S.C.B.
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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 $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\... View answer Accepted answer 35 votes As I don't fully understand your question, I will divide my answer into three parts: the definition of a square root, how to understand it, and algorithms to calculate it. DEFINITION OF A SQUARE ... View answer 35 votes Note that 1010101....10101 is \frac{10^{4034}-1}{99}. Also, 10^{4034}-1 is (10^{2017}-1)(10^{2017}+1), both of which are larger than 99. This implies that the number is not prime. View answer 25 votes This is overkill, but by Lucas's Theorem$$\frac{1000!}{(500!)^2}=\binom{1000}{500} \equiv \binom{2}{1}\binom{6}{3}\binom{2}{1}\binom{6}{3} \equiv 4 \pmod{7}$$As 1000=2626_{7} and 500=1313_{7}. ... View answer Accepted answer 20 votes Let T_{m} be a^m+b^m+c^m. Let k=-ab-bc-ca, and l=abc. Note that this implies a,b,c are solutions to x^3=kx+l. Using Newton's Identity, note the fact that T_{m+3}=kT_{m+1}+lT_{m}(... View answer 18 votes You're assuming that the limit of the sum of the first n Fibonacci numbers exists as n \to \infty, which it doesn't. Which is to say that in order to apply your method, the series must be ... View answer 17 votes HINT Divide the cases for the number of zeros, namely 0,2,4,6,8, which gives us$$\binom{8}{0}+\binom{8}{2}+\binom{8}{4}+\binom{8}{6}+\binom{8}{8}=\frac{2^8}{2}$$From the Binomial Theorem. You ... View answer Accepted answer 15 votes You can only use the fact that$$\lim_{x \to \infty}f(x)g(x)=\lim_{x \to \infty} f(x) \lim_{x \to \infty}g(x)$$When it is given both limits exist. So your method of saying this is undefined is ... View answer Accepted answer 13 votes I am assuming that there is a condition that \angle DOE=45° from your graph. HINT Note that the set of O such that \angle DOE=45° forms a circle. And a point on a circle that is farthest ... View answer Accepted answer 11 votes CLAIM For all n \ge 1, then let us prove$$\sqrt{2n+25+\frac{1+\ln (n-1)}{2}} > x_{n} > \sqrt{2n+25}$$PROOF This holds for n=1. This can be checked numerically. Assume this holds for ... View answer 11 votes Note that from the given conditions, we have that P(x)(x+1)-1 is a twelfth degree polynomial with solutions 0,1,2, \dots 11. In other words,$$P(x)(x+1)-1=ax(x-1)(x-2) \dots (x-11)$$If x=-1, ... View answer Accepted answer 11 votes Considering that the sequence in which he visits these n cities, is$$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]$$... View answer Accepted answer 11 votes Your mistake can be seen in your first line: you should not divide by 2 as you did not count the handshakes between men and women twice. Instead, the ways to pick a man is 20. The number of men ... View answer Accepted answer 10 votes Note that x^2 > 0 for x \neq 0, so if set 1-x^2=t, we have t < 1. So we have that$$\lim_{x \to 0} f(1-x^2)=\lim_{t \to 1^{-}}f(t)=\lim_{t \to 1^{-}}(t^2+2)=3$$As f(t)=t^2+2 for t&... View answer 10 votes This is because Wolframalpha is plotting y=(x-2)(x-3), which is a parabola. As you have entered (x-2)(x-3)=0, it is merely indicating where the intersection is between y=(x-2)(x-3) and y=0, ... View answer Accepted answer 10 votes Note that the problem is equivalent to finding integer solutions to$$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=...

$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+... View answer 9 votes Let us assume there exists such naturals a,b that b=5a. Note that b cannot have more digits than a from the definiton of b. We have from that b=5a that b\equiv 0 \pmod {5}, or the ... View answer Accepted answer 9 votes Let x=t^6. Note that$$\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}$$So the limit is \frac{2}{3}. View answer Accepted answer 9 votes No, it cannot. Note that for all x \in \mathbb{Z},$$x^2 \equiv 0, 1 \pmod {4}$$Thus a sum of two squares can only be 0,1,2 modulo 4. However, note that$$...

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 $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$ ...

$$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$ Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$...
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 $$\left(\sum_{k=1}^n p_k^2\right)\left(\sum_{k=1}^n 1\right) \ge \left(\... View answer 9 votes As you have already observed, d must divide 11, thus d is either 1 or 11. However, one thing to note that is both are possible-for example, for n=2, d=1, and for n=12, d=11. A ... View answer Accepted answer 8 votes Note that$$\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$$This follows elementarily ... View answer 8 votes Note that it says on Wikipedia that [...] in Euclidean geometry, a point is a primitive notion upon which the geometry is built. Being a primitive notion means that a point cannot be defined in ... View answer Accepted answer 8 votes Assume that a,b,c are distinct. Let a-b=x \neq 0,\; b-c=y \neq 0. Note that a-c=x+y \neq 0 Note that the equation becomes$$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$$... View answer Accepted answer 8 votes Assume k \ge 9. Then we have$$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$$As follows ... View answer Accepted answer 8 votes Note that if A,B,C are angles of a triangle, we have that$$\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$$Then we have that for at ... View answer Accepted answer 8 votes You have figured out that it is a linear function. Now you just need to note that by putting f(t)=at+b back into the original equation, we get$$a= 1+\int_{0}^{1} (at^2+bt) \mathrm{d}t=1+\frac{a}{3}+...