S.C.B.
  • Member for 5 years, 11 months
  • Last seen more than a month ago
Multiple-choice: sum of primes below $1000$
44 votes

We have to decide among $\text{(A)}$ and $\text{(B)}$. Note that the $26$th prime is $101$. This implies that if $p_{n}$ denotes the $n$ th prime, then $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\...

View answer
What is a square root?
Accepted answer
35 votes

As I don't fully understand your question, I will divide my answer into three parts: the definition of a square root, how to understand it, and algorithms to calculate it. DEFINITION OF A SQUARE ...

View answer
Prove that 10101...10101 is NOT a prime.
35 votes

Note that $1010101....10101$ is $\frac{10^{4034}-1}{99}$. Also, $10^{4034}-1$ is $(10^{2017}-1)(10^{2017}+1)$, both of which are larger than $99$. This implies that the number is not prime.

View answer
How to prove that $\frac{1000!}{(500!)^2}$ is not divisible by 7?
25 votes

This is overkill, but by Lucas's Theorem$$\frac{1000!}{(500!)^2}=\binom{1000}{500} \equiv \binom{2}{1}\binom{6}{3}\binom{2}{1}\binom{6}{3} \equiv 4 \pmod{7}$$ As $1000=2626_{7}$ and $500=1313_{7}$. ...

View answer
Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$
Accepted answer
20 votes

Let $T_{m}$ be $a^m+b^m+c^m$. Let $k=-ab-bc-ca$, and $l=abc$. Note that this implies $a,b,c$ are solutions to $x^3=kx+l$. Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(...

View answer
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$?
18 votes

You're assuming that the limit of the sum of the first $n$ Fibonacci numbers exists as $n \to \infty$, which it doesn't. Which is to say that in order to apply your method, the series must be ...

View answer
Finding how many 8-bit bytes contain an even number of zeros . . .
17 votes

HINT Divide the cases for the number of zeros, namely $0,2,4,6,8$, which gives us $$\binom{8}{0}+\binom{8}{2}+\binom{8}{4}+\binom{8}{6}+\binom{8}{8}=\frac{2^8}{2}$$ From the Binomial Theorem. You ...

View answer
Evaluate the limit at infinity
Accepted answer
15 votes

You can only use the fact that $$\lim_{x \to \infty}f(x)g(x)=\lim_{x \to \infty} f(x) \lim_{x \to \infty}g(x)$$ When it is given both limits exist. So your method of saying this is undefined is ...

View answer
Simple method to solve a geometry question for junior high school student
Accepted answer
13 votes

I am assuming that there is a condition that $\angle DOE=45°$ from your graph. HINT Note that the set of $O$ such that $\angle DOE=45°$ forms a circle. And a point on a circle that is farthest ...

View answer
Show that $45<x_{1000}<45.1$
Accepted answer
11 votes

CLAIM For all $n \ge 1$, then let us prove $$\sqrt{2n+25+\frac{1+\ln (n-1)}{2}} > x_{n} > \sqrt{2n+25}$$ PROOF This holds for $n=1$. This can be checked numerically. Assume this holds for $...

View answer
$P(x)=\frac{x}{x+1}$ for $x=1,2,...,n$, find the value of $P(n+1)$.
11 votes

Note that from the given conditions, we have that $P(x)(x+1)-1$ is a twelfth degree polynomial with solutions $0,1,2, \dots 11$. In other words, $$P(x)(x+1)-1=ax(x-1)(x-2) \dots (x-11)$$If $x=-1$, ...

View answer
Number of ways of visiting N places
Accepted answer
11 votes

Considering that the sequence in which he visits these $n$ cities, is $$a_1, a_2, \dots, a_n$$ then $a_i \neq i$. This is a derangement, whose formula is given by $$!n=\left[\frac{n!}{e}\right]$$ ...

View answer
Handshakes in a party
Accepted answer
11 votes

Your mistake can be seen in your first line: you should not divide by $2$ as you did not count the handshakes between men and women twice. Instead, the ways to pick a man is $20$. The number of men ...

View answer
Composition of Piecewise Function Limit
Accepted answer
10 votes

Note that $x^2 > 0$ for $x \neq 0$, so if set $1-x^2=t$, we have $t < 1$. So we have that $$\lim_{x \to 0} f(1-x^2)=\lim_{t \to 1^{-}}f(t)=\lim_{t \to 1^{-}}(t^2+2)=3$$ As $f(t)=t^2+2$ for $t&...

View answer
Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola?
10 votes

This is because Wolframalpha is plotting $y=(x-2)(x-3)$, which is a parabola. As you have entered $(x-2)(x-3)=0$, it is merely indicating where the intersection is between $y=(x-2)(x-3)$ and $y=0$, ...

View answer
Find $x$ positive integer such that $x^4+x^3+x^2+x+1$ is a perfect square
Accepted answer
10 votes

Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$ Now proceed to note that if $x>3$, we can find $$(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=...

View answer
Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$
Accepted answer
10 votes

$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+...

View answer
Prove that it is impossible to have integers $b=5a$ under a digit re-ordering
9 votes

Let us assume there exists such naturals $a,b$ that $b=5a$. Note that $b$ cannot have more digits than $a$ from the definiton of $b$. We have from that $b=5a$ that $b\equiv 0 \pmod {5}$, or the ...

View answer
How to solve this limit without L'Hospital?
Accepted answer
9 votes

Let $x=t^6$. Note that $$\lim_{x \to 1 }\frac{\sqrt[3]{x}-1}{\sqrt[]{x}-1}=\lim_{t \to 1} \frac{t^2-1}{t^3-1}=\lim_{t \to 1}\frac{t+1}{t^2+t+1}$$ So the limit is $\frac{2}{3}$.

View answer
Could be this : $7131372917538397234773191167617941438959$ written as $x^{2}+y^{2}$ with $x, y$ integers?
Accepted answer
9 votes

No, it cannot. Note that for all $x \in \mathbb{Z}$, $$x^2 \equiv 0, 1 \pmod {4}$$ Thus a sum of two squares can only be $0,1,2$ modulo $4$. However, note that $$...

View answer
Can both $n+3\; \text{and}\; n^2+3$ both be cubic numbers at same time?
Accepted answer
9 votes

Your question is asking if there is $x,a \in \mathbb{Z}$ such that $$x^6-6x^3+12=a^3$$ Note that if we have $x \ge 2, x \le -2$, then we have that $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$ ...

View answer
Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$
Accepted answer
9 votes

$$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$ Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$...

View answer
Proof of an inequality using Cauchy's inequality: $\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}$
9 votes

Note that $$\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 =\sum_{k=1}^n p_k^2+2n+\sum_{k=1}^n \frac{1}{p_k^2}$$ Now note that $$\left(\sum_{k=1}^n p_k^2\right)\left(\sum_{k=1}^n 1\right) \ge \left(\...

View answer
What is $\gcd(n-1,n+10)$, for $n$ a natural number?
9 votes

As you have already observed, $d$ must divide $11$, thus $d$ is either $1$ or $11$. However, one thing to note that is both are possible-for example, for $n=2$, $d=1$, and for $n=12$, $d=11$. A ...

View answer
Show that $\prod_{k=1}^n \left(\prod_{j=1}^k\frac{k}{j}\right)$ is always an integer.
Accepted answer
8 votes

Note that $$\prod_{k=1}^{n} \binom{n}{k}=\prod_{k=1}^{n} \frac{n^{\underline{k}}}{k!}=\prod_{k=1}^n \frac{k^k}{k!}$$ As $$\prod_{k=1}^{n}n^{\underline{k}}=\prod_{k=1}^n k^k$$ This follows elementarily ...

View answer
How is "point" in geometry undefined? And What is a "mathematical definition"?
8 votes

Note that it says on Wikipedia that [...] in Euclidean geometry, a point is a primitive notion upon which the geometry is built. Being a primitive notion means that a point cannot be defined in ...

View answer
Does there exist a triple of $distinct$ numbers $a,b,c$ such that $(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$?
Accepted answer
8 votes

Assume that $a,b,c$ are distinct. Let $a-b=x \neq 0,\; b-c=y \neq 0$. Note that $a-c=x+y \neq 0$ Note that the equation becomes $$x^5+y^5=(x+y)^5$$ So $$(x+y)^5-x^5-y^5=5xy(x^3+2x^2y+2xy^2+y^3)=0$$ ...

View answer
Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$
Accepted answer
8 votes

Assume $k \ge 9$. Then we have $$9=1^6+1^5+1^4+1^2+1+4>k \times 1^3 \ge 9$$ Contradiction. So we have $k<9$. Note that $$x^6+x^5+x^4+x^3+x^2+x+4=(x^6+4)+(x^5+x)+(x^4+x^2)>8x^3$$ As follows ...

View answer
If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle
Accepted answer
8 votes

Note that if $A,B,C$ are angles of a triangle, we have that $$\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ As seen here. If $$0=\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ Then we have that for at ...

View answer
If $f(x)=x+ \int_{0}^{1} t(x+t)f(t)dt$, find $\int_{0}^{1} f(x).dx$
Accepted answer
8 votes

You have figured out that it is a linear function. Now you just need to note that by putting $f(t)=at+b$ back into the original equation, we get $$a= 1+\int_{0}^{1} (at^2+bt) \mathrm{d}t=1+\frac{a}{3}+...

View answer
1
2 3 4 5
17