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0

For $\forall x\in R, h>0$, by the Mean Value Theorem for integrals, there is $c\in(x-h,x+h)$ such that $$ \int_{x-h}^{x+h}f(t)dt=f(c)\cdot2h $$ and hence $f(c)=0$. Letting $h\to0$ implies $c\to x$. by the continuity of $f$ at $x$, we have $$ f(x)=\lim_{c\to x}f(c)=0. $$ Since $x$ is arbitrary, $f(x)\equiv0$ for all $x\in R$.


0

If $f$ is not zero, there is an $x_0 \in \mathbf{R}$ ($f$ is necessarily defined on $\mathbf{R}$) such that $f(x_0) \not = 0$. Switching to $-f$ if needed we can assume that $f(x_0)>0$. As $f$ is continuous, with $\varepsilon = f(x_0) / 2$ we can find $\eta>0$ such that $|f(x) - f(x_0)| < \varepsilon$ whenever $|x-x0| \leq \eta$. Then for $x\in [x_0 ...


2

Let $a$ a fixed real and $$F(b)=\int_a^bf(x)dx$$ then by the fundamental theorem of analysis we have $$0=F'(b)=f(b),\quad \forall b$$


1

This was my submission. Say, for sake of contradiction, that $f(x)$ is non-zero for some real-valued $x=t$. That is, $ f(t) \ne 0 $. If $f(x)=c$ for all $x$, then $$ \displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x = \displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x = \displaystyle\int_{a}^{b} c \, \mathrm{d}x = c \cdot \left( b - a \right), $$which is non-zero ...


1

Let $$ F(x) = \int_0^x f(t) \, dt. $$ By assumption, $F \equiv 0$. By the fundamental theorem of calculus, $F$ is an antiderivative of $f$, i.e. $$ f = F' \equiv 0. $$


3

Hint: Suppose by contradiction that $f(c) >0$, continuity provides you a $\delta >0$ such that $f(]c-\delta,c+\delta[)\subset \{t \in \Bbb R\mid t> \epsilon\}$ for some $\epsilon$ small enough. Then get a contradiction.


1

Assume for a contradiciton that $f$ is not identically zero, so we have an $x$ such that $f(x)\neq 0$, without loss of generality say $f(x)>0$. Since $f$ is continuous you can take a small ball of radius $\epsilon$ around $x$ where $f$ takes strictly positive values. Integrate over $[x-\epsilon, x+\epsilon]$ and the integral must be strictly positive. ...


2

Hint: for any fixed $x$, take $a=x$, $b=x+h$, and take the limit as $h\to 0$ (applying the fundamental theorem of calculus)..


3

You can use Vieta's but not right away! Here is my solution, which also is the official one. First, note that $(a,b)=(1,-1)$ is an easy solution. Then, let $z=a-bi$. Multiply the second equation by $-bi$ and multiply by the first and we find $z^3+z^2+z=-1+5i$. Since $(1,-1)$ is a solution, this motivates us to factor out $ z - \left( 1 + i \right) $. We find ...


2

Several commenters say you’ve misquoted a contest problem. But regardless, you asked a question that has an answer, and the answer is no. Call the bits of the original sequence $b_i$, so the original word is $b_1b_2\dots b_{2015}$. A transposition $b_i\leftrightarrow b_j$ turns the sequence into the sequence $b_{\tau(1)}b_{\tau(2)}\dots b_{{\tau(2015)}}$, ...


2

Assume that the bits are given to you already sorted (This problem is not harder). label the bits $b_1,b_2\dots b_{2015}$ where $b_1\leq b_2\dots \leq b_{2015}$ So that the sequence is $b_1b_2\dots b_{2015}$ To make a move the mathematician must select position $A$ and position $B$,so that if $A$ is $1$ and $B$ is zero they swap, and nothing happens ...


12

For your question, the answer is no. The output string is a known permutation of the input string. If we could say two given bits were equal at the end, we could reverse the permutation and say two given bits were equal at the beginning, which we cannot.


7

You have translated the problem incorrectly. If A has an electron and B does not, then the electron jumps from A to B. This is completely different from just swapping two bits! The swap only happens when Bit A is 1, and bit B is 0, and you can designate which bit is A and which bit is B. For that, seems like a bubble sort type of approach will do (just ...


1

One way of doing it: long long and(long long a, long long b) { long long x = a^b; long long s = x>>1; while (s) { x = x|s; s >>= 1; } return a&b&~x; } First, the idea is exactly Ross'; starting from the highest bit, check if $\operatorname{bit}_k(a) = \operatorname{bit}_k(b)$ in which case the ...


4

The only bits that will be $1$ will be bits that are common to the upper bits of $A$ and $B$. Everything else will have at least one instance of a $0$ in that range. So just start from the high order bit downwards. Output the matching bits. As soon as you hit a disagreement between the binaries of $A$ and $B$ (which will be $0$ in $A$ and $1$ in $B$) ...


0

It's not that complex. Hammer and Nail = $1.10. Nail = .05 Hammer = 1.05 (1 dollar more than nail) Total: $1.10


2

Just another way: sum the AM-GMs... $$2x^2+2k^2 \ge k\cdot 4 x,\quad 3y^2+\frac34k^2 \ge k \cdot 3 y, \quad 4z^2+ \frac14k^2 \ge k\cdot2z$$ to get $\dfrac1k+3k \ge (4x+3y+2z)$, and the maximum is achieved when $\frac1k=3k\implies k = \frac1{\sqrt3}$, for a maximum of $2\sqrt3$.


5

Apply Cauchy-Schwarz Inequality: $\left(4x+3y+2z\right)^2 = \left(2\sqrt{2}\cdot \sqrt{2}x+\sqrt{3}\cdot \sqrt{3}y+1\cdot 2z\right)^2 \leq \left((2\sqrt{2})^2+(\sqrt{3})^2+1^2\right)\cdot\left(2x^2+3y^2+4z^2\right) = 12 \Rightarrow 4x+3y+2z \leq 2\sqrt{3} = \text{Maximum Value}$.


0

This problem is same as the problem SGARDEN of codechef July Long Challenge 2014.


0

WLOG, we can set $\sqrt2x=\cos A,\sqrt3y=\sin A\cos B,2z=\sin A\sin B$ $\implies4x+3y+2z=2\sqrt2\cos A+\sqrt3\sin A\cos B+\sin A\sin B$ $=2\sqrt2\cos A+\sin A(\sqrt3\cos B+\sin B)$ $=2\sqrt2\cos A+2\sin A\cos\left(B-\dfrac\pi6\right)$ If $\sin A\ge0,$ $4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\le1$ The equality occurs if ...


0

It is obvious that the first equation stands closed surface, so when $4x+3y+2z=k$ gets the minimum, the line just touches the closed surface. Let me put the tangent point $A(x_0,y_0)$, and then I get the partial derivative of $2x_2+3y_2+4z_2=1$ about $X$ and $Y$ at the point $A$. I know I can find a line that coincides with the given line. After this I can ...


2

For me, $4x + 3y + 2z = \langle\begin{pmatrix} \sqrt{2} x \\ \sqrt{3}y \\ 2z\end{pmatrix}, \begin{pmatrix} 2\sqrt{2} \\\sqrt{3} \\\ 1\end{pmatrix}\rangle \leq 1\times 2\sqrt{3}=2\sqrt{3}$


2

First, change the ellipsoid to a sphere by writing $v = x \sqrt{2}, w=y\sqrt{3}, u = 2z$ THe function to be maximized will still be linear, now in $u, v, w$. $$ a_u u + a_v v + a_w w $$ Now proceed in the direction perpendicular to that plane, and lets look at a lihne thru the origin in that direction, which could be parameterized as $$ u = s/a_u \\ v = ...


1

Let $K$ be the area of the triangle, let $a,b,c$ be the sides with altitudes of $2014,1,h$ respectively. Then, $K = \dfrac{1}{2} \cdot a \cdot 2014 = \dfrac{1}{2} \cdot b \cdot 1 = \dfrac{1}{2} \cdot c \cdot h$, so $a = \dfrac{2K}{2014}$, $b = 2K$, and $c = \dfrac{2K}{h}$. Thus, this triangle has a semiperimeter of $s = \dfrac{1}{2}(a+b+c) = ...


1

Each move changes the parity of all three colors and reduces the total number of chips by one. Since the final state has the numbers of colors disagreeing on parity the initial state must not have the colors all agree. The final color will be the odd color out in parity at the start. In fact, the total number of moves is one less than the starting number ...


8

As mentioned on Kiran Kedlaya's webpage: Those [official solutions] appear in the official exam summary, along with results and statistics, in the American Mathematical Monthly sometime in the year following the competition, usually in October. (One typically also finds solutions in Mathematics Magazine in early spring.) A set of unofficial solutions, ...


1

Notice $A=1$ ,there are $3$ possible values for $E$ since $E$ is odd and between $0$ and $5$. once $E$ has been fixed we can easily obtain $D$. once $D$ has been obtained we have the last two digits so we can obtain $C$. Also notice from $E$ you can obtain $B$ immediately. Thus there are three cases(three possible values of $E$), they are: $11311$ This one ...


0

$A=1$, because it won't be a five digit number if $A=0$ and those are the only remainders when dividing by $2$. $C$ and $E$ are then odd, $C=1,3$ and $E=1,3,5$. $B\lt 3$ and $D\lt 5$ We are down to $90$ possibilities, within range of brute force on a contest.


1

Noting that $\;\displaystyle\sum_{j=1}^k j^2=\frac16k(k+1)(2k+1)=\frac1{24}(2k)(2k+1)(2k+2)\;$ may help. The cubic root of $24\,N$ will thus be near of the odd number $2k+1$ (alternative formulation : the cubic root of $3\,N$ would be nearly $\,k+\frac 12$). To translate this in a numerical algorithm we may use (for $x:=2k+1$) : $$\sqrt[3]{x^3-x}\approx ...


0

Here's an $O(n \log n)$ algorithm that works for all possible sets $A$ and $B$. Let $n = |A| + |B|$. This question has been asked in other platforms (here or here), but none of them give satisfactory answers (...at least to me. I cannot verify that they work in the claimed time for all possible inputs. For example, using KD tree does not guarantee $O(n ...


0

The sum of $k$ first squares is $$\frac16k(k+1)(2k+1)$$ Since this increases as $N$ does, you "only" have to solve $$\frac16k(k+1)(2k+1)=N$$ Alas, this is a cubic equation. This makes things a bit difficult, but not impossible. Can you manage?


1

The Putnam competition is offered every year on the first Saturday in December. It is limited to undergraduates, but not limited to math majors - any undergraduate in the US is eligible to take it. However, if you intend on taking the exam you need to let your math department know, preferably sometime in October or so; the Putnam committee only sends as ...


1

You are right. We can assume that our polygon $P$ is convex, since if $P\subset\Gamma $ for some circle $\Gamma$, then $\operatorname{Conv}(P)\subset\Gamma$, too. Assuming that $\Gamma_1$ is a minimal solution with its center outside the axis of symmetry, then the symmetric of $\Gamma_1$ with respect so such axis, $\Gamma_2$, is also a minimal solution, ...


1

The ones digit has to be even, which will make the final number divisible by $2$. If the ones digit is $2$ or $6$, the tens digit needs to be odd so the final number is divisible by $4$. If the ones digit is $0,4,8$ the tens digit needs to be even. Then pick the hundreds digit (you can find a similar rule) so the whole thing is divisible by $8$ Of ...


0

There are three non-symmetries between the LHS and RHS of your equation: $x\to-x$ , $y\to-y$, and $x^{2}\leftrightarrow2y^{2}$. From $y\to-y$ after you subtract and make $x=0$ you get $f^{2}(y)=f^{2}(-y)$ for every y (see barto's comment). From $x\to-x$ after you subtract and make $y=0$ you get $f(0)(f(x)-f(-x))=0$. This implies that $f$ is always even, ...


4

You’re assuming that $\Delta x,\Delta y,\Delta z\ge 0$ but that doesn’t cover all of the possibilities. The diagonal that runs from $\langle 1,4,4\rangle$ to $\langle 4,1,1\rangle$, for instance, isn’t covered by any of your cases.


1

Just reading your solution, when I got to the part about $x = 4$, then $\Delta x = -1$ is a permissible condition and I stopped right there. Reading a bit further, we can also immediately see that your enumeration is flawed in a second way, which is that the choices of $(x, \Delta x)$, $(y, \Delta y)$, and $(z, \Delta z)$ are not independent of each other, ...


1

You can think of it in a physical sense instead of math. For each pair of opposite faces, there are 16 lines that can go through the $4\times4$ cube perpendicularly. So that's $3\times 16=48$. And for each pair of diagonally opposite edges, there are 4 lines that will go through 4 lattice points. So that's $6\times 4 = 24$ Then for each opposite pair of ...


0

It's 28. a+24b=n for all n has solutions since gcd(1,24)=1.Thus,a+24b=143z for all z has solutions in integers. For non- positive z obviously the equation cannot have solution in postive integers.Thus z is positive. For z=1,selecting the largest b such that 24b<143 we have b=5 and thus 23 + 24(5)= 143.For z>=5 we have 1+24(27)< 143(5)=715. Thus all ...


-1

Here's a brute force approach: start with $a+11b=13k$ and $a+13b=11m$, with $k,m\in \mathbb N$. Solve it in $a,b$ and you get: $a=84.5k-60.5m$ $b=-6.5k+5.5m$ $a+b=78k-55m$ So $a$ and $b$ are integers iff $k$ and $m$ are both odd or both even; in this subspace for $(k,m)$, just minimize $78k-55m$.


-1

Given that H+N = 1.1 and H = N+1 we can solve for N for both. H = 1.1-N AND H = 1+N so 1.1-N = 1+N, then 1.1 = 1+2N, so 0.1 = 2N, therefore .1/2 = N, thus .05 = N. Substitute .05 for N either original equation and H = 1.05.


1

Define a sequence $(b_n)$ by $b_0=1$ and $b_n=b_{n+1}+b_{n+1}^3$ for every $n\geqslant0$, then $(b_n)$ is decreasing to zero and, for every $n\geqslant1$, $$N(x)=n\iff x\in(b_n,b_{n-1}].$$ Furthermore, $$\frac1{b_{n}^2}-\frac1{b_{n-1}^2}=\frac1{b_{n}^2}\left(1-\frac1{(1+b_n^2)^2}\right)=\frac1{b_{n}^2}\left(1-(1-2b_n^2+o(b_n^2))\right)=2+o(1),$$ when ...


0

HINT: Let $a,b,c$ be the roots of the cubic $f(x)=x^3-px^2+qx-r=0$ and let $s_k=a^k+b^k+c^k$ Then evaluate $x^kf(x)$ at $a,b,c$ and add to obtain the recurrence $s_{k+3}-ps_{k+2}+qs_{k+1}-rs_k=0$ Note that $s_0=3$. Now you need to work out how to use the data you have and keep the calculations under control.


1

The polynomial $f(x)=x^6-x^3+1$ has six distinct complex roots, and annihilates $A$. Hence the minimal polynomial of $A$ has distinct complex roots (we don't know how many, but it it is at most six). Hence $A$ is diagonalizable, because having only linear terms in a minimal polynomial is a characterization of diagonalizability. Addendum: ...


4

One may adopt the approach as in Pranav Arora's comment. But this approach involves a double integral whose calculation seems painful. So here is an indirect approach that makes calculation slightly easier (at least to me): Let us consider the following integral: for $\alpha, \beta \in \Bbb{C}\setminus(-\infty, 0]$ and $0 < s < 1$, $$ I = ...


5

If they are all odd, then their remainders mod 4 must alternate 1 and 3, and it is impossible for an odd cycle to alternate. So $p=2$ must appear. It can appear $n$ times; or $n-1$ times and 7 appear once. If they must all be different, then the neighbours of 2 must both be $3\mod 4$, and again alternation is impossible.


2

$$\int f f' dx=\int f df=1/2f^2$$ Thus $$\int_0^1f(x)f'(x)dx=f(x)|_0^1=1/2f(1)^2$$ since $f(0)=0$. Let $g(x)=1$ then by Cauchy's theorem $$\left|\int (f')^2 dx\right|\left|\int g^2 dx\right|\geq\left(\int f'g\,dx\right)^2=(f)^2$$ Thus since $\int_0^1 g(x)dx=1$ $$1/2\int_0^1 (f')^2 dx\geq1/2f(1)^2=\int_0^1 f f' dx$$ The only part I did not get was the ...


0

Hint: Let's $k$ is the number of columns with product equal $-1$, then $n-k$ is the number of columns with product $1$. Now if $\sum_{j=1}^n r_j+\sum_{k=1}^{n} c_k = 0$, then there are $k$ rows with product $1$ and $n-k$ rows with product $-1$. Now try to find number of $-1$ in the matrix with two different ways. Two results will have different parity.


3

Hint: For a matrix $M$ let $K(M)=\sum r_j+\sum c_k$. Then if $M'$ is identical to $M$ except that entry $ab$ is flipped, try to prove that $$K(M')\equiv K(M) \!\!\mod 4$$ Can you see how to finish this problem?


0

Very partial solution. I show that $f$ is even when $f(0)=0$. I use a comment of @soulEater: in this case we have $$(f(x+y)+f(x-y))(f(y)-f(-y))=0$$ Now suppose that there exists an $y_0$ such that $f(y_0)-f(-y_0)\not =0$, put $L=2y_0\not = 0$. We have $f(x+y_0)=-f(x-y_0)$ for all $x$, hence $f(x+L)=-f(x)$ and $f(x+2L)=f(x)$ for all $x$ (and $f$ is ...



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