comment Prime Reflections
@JeffreyYoung "Why $120$?" is entirely relevant to this question. You asked us to suggest a description for "Left Mirror Center", but none of your examples serve to pin down what you mean. Just saying "$119$ is divisible by $7$ and $121$ is divisible by $11$" is not enough, because $42$ is divisible by $7$ and $44$ is divisible by $11$. So I repeat, why not $43$ (and its mirror pair $2267$)?
Oct
8
comment Prime Reflections
@JeffreyYoung Why $120$ and not $43$ or $197$? I presume because $120$ is divisible by $3$ and $5$. That would make it just a basic artifact of en.wikipedia.org/wiki/Chinese_remainder_theorem. In fact there is nothing special about primorials here; you could replace the primes with any sequence of relatively prime integers.
Oct
5
comment Probing a particular function
To know the number of digits in base $B$ is equivalent to knowing the approximate value of $\log_{B} Q(n)$. See Stirling's approximation.
Oct
4
comment How can one handle very large numbers such as ${1,000,000 \choose 500,000}$ using binomial formula and very tiny numbers such as $0.5^{1,000,000}$?
@David The great thing about Stirling's approximation is that, even its most basic form, the relative error in $\log n!$ is $O(1/n)$, so for really large $n$ the approximation is superb.
Oct
4
comment How can one handle very large numbers such as ${1,000,000 \choose 500,000}$ using binomial formula and very tiny numbers such as $0.5^{1,000,000}$?
@David So is your question specifically about exact values or isn't it? I meant that there is almost no context where you would need the exact value of $C(100000000,50000000)$, a number with millions of digits.
Oct
3
comment How can one handle very large numbers such as ${1,000,000 \choose 500,000}$ using binomial formula and very tiny numbers such as $0.5^{1,000,000}$?
I'm not sure if this is a question about math so much as it is about why free online tools don't expend extra effort to handle numbers that users almost never care about.