New answers tagged

0

If $N$ ends in $k$ digits of $9$, then $N+1$ has a digit sum equal to the digit sum of $N$ minus $9k-1$. (That is true for $k=0$, too, of course - then subtracting $-1$ is adding one.) So you need $9k-1$ to be divisible by $7$. And then you have to pick the smallest number possible for the digits left of the ending nines.


0

Clearly $N$ must have a $9$ in its ones place or else the sum of digits from $N$ and from $N+1$ will differ by exactly $1$, and therefore could not both be divisible by $7$. Now when $d_k\cdots d_29$ has $1$ added to it, you maybe get $d_k\cdots (d_2+1)0$ if there is no carrying into the hundreds place. But this $N$ and $N+1$ have digits summing to values ...


0

This is essentially the 3D equivalent of the Radical Center of 3 circles in 2D, and the proof works out in a similar way. The plane through the intersection circle of two spheres $A, B$ is their radical plane, the locus of points with equal power $h_A = h_B$ with respect to the two spheres. If the centers of spheres $A, B, C, D$ are all distinct, then all ...


0

$$\frac{26}{672-x} + \frac{24}{372-x} = \frac{50}{480-x}$$ We could just go ahead and start solving this, but it will make the terms smaller if we let $2u=372-x$ $$\frac{26}{300+2u} + \frac{24}{2u} = \frac{50}{2u+108}$$ factoring a two out of top and bottom, we get $$\frac{13}{u+150} + \frac{12}{u} = \frac{25}{u+54}$$ We then want to get common fraction on ...


2

Hint: There is a four-coloring of the board so that Any straight tetranimo covers one of each color, and Any zig-zag covers either two of two colors or one each of all four coors. The square must color two colors exactly once, and another twice.


0

Yes, that is correct. The closed-form solution is $f(n)=2^n-n-1$


1

Here's a quite straightforward method with two coins, bounded number of throws, and both coins have rational values. First coin is unbiased; second coin has the following probability for a Head: $$p = \frac{2^m}{n!}; m = \lfloor{\log_2(n!)}\rfloor$$ With the unbiased coin you can emulate a biased coin which has $\frac{k}{2^m}$ probability for a Head ($k$ is ...


2

No, you can't. For example, the pairs $$4 + 3 + 2 + 1 = 10, \; \; 4 \cdot 3 \cdot 2 \cdot 1 = 24$$ and $$12 + 1 + (-1) + (-2) = 10, \; \; 12 \cdot 1 \cdot (-1) \cdot (-2) = 24$$ give different results for $\frac{1}{\pi_1} + \frac{1}{\pi_2} + \frac{1}{\pi_3} + \frac{1}{\pi_4}.$


3

Link $AC$, $BD$ and denote $O$ as their intersection point. Since $PQ \bot QR$ ,$PQ//AC,AC = 2 PQ $ and $QR//BD,BD = 2QR\Rightarrow AC \bot BD,AC = 6,BD =8 $ then the quadrilateral is divided into to triangle $ABC$, $ACD$ which share the same edge $AC$.Then the area of the quadrilateral equals to the sum area of these two triangle. Solve the problem using ...


0

Alright, lets bring this down with some Modular-Arithmetic-fu. We can see that the value of A(n) is uniquely determined by the number of digits of A(n-1) mod 9. We will use D(n) to mean 'the number of digits of A(n) mod 9'. Clearly, $D(n) = D(n-1)*2 (mod\ 9)$. We can transform this recursive formula into a direct one: $D(n)= D(1)*2^{n-1} (mod\ 9)=2^{n-1} ...


4

We know that $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$ So now $$\gcd(2^{21}-1,2^{27}-1)=2^{\gcd(21,27)}-1=7$$ We'll now prove the theorem I used here. We can do the following. Assume $n>m$, then: \begin{align} \gcd(a^n-1,a^m-1)&=\gcd((a^n-1)-(a^m-1),a^m-1)\\ &=\gcd(a^m(a^{n-m}-1),a^m-1)\\ \end{align} since $\gcd(a^m,a^m-1)=1$, we now know ...


2

The generating function for the central binomial coefficients is \begin{align*} \sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad |z|<1 \end{align*} This is an application of the binomial series \begin{align*} (1+z)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n\qquad |z|<1, \alpha\in\mathbb{C} \end{align*} and the ...


0

Let's say $2014=a+(a+1)+\cdots+(a+k)$. If we denote by $m$ the average of this sequence, then we know that $2014=k\cdot m$. Now $m$ is either an integer or of the form $n+\frac{1}{2}$ for some integer $n$. If $m$ is an integer, then $k$ is odd and we need $k\mid 2014$. Since $2014=2\cdot 19\cdot 53$ the only options for $k$ are $1$, $19$, $53$ and $19\cdot ...


5

Let the first number in the sequence of positive integers be $m+1$ and the last number be $n$. Then $$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}=2014$$ $$n^2+n-m^2-m=4028$$ $$(n-m)(n+m)+(n-m)=4028$$ $$(n-m)(n+m+1)=4028$$ As $n$ and $m$ are positive integers, we have $n+m+1$ greater than $n-m$. Now $$4028=2*2014=4*1007=19*212=38*106=53*76$$ For the sequence to ...


1

An alternate way to think about your function $f(x)$ is as: Double the $x$ value. If it is less than 1 (i.e. $2x\leq1$ from $x\leq\frac12$) then leave it. If it is more than one then reflect it in the line $y=1$ (from second half being $2(1-x)$ ). So double your input and fold it down if its over 1. Using this definition and having knowledge of the first ...


4

In the diagram below (borrowed from this answer), we have: $\sin(\theta) < PQ$, because $PQ$ is the hypothenuse of the right triangle $PQR$, $PQ < \theta$, because $PQ$ is the shortest distance from $P$ to $Q$, and so $\sin(\theta) < PQ < \theta$.


7

A simple draw : $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$


1

Let $l(x)$ be the length of the decimal representation of $x$. Use induction on $l(x)$. Suppose the identity holds for $l(x)\le n$. Let $x=10y+r$ where $l(y)=n$ and $r\in\{0,1,\cdots,9\}$. $S(10z)=z\;\forall\;z$. If $r<5\implies N(x)=N(y)$ $$ S(2x)=S(20y+2r)=S(20y)+2r=S(2y)+2r\\ =2S(y)-9N(y)+2r=2S(10y)-9N(x)+2r\\ =2S(10y+r)-9N(x)=2S(x)-9N(x) $$ ...


0

Keeping in mind that the number of digits greater than $4$ is $N(x)$, we can look at $S(2x)$. Let the number of digits of $x$ be $n$. Then the number of digits less than or equal to $4$ is $n-N(x)$, and those digits multiplied by $2$ are less than $10$, so the sum of these digits of $2x$ is simply twice the sum of these digits of $x$. If a digit is greater ...


0

You can start by setting up two equations to get an idea of the neighborhood where your solutions appear. $$2C=5D, C-D=19$$


0

I think they are not using Gauss sign (or floor function). The symbols $\lceil\ \rceil$ refer to the ceiling function; i.e. $\lceil x\rceil = \min\{n\in\mathbb{N}\ |\ x\le n \}$.


1

Take the Young diagram of a partition of $n$ into at most $r$ parts, and add an extra box to the end of each row. If there were less than $r$ rows, then add additional rows of length one so that the diagram has $r$ rows. This way, we get the Young diagram of a partition of $n + r$ with exactly $r$ rows. To see that this is a bijection, it is enough to show ...


3

You can find a bijection between the set of partitions of $n$ into $r$ non-negative integers and of $n+r$ into $r$ positive integers. Let $(\alpha_1, \alpha_2, \dots, \alpha_r)$ be a partition of $n$ such that $\alpha_i \geq 0, \forall i \in \{1,2,\dots,r\}.$ Then $(\alpha_1 + 1, \alpha_2 + 1, \dots, \alpha_r + 1)$ is a partition of $$ (\alpha_1 + 1) + ( ...


3

The flaw in your reasoning is that as the disk rotates along the edge of the clock face, there are two components to its orientation: the disk's own rotation, which you accounted for, but also a second rotational movement corresponding to its changing position relative to the clock. To understand this, take two coins of equal size, and roll one around the ...


0

Hint: You must look at the distance travelled by the center of the circle.


0

These pamphlets are aimed at Oxford and Cambridge entrants, so are slightly off-topic, but they are aimed at moving from high-school to proof-based, deeper maths. I use them in teaching maths clubs here in the UK. They both contain many many commented-upon problems, which is one of the best ways to approach Olympiad material, and are free! ...


0

Hint: you will retransmit if there are $1$ or $3$ errors, not if there are $2$ or $4$ errors. Compute the chance of $1$ or $3$ errors and divide by the chance that there is at least one error.


1

The probability of having at least one error in one block is equal to (1- probability of having no errors) = $ 1- 0.9^3$ Assuming the errors independent. But I am not sure why you described the parity bit. I had to update it, I mistakenly used 0.1 instead of 0.9.


1

Since $\cos^{-1}\sin\theta = \pi/2 - \theta$ and $\tan(\pi/2 - \theta) = 1/\tan \theta$ for $0 < \theta < \pi/2$, we have for any $x > 0$,$$\tan\cos^{-1}\sin\tan^{-1}x = \tan(\pi/2 - \tan^{-1}x) = 1/x.\tag*{$(1)$}$$Also for $x \ge 0$,$$\cos\tan^{-1}\sqrt{x} = 1/\sqrt{x + 1}.\tag*{$(2)$}$$By $(1)$ and $(2)$, we can obtain $\sqrt{r}$ for any ...


0

Consider a Ferrer's diagram of $n$ into at most $r$ parts. $ooooooo\\ oooooo\\ ooo\\ o$ For example this is a partition of $17$ into $4$ parts - $17=7+6+3+1$. So in this example we know $r\ge4$. This means we can insert a first column of size exactly $r$ to get a partition of $n+r$ with exactly $r$ parts. We also need to establish a bijection, and we can ...


3

Observe that $5^2+12^2=13^2$ and $12^2+16^2=20^2.$ Therefore in $\triangle A B C, \;$ $\angle B A C=90^o$ and in $\triangle A C D, \;$ $\angle A C D=90^o$. So the area of $\triangle A B C$ is $5.12/2=30$ and the area of $\triangle A C D$ is $12.16/2=96$, for a total area of $30+96=126.$


3

By the Law of Cosines: $$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos \angle ABC$$ $$\iff \cos \angle ABC=\frac{5^2+13^2-12^2}{2\cdot 5\cdot 13}=\frac{5}{13}$$ Analogously: $$\cos \angle CDA=\frac{16^2+20^2-12^2}{2\cdot 16\cdot 20}=\frac{4}{5}$$ Now, we know that $\angle ABC$ and $\angle CDA$ are both acute, because $\angle ABC<\angle BAC$ and $\angle ...


2

HINT.-There is the formula $$A=\frac{\sqrt{4(l\cdot l_1)^2-(a^2-b^2+c^2-d^2)^2}}{2}$$ where $A$ is the area, $a,b,c,d$ the sides, $l$ and $l_1$ the two diagonals. You have to calculate the second diagonal and apply the formula.


4

Hint: The quadrilateral is composed by two triangles for which you know the three sides. So you can calculate the area of each triangle with the Heron's formula.


2

Suppose that every element of $A$ is less than $\frac2n\binom{n}{n/2}$. If $S\subseteq A$, and $|S|=\frac{n}2$, then $\sum S<\binom{n}{n/2}$. Thus, there must be $S_0,S_1\subseteq A$ such that $S_0\ne S_1$, $|S_0|=|S_1|=\frac{n}2$, and $\sum S_0=\sum S_1$. Show that $S_0\setminus S_1$ and $S_1\setminus S_0$ are disjoint subsets of $A$ with the same sum.


1

One way is to consider the Ferrers diagrams of the partitions. It’s easiest to start with a partition of $n+r$ into $r$ parts. Its Ferrers diagram will have exactly $r$ rows. If you remove the first column of the diagram, you’ll have the Ferrers diagram of a partition of $n$. That diagram might have fewer than $r$ rows (why?), but it can’t have more, so ...


1

Let $\angle AOB=2\gamma$, $\angle BOC=2\alpha$, $\angle COA=2\beta$, so that: $$ [ABC]={1\over2}r^2(\sin2\alpha+\sin2\beta+\sin2\gamma), $$ where $r$ is the radius of $\omega$. On the other hand it follows from the definition of $DEF$ that $$ [DEF]={1\over2}r^2(\sin(\alpha+\beta)+\sin(\beta+\gamma)+\sin(\gamma+\alpha)). $$ As $[A'B'C']=[DEF]$, we must prove ...


0

Lets create spaces the consonants leaving 1 space can be arranged in $7!/2!$ ways and these $7$ create 8 spaces ends inclu also so these $8$ can be filled by $4$ of them in ${8\choose 4}.\frac{4!}{2!}.2520 $ ways


2

Firstly, write all the consonants of CHICAGOLAND, in the following way. $$-C-H-C-G-L-N-D-.$$ This can be done in $ 7!/2!=2520$ ways. Now place the vowels in the gaps, can be done in $\binom{8}{4}\times 4!/2!=\binom{8}{4}\times 12$ ways, Since $a$ comes twice. So, the total number of ways in which we can make a $11$ letter string, is $$\bbox[border:2px ...


1

I do not agree. It looks correct to me. First of all, if variables $x$ and $y$ are unconstrained, it is correct to have $=$ type constraints in the dual. Authors are not required to specify that a variable is unconstrained, they only have to specify which ones are not, and how. This makes sense. If your primal is a minimization problem and constraints are ...


4

Given a sum with summands $>1$, syntactically replace a summand $k$ with a sequence of $k$ times "1", then replace any occurance of "1+1" with "2". Remove the first and last "1" and insert "+" signs. For example $$3+3\mapsto 111+111\mapsto 11211\mapsto 121\mapsto 1+2+1 $$ $$2+2+2\mapsto 11+11+1\mapsto 1221\mapsto 22\mapsto 2+2 $$ $$4+2\mapsto ...


0

Find biggest bead call it xp After removing it we have n-1 beads, the consecutive substring chosen starts with xp+1 ( the bead next to the one we cut ) this substring ends at xp-1 and must be less then n-2 since xp was at least positive 1. Next take the second largest bead where ever it is and cut again call it xp2. Reform the necklace .the same happens as ...


3

Hint Pick $n-1$ objects in order from $m$. Send them via $f$ to $1,2,3,..,n-1$. Send all the remaining object to $n$. In how many ways can this be done?


3

Hint: choose single elements to map to each of $1, 2, \ldots, n-1$, and map everything else to $n$.


6

We have $13^3 > 2016$. So you want to count the numbers $\leq 2016$ that are divisible by at least one of: $2^3, 3^3, 5^3, 7^3, 11^3$. There is no overlap among these, except for those numbers that are multiples of both $2^3$ and $3^3$ (i.e., numbers divisible by $6^3$), and those numbers that are multiples of both $2^3$ and $5^3$ (i.e., numbers divisible ...


0

Let $p$ be a positive integer. We answer a more general question. Is the sum \begin{equation} S_p(n) = \sum\limits_{k=0}^{n-1} k \cdot (k+p)! \end{equation} given as a hypergeometric term plus a constant. We will be using Gosper's algorithm . Denote $t_n := n \cdot (n+p)!$. Calculate the ratio of the terms in the sum: \begin{equation} r_k = ...


3

You asked two different questions. The question in the title appears to be: "How many three digit numbers exist such that the third digit is the geometric mean of the other two?" I will answer the question in the statement of the problem: "How many three digit numbers exist such that one of the digits is the geometric mean of the other two?" ...


-1

Third digit is $\sqrt{ab}$ implies square root should yield single digit so it can be $[0,9]$ thus now digits whose product is 0 so $100,..900$ , for 1 is only $1,1$ so number $111$ then for 2 so product is $4$ thus numbers are $222,142,412$ then $3$ so $333,193,913$ now $4$ so $284,824,444$ then $5$ so $555$ then $6$ so $496,946,666,$ for $7$ so $777$ then ...


0

By finding the gcd of 13 and 17 you can find a solution to $13x-17y=1$. In this case $x=4,y=3$. Consider $N=3\cdot 4 \cdot 50$ $$13 \cdot 4 =52,17\cdot 3=51$$ $$x=50 \rightarrow x(x+1)(x+2)=50\cdot 51\cdot 52 = 13\cdot 17 \cdot (3\cdot 4\cdot 50)=13\cdot 17 \cdot N$$ Hence smallest value of $N=3\times 4\times 50 = 600$ Solutions for $13x-17y=2$ would be ...


1

Count how many nonprimes there are between $1$ and $50,$ then the minimum number of cards to pick is 3 more.



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