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2

There's one more solution (it isn't mine). One can even prove that $(a + b + c) \mid (a^n + b^n + c^n)$ for all $n=3k+1$ and $n=3k+2$. It's enough to prove that $a + b + c \mid a^n + b^n + c^n$ => $a + b + c \mid a^{n+3} + b^{n+3} + c^{n+3}$. The proof is here: https://vk.com/doc104505692_416031961?hash=3acf5149ebfb5338b5&dl=47a3df498ea4bf930e ...


-1

Here $34!$ contain $2^{32}$ and $5^7$. So $2^{25}$ remaining and $2^{7}\cdot 5^7 = (10)^7$ form $7$ zeros at the end. So $34! = 295232799cd9604140809643ab \times 10^7$ now after deleting $7$ zero,s , So we get $34! = 295232799cd9604140809643ab$ Now we use divisibility test for last $7$ digits using $2^7$. Here $(a,b)\in \{0,1,2,3,4,5,6,7,8,9\}$ So ...


4

You know that $34!$ is divisible by $9$, so, because the sum of all other digits is $141$, you know that $c+d=3$ or $c+d=12$. Now do the same for divisibility by $11$ (remember the alternating sums criterion?).


13

HINT: The number is a multiple of $9$ and a multiple of $11$.


1

Just another way to look at it, consider the Diophantine equation $x^2=3y^2-2$, or in a more Pell-like form $x^2-3y^2=-2$. From Pell theory, if you start with any integer solution $(x_1, y_1)$, you generate another integer solution by the recursion $x_2+\sqrt3y_2 = (x_1+\sqrt3 y_1)(2+\sqrt3)$. Thus $y_2 = x_1+2y_1 = 2y_1 + \sqrt{3y_1^2-2}$, which is ...


2

what you are missing is the automorphism group of the quadratic form $x^2 - 4 x y + y^2.$ On this site, the relevant observation is usually referred to as Vieta Jumping, which is a special case. The point is that a solution $(x,y)$ to $x^2 - 4 x y + y^2= k$ (in this case $k=-2$) can be replaced by a new solution $$ (y, 4y - x). $$ CHECK!!!!!!!! This means ...


2

REVISED SOLUTION: As you remarked, we have $$a_{n+1}^2-4a_{n+1}a_n+a_n^2=-2$$ Of course, we also have: $$a_n^2-4a_na_{n-1}+a_{n-1}^2=-2$$ Thus, $a_{n+1}$ and $a_{n-1}$ are roots of the quadratic $$A^2-4Aa_n+a_n^2+2$$ They are distinct because the $a_n$ are increasing. It follows that $$a_{n+1}+a_{n-1}=4a_n$$ Your desired result follows instantly. Note ...


0

This might not be much help, but I would show that if $a_n$ is positive, then $2 a_n$ is positive and then also show $\sqrt{3a_n^2-2}$ is positive. Since $a_1$ is positive, then all proceeding $a_n$ will also be positive.


2

The sum in question is a Riemann sum approximating $\int_0^1x\,dx$. But you don't need to know that. Draw a picture and you see the sum is the area of a bunch of rectangles. The area of the rectangles equals $T + P - N$, where $T=1/2$ is the area of the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$, $P$ is the green area, and $N$ is the red area. ...


1

$10$ is an element of the group $\left(\mathbb{Z}_{/19\cdot 53\mathbb{Z}}\right)^*$ whose order equals $18\cdot 52=936$. Lagrange's theorem for groups hence gives: $$ 10^{936}\equiv 1\pmod{19\cdot 53} $$ so if we take $936$ ones followed by a zero, we have for sure a multiple of $2014$.


0

Here's a more intuitive way to get the idea of considering powers of $2$. Note that $a+b+c\mid(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$. By The Fundamental Theorem of Symmetric Polynomials (FTSP), $a^n+b^n+c^n$ is an integer polynomial in $a+b+c$, $ab+bc+ca$ and $abc$. If $3\nmid n$, no term has degree divisible by $3$ so each term has at least one factor ...


6

Hint. Consider the remainders when dividing $1$, $11$, $111$, and so forth upto $\underbrace{111\cdots111}_{2015\text{ ones}}$, by $2014$. Apply the pigeonhole principle. Subtract.


3

Hint: Pigeon hole and the fact that the difference of two such numbers of same number of digits is of that form should do it.


2

Hint: The prime factorisation of 2014 is $2 \cdot 19 \cdot 53$, so it is enough to show that there is a number of the form 11111...11111 that is divisible by $19 \cdot 53$. Try to show that for every prime $p\geq 7$, we have that $p$ is a divisior of $$ \frac{10^{p-1}-1}{9} = \underbrace{11111...11111}_p$$


12

Claim. $a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$. Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that ...


1

As noted in the comments, the problem has been misread. Here's an (inelegant) solution the problem. This is one of those sad coordinate-geometry arguments that illuminates very little about the geometry going on here, but can be done with bullet-headed algebraic computation. If the statement of the problem is true, the constant value must be: ...


2

It seems that there's a partial solution. Suppose that $\mathrm{gcd}(a,a+b+c)=\mathrm{gcd}(b,a+b+c)=\mathrm{gcd}(c,a+b+c)=1$. Then for $n=k\cdot \phi(a+b+c)+1 \, (k=1,2, \ldots )$, where $\phi$ is Euler's function, we have: $$ (a^n+b^n+c^n)-(a^2+b^2+c^2)=a^2 (a^{n-1}-1) + b^2 (b^{n-1}-1) + c^2 (c^{n-1}-1), $$ where all round brackets are divisible by ...


4

In the following $p$ will always denote a prime numer. $p_k$ will be the $k$-th prime. Clearly it suffices to show that, given $0\le x_i<y_i\le\alpha$ ($i=0,\dots,n$), we can always find $x$ s.t. $\phi(x+i)\in(x_i,y_i)$ for any $i$. Here $\alpha:=\prod_{p\le n+1}\frac{p-1}{p}$. We first prove this crucial lemma, which in particular gives the thesis for ...


0

A solution by Sumit Ray ac=-bd a/b = -d/c = k a=bk and d=-ck a^2+b^2=1 => b^2 = 1/(k^2+1) => c^2 = 1/(k^2+1) Thus b^2 - c^2 =0 Now ab+cd = b^2*k-c^2*k = k(b^2-c^2) = 0


2

Let me try to cover the motivation for writing something like $$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}8 + \frac{1+z}8$$ for doing AM-GM, seemingly out of the blue. We obviously start with the first term. Now clearly, to get rid of the denominator, it would be great to consider the terms $(1+y)$ and $(1+z)$. As there is $x^3$ in the numerator, we can use these ...


2

I hope one of the $3$ parts below would help you understand the proof above. Part 1: the AM-GM ineq is: $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{1+y}{8}+\dfrac{1+z}{8} \geq 3\sqrt[3]{\dfrac{x^3(1+y)(1+z)}{(1+y)(1+z)\cdot 8\cdot 8}} = \dfrac{3x}{4}$. Part 2: $\sum_{cyclic} \left(\dfrac{1+y}{8}+\dfrac{1+z}{8}\right) = 2\sum_{cyclic} \dfrac{1+x}{8}= ...


0

The AM-GM inequality for three terms can be written $$a+b+c\ge3\sqrt[3]{abc}\ .$$ Just substitute your three terms on the LHS into this. That is, take $$a=\frac{x^3}{(1+y)(1+z)}\ ,\quad b=\frac{1+y}8\ ,\quad c=\frac{1+z}8\ .$$


4

Hint : Every natural number can be written in the form $2^{m-1}(2n-1)$ with unique positive integers $m,n$.


1

HINT : We have $$\frac 1n\lt\frac mk\lt n\iff \frac mn\lt k\lt mn\tag 1$$ Setting $\lfloor\frac mn\rfloor=s$ gives $$(1)\iff s+1\le k\le mn-1$$with $s\le\frac mn\lt s+1$. Hence, one has $50=(mn-1)-(s+1)+1$.


0

Perhaps it will help to see $$|\log m-\log k| < \log n \implies \log m-\log n < \log k < \log m+\log n \implies \log\frac{m}{n} < \log k < \log mn$$ Where $\log$ is monotonic.


2

Now that the question has finally been quoted properly, it can be answered. Let $x$ and $y$ denote the total amount of milk and coffee, respectively, in ounces. Then we have the linear system of equations $$ \frac17x+\frac2{17}y=8\;,\\ x+y=8n\;, $$ where $n$ is the number of family members. Solving for $x$ and $y$ yields $$ ...


-3

For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$. From these 2 inequalities: $a + b + c = ...


3

I think there is a typo as I read that book and own one and some typos were spotted by me as well. It should be: $a^2+\sqrt{a}+\sqrt{a} \geq 3\sqrt[3]{a^2\sqrt{a}\sqrt{a}} = 3a$


2

A classical problem in simultaneous approximation. Let $a_n=(\lambda n\pmod{\pi},\mu n\pmod{\pi})\in\mathbb{T}$ and let we fix some huge $N$. Let $Q_n\subset\mathbb{T}$ be a square centered in $a_n$ having side length $\frac{\pi}{\sqrt{N}}$. Among $Q_1,\ldots,Q_N$, at least two squares have to overlap, since the sum of their Lebesgue measures equals ...


6

$$\sum_{n=2}^{+\infty}\frac{1}{j^n}=\frac{1}{j(j-1)},\tag{1}$$ hence the problem boils down to computing: $$\begin{eqnarray*} \sum_{j\geq 2}\frac{1}{j(j-1)j!}&=&\sum_{j\geq 2}\frac{1}{(j-1)j!}-\sum_{j\geq 2}\frac{1}{j\cdot j!}\\&=&\frac{1}{2}+\sum_{j\geq 2}\left(\frac{1}{j\cdot(j+1)!}-\frac{1}{j\cdot ...


3

Let $D, L$ be the obvious random variables. Then the desired probability is \begin{align} P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\ & + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\ & + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\ & + \cdots \end{align} Each row constitutes an element in a ...


0

HINT: $$ac+bd=0\iff\dfrac ad=\dfrac b{-c}=\pm\sqrt{\dfrac{a^2+b^2}{d^2+(-c)^2}}$$ But if $a^2+b^2=d^2+c^2,$ not necessarily $=1$ $$\dfrac ad=\dfrac b{-c}=\pm1$$ So, either $a=d,b=-c$ or $a=-d,b=c$ The result should follow immediately.


0

Now since $a^2+b^2=c^2+d^2=1$ (a,b) and (c,d) are points on unit circle. You can think of these ordered pairs as vectors, since their dot product is zero. Then this means they are orthogonal. Now A x B gives area of triangle under first vector and C x D gives area under second vector. These triangles are similar since angles are all same, so we can use ...


1

If $b=0,ac=0\implies c=0\implies ab+cd=0$ Else $ac+bd=0\iff ac=-bd\iff\dfrac ab=\dfrac{-d}c=k$(say) $\implies a=bk, d=-ck$ If $a^2+b^2=c^2+d^2,$ not necessarily $=1$ $b^2(1+k^2)=c^2(1+k^2)\implies b^2=c^2$ if $1+k^2\ne0$ Now $ab+cd=(bk)b+c(-ck)=k^2(b^2-c^2)=?$


23

$$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$


2

If $D$ and $L$ denote the number of rolls needed by Dave and Linda and $E$ denotes the event then:$$P(E)=P(D=L)+P(D=L-1)+P(D=L+1)$$ This with $P(D=L-1)=P(D=L+1)$ on base of symmetry, and with: $$P(D=L)=\sum_{k=1}^{\infty}P(D=k\wedge L=k)=\sum_{k=1}^{\infty}P(D=k)P(L=k)$$For $P(D=L-1)$ you can also find such an expression. edit: ...


10

I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$. We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = \cos(\alpha - \beta) = 0$. You have ...


5

You can interptet $(a, b)$ and $(c, d)$ as two orthogonal vectors that lie on the unit circle. Converting this into polar coordinates, this means there are angles $\phi$, $\theta$ such that $(a, b) = (\cos(\phi), \sin(\phi))$, $(c, d) = (\cos(\theta), \sin(\theta))$ and $|\phi - \theta| = \frac{\pi}{2}$. Now observe that $$ab + cd = \cos(\phi)\sin(\phi) + ...


2

Small variation of the answer of loup blanc that does not mention minimal polynomials, and produces an actual annihilating polynmial of degree $r+1$. Note that is is irrelevant that the vector space is over$~\Bbb C$. The image space $W$ of $M$ (or column space if you prefer) has dimension $r$ by definition of the rank. The characteristic polynomial $Q$ of ...


3

Let $D$ denote the random variable telling how many rolls Dave took to roll a $6$, and let $L$ denote the variable telling how many rolls Linda took. Then what you calculated were $P(D=k)$ for certain values of $k$. For example, the probability that Dave took $5$ rolls to roll a $6$ is $$\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac16$$ What you ...


1

Let the vectors be $u_k = (x_k,y_k)$ for $1\leqslant k \leqslant 2n$. Consider first the case where $x_k \geqslant 0$ for all $k$. Using $\lvert y_k\rvert \geqslant 1 - \lvert x_k\rvert$ for unit vectors, we have $$\sum_{k = 1}^{2n} y_k \geqslant \sum_{k = 1}^{2n} (1 - x_k) = 2n - \sum_{k = 1}^{2n} x_k,$$ so $\sum\limits_{k = 1}^{2n} y_k$ is a ...


0

$$m=n=1 \Rightarrow f(1)=1$$ so we have $f(3)\ge 2,f(37)\ge 2$ since $f(3)\neq f(37)$ $$f(999)=f(27)\cdot f(37)=(f(3))^3 \cdot f(37)\ge 2^3\cdot 3=24$$ Let us define $f(2)=37,f(3)=2,f(37)=3$ and $f(p)=p$ for all primes $p\neq 2,p\neq 3,p\neq 37$ Let us show the function is injective. The condition implies $f(\prod_{i=1}^k p_i^{e_i})=\prod_{i=1}^k ...


2

f(999)=f(37)*f(3)*f(3)*f(3) Now, f(3) or f(37) cannot be 1. Because, if, for example, f(3)=1, then f(999)=f(37) which implies 999=37 (because the function is one-to-one). Therefore, f(3) can have lowest value 2. And hence f(37) can have lowest value 3. Thus, f(999)=3*2*2*2=24.


0

The probability is going to be: $\frac{\binom{N}{k}{m\brace k}k!}{N^k}$. Where $\binom{N}{k}$ is a binomial coefficient, $n\brace k $ is a stirling coefficient of the second kind and $k!$ is a factorial. You can compute each of these using their own recursions in a $100\times 100$ matrix and a $100$ array for the factorial. I recommend java BigIntegers. ...


4

Borrowing @Rory Daulton's notation, construct a second square directly beneath the original, and extend the line $\overleftrightarrow{DC}$ to meet the lower square at vertex $K$. The circle centered at $B$ with radius $BE$ passes through points $E$, $G$ and $K$. Therefore the inscribed angle $\angle EKG$ is half the central angle $\angle EBG$, since both ...


5

Place the square on a Cartesian coordinate grid. We can choose the units so the square is a unit square. The coordinates of the vertices $B,F,D,E$ are then obvious. (I limited this diagram to only what is necessary for my solution: your diagram has unneeded line segments and not enough labels for the points.) Point $C$, the midpoint of $\overline{BF}$, ...


2

You have to look for a number $k$ such that between $k^3$ and $(k+1)^3$ there are exactly $70$ multiples of $k$. Since $k^3$ and $(k+1)^3-1$ are multiples of $k$ this gives $$[(k+1)^3-1]-k^3=69k$$


2

Consider the student with the most friends. If he's friend with at least 12 people, your problem is solved. If he is not, there is another student who's not his friend. Let's name those two students A and B. A having strictly less than 12 friends, there is at least 12 people different from B he's not friend with. Any of these people are not friend with A, ...


2

Here is perhaps a quick way. By the AM-GM-HM inequality, it must always be the case that $H_n < H_{n+1} < G_{n+1} < A_{n+1} < A_n$, which resolves the sequences of $H_n$ and $A_n$. It is also easy to see that $A_{n+1} H_{n+1} = A_n H_n$, so we also readily get $G_{n+1} = G_n$.


1

Let $2$ distinct positive no. be $a$ and $b\;,$ Here $A_{1}\;,G_{1}$ and $H_{1}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$ of these two positive no., So $$\displaystyle A_{1} = \frac{a+b}{2}\;\;,G_{1} = \sqrt{ab}\;\;,H_{1} = \frac{2ab}{a+b}$$. And given $A_{n}\;,G_{n}$ and $H_{n}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$ of no.,s $A_{n-1}$ and $H_{n-1}.$ ...



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