# Tag Info

10

Set $r=\sqrt{7+\sqrt{14}}$; then $r^2=7+\sqrt{14}$ and so $$14=r^4-14r^2+49$$ or $$r^4-14r^2+35=0$$ The polynomial $X^4-14X^2+35=0$ is irreducible over the rational numbers by Eisenstein's criterion (with $7$), so the degree of $r$ over the rationals is $4$. A number of the form $a+b\sqrt{c}$ with rational $a,b,c$ has degree $2$ over the rationals. ...

9

Use the fact that $a^2-b^2 = (a+b)(a-b)$ Multiplying $(6-5)$ on your LHS, we obtain: $$\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})=(6-5)\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})$$ $$=(6^2-5^2)\prod\limits_{k=1}^5(5^{2^k}+6^{2^k})=(6^4-5^4)\prod\limits_{k=2}^5(5^{2^k}+6^{2^k})=...$$ Iterating for all the terms in the products, you should get $x=y=2^6=64$, so $x-y=0$

8

This problem is "isomorphic" to the one where you have the $n$ points on a circle and must show that two of them are as near as $1/n$ to each other. This is because you can take away an integer from $x_i - x_j$, so it always lands on $[0, 1]$ and zero and one are identyfied.

7

$x^3+x^2=2^y+16$. The RHS is positive, so $x^2(x+1)>0\iff x\ge 1$. Since $2^y$ is an integer, we have $y$ is a positive integer too ($y=0$ won't give a solution). $x,y$ are positive integers. $x^3+x^2-16=2^y$. You see a cubic polynomial on the LHS that could easily be strictly bounded between two consecutive cubes (namely $x^3$ and $(x+1)^3$) for most ...

5

I think my method is essentially equivalent to Archaick's, but it might make it more clear how you can count without it being brute force per se. Note that the number we are squaring is $\frac{10^{1992}-1}{9}$, so we are counting the digits of $$\left(\frac{10^{1992}-1}{9}\right)^2 = \frac{1}{81}(10^{3984} - 2\cdot 10^{1992} +1)$$ To write this out as an ...

5

Reformulate: Let $a,b \in [0,1)$ and $m,n \in \mathbb Z$. Solve $$(m+a)n = 7\\ (n+b)m = 8$$ Rearrangement gives $$nm = 8 - bm = 7 - an$$ The product on the left is an integer so we already know that $bm, an\in \mathbb Z$ or in other words $$a = \frac kn; \quad b = \frac lm$$ With $k,l\in\mathbb Z$. Substituting this back gives us $$mn = 8-l = 7-k$$ ...

5

It is sometimes called a slack variable. If $x_1+x_2+x_3<50$, then there must exist a $y>0$ such that $x_1+x_2+x_3+y = 50$, e.g. if $x_1+x_2+x_3 = 46$ then $y = 4$. If $x_1+x_2+x_3 = 50$ then $y = 0$, so $$x_1+x_2+x_3 \leq 50$$ is the same as, for some $y\geq 0$, that $$x_1+x_2+x_3+y = 50.$$

5

If $P(x)$ is irreducible $\implies$ gcd(P,P')=1 Now we can apply Bézout identity which give us two polinomials $A(x),B(x)$ such that: 1=$A(x)P(x)+B(x)P'(x)$ evaluating this in $x_0$ 1=$B(x_0)P'(x_0)$ and we get the desired result.

4

You have $50$ identical candies, and $3$ kids. You want to give $50$ or fewer candies to the $3$ kids (you might choose to distribute all $50$ of the candies, or just $22$, or even (poor kids) a total of $0$ candies. We want to know how many ways there are to do the job. You will be possibly keeping some or all of the candies. So in effect you are ...

4

Dan Schwarz (one of the major problem proposers for EGMO, RMM, Balkan...) has posted a solution at here. I'll briefly sketch it here. First, one takes the midpoints $M_1$, ..., $M_5$ of $A_1A_2$, ..., $A_5A_1$. Then at each angle $A_i$, one takes the circumcircle of triangle $M_{i-1}A_iM_{i+1}$ (which has radius $\frac 12 R_i$) and the point diametrically ...

4

Let's allow leading zeroes in the numbers, but require that they have exactly $2015$ digits ($10^{2015}$ is the smallest number with $2016$ digits). Then, following user MJD's hint, we realize that the number of such numbers with digits in non-decreasing order is the same as the number of ways to put $2015$ balls into $10$ boxes (the number of balls in the ...

4

I can help find a generating function, but judging from my first impressions this is not a simple problem. Let us consider an infinite sum. We treat $x$ as a "dummy" variable, and consider only the coefficient in front of this. $$\sum_{n=0}^\infty s(n)x^n$$ Where $s(n)$ denotes the value of the amount of ways to create a number as a sum of 3's and 2's. ...

4

note when $x=0$, the three items are equal which hints the min is $\sqrt{3}$. first take $\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}} \ge \sqrt{\dfrac{8}{a+b}}$ which is easy to prove by AM-GM. $P \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{4x^2+6x+6}} \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{6x^2+6x+6}}$, when $x=0$ get "=" let ...

3

The problem is that your PIE formula is wrong; it should be \begin{align*} n(A\cup B\cup C)&=n(A)+n(B)+n(C)\\ &\quad-\big(n(A\cap B)+n(A\cap C)+n(B\cap C)\big)\\ &\quad+n(A\cap B\cap C)\;. \end{align*}

3

Here is a way to use CS to solve the inequality. First, we re-write what we want to prove as $$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge 2-\frac{(z+1)^2}{z^2+1} = \frac{(z-1)^2}{z^2+1}$$ Now the constraint gives $z = \dfrac{x+y-xy}{x+y-1}$. Using this, we need to only show $$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge ... 3 You will indeed use the pigeonhole principle: if n = km + 1 objects are distributed among m boxes, then the pigeonhole principle asserts that one of the box will contain at least k + 1 objects. To expand a bit about my comment, you can even actually use the result for 6 vertices an 2 colors, to solve for 17 vertices and 3 colors. The simpler ... 3 Using (Like) pigeonhole principle Let's assume that every x_i=\epsilon_i+h_i with h_i \in \Bbb Z and \epsilon_i\in [0,1] and assume that (we can alawys find an increasing order for \epsilon_i):$$0\leq \epsilon_1\leq \epsilon_2\leq \cdots \leq \epsilon _n<1$$if for every i we have \epsilon_{i+1}-\epsilon_i>\frac{1}{n} take (i,j)=(1,n) ... 3 This is only a solution for even values of x: For x even, x^2 is even and x+1 is odd. So \gcd(16, x+1) = 1. So 16 \mid x^2. So 4 \mid x. So x = 4k for some k. 16k^2(4k+1)=16(2^z+1) k^2(4k+1)=2^z+1 If k is even then the RHS is even and we have a contradiction. So k must be odd. So k = 2t+1 for some t. (2t+1)^2 (8t+5) = 2^z ... 2 Then... from this equation you can easily get that: f(n) = (2 / (n(n+1))).f(1) How? Just express f(n) by f(n-1), then f(n-1) by f(n-2) and so on until you reach the bottom. 2 First of all note that if I-AB is invertible then I-BA is also invertible, this implies that If I-A_1A_2\cdots,A_m is ivertible then any circular permutation of the product I-A_iA_{i+1}\cdots A_mA_1\cdots A_{i-1} is invertible. For example if I-ABC is invertible then I-BCA,I-CAB are also invertible, but this does not cover all matrices in fact ... 2 A simple trick to compute k such that 2^k|n! is to compute \sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor, this is because n has [n/2] numbers divided by 2, if we pick out these numbers and find out that there're [n/4] numbers divided by 4.. If we continue this procedure, we see that ... 2 You have a_{n+1}=f(a_n) where f(x)=2x^2-1. Since f(x)>x whenever x>1, we must have a_n\leq 1 for every n. Since f(x)>1 whenever x<-1, we must have a_n\geq -1 for every n. So we have a_n\in [-1,1] for every n. In particular, a_0\in [-1,1], so there is a \theta\in{\mathbb R} such that a_0=\cos(\theta). By elementary ... 2 Notice the pattern among last digits of ascending powers of 2 Last digits of: 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 6, 2^5 is 2, 2^6 is 4... etc Also notice that the last digits of 2^4, 2^8, 2^{12}, 2^{16}... will all be 6 We note that as 2004 is also a multiple of 4, therefore 2^{2004} will have a last digit of ... 2 This is not particularly pretty but I figured I'd share the complications which arise in using this method. If you multiply this out the way that most elementary students are taught to multiply numbers (at least in the US), you'll have a sum of the form (11\ldots11)\sum_{i=1}^{1992}10^i, lined up into 1991*2+1=3983 columns. In the first column, there ... 2 Your claim :"If the original numbers are all pairwise distinct, then we always need n steps." is not correct, in fact take for example the following sequence:$$(2,14,4,12,6,10)\xrightarrow{a=8}(6,6,4,4,2,2)\xrightarrow{a=5} (1,1,1,1,3,3)\xrightarrow{a=2} (1,1,1,1,1,1)\xrightarrow{a=1} (0,0,0,0,0,0)$$which requires only 5 steps, this can be preformed to ... 2 We have to assume also that P is a nonzero polynomial. The thesis should be that Q has at least n+1 non zero coefficient (otherwise (x-1)(x-2)=x^2-3x+2 would be a counterexample). We can also assume that P(0)\ne0, because otherwise we can divide P by the maximum possible power of x and the number of nonzero coefficients is unaffected. If ... 2 Let's slightly generalize the problem by considering clocks where the hour hand makes one round in n hours (for our normal clocks we have n=12; however e.g. for this clock you have n=24). Let's measure the time in hours. The minute hand makes one cycle every hour, so it gives the position modulo 1; that is, the minute hand at times t_1 and t_2 ... 2 Let's measure time in units of 12 hours, measure angles as a fraction of 2\pi and consider a period from midnight to noon. Time will run from 0 to 1. The hour hand position is t. The minute hand position is 12 t \pmod 1, the fractional part of 12t The times when you can't tell the time is when you can interchange the hands and get a legal ... 2 Setting f(x) = 4x^5 + 3x^3 -5x^2 + 7x - 12, we have that$$f'(x) = 20x^4 + 9x^2 - 10x + 7 = 20x^4 + (3x-5/3)^2 + 38/9 > 0 Hence, $f(x)$ is an increasing function with odd degree. Hence, it has only one root. Further, $f(0) = -12$, which implies that the lone root has to be positive. Hence, the difference between the largest positive and smallest ...

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