# Tag Info

1

If there are $n$ question is the bank and we have already picked our 8 for the first time, then the $n$ questions are split into two categories: the 8 chosen the first time and the remaining $n-8$ not chosen the first time. Thus the question can be restated as: what is the smallest $n$ for which $$... 0 Hint: In the second draw, the four repeated questions should be drawn from the first 8. This you can draw it in {8\choose4}. Assume there are n questions in the test bank. The other 4 questions in the second draw should come from n-8. The number of ways that could happen is{{n-8}\choose4}. Extending the same to i = 5,6,7,8 and the total number of ... 0 In a game context, we should probably interpret the question as asking that the probability of 4 or more duplicates is less than 5\%. And in a game context, for successive games, we would presumably not choose questions at random. But the problem seems to ask us to assume that for each new game, the 8 questions are chosen at random from the question ... 1 Let n be the number of questions. We will calculate the chance that exactly four questions are repeated out of eight. There are {n-8 \choose 4}{8 \choose 4} ways to choose four matching and four non-matching, and n \choose 8 ways to choose the questions overall. So we want$$\frac {{n-8 \choose 4}{8 \choose 4}}{n \choose 8}=0.05$$I find 36 gives ... 0 I think 57 or more and at least 100 chocolates is enough to get the conclusion, if there is at least one chocolate eaten per day. At the end of every day there is some total number of chocolates that has been eaten. The set of all totals includes 0 and 100, and 57 other distinct integers from 1 to 99. In terms of this set, the question is... ... 0 This is only the solution to part 4. I assume that you have seen solutions to the rest. If not, I strongly recommend reading Ewan's answer. Solution 1: Apply Vieta's root jumping technique with a base case of (4, -1). The result is immediate. This also explains why in Will's answer, you see that the b and a have the same (absolute) value. Note that ... 2 Or, solutions given as \frac{a}{b}, beginning with a fake one with denominator zero,$$ \frac{1}{0},\frac{0}{1},\frac{-1}{4},\frac{-4}{15},\frac{-15}{56},\frac{-56}{209},\frac{-209}{780},\frac{-780}{2911}, $$As you can see, both the sequence of numerators and the sequence of denominators satisfy$$ x_{n+2} = 4 x_{n+1} - x_n. $$The way I found this ... 3 A clever trick (or a standard trick if you're familiar with using number fields to solve problems like these) to observe that$$ a^2 + 4ab + b^2 = (a - b \alpha)(a - b \beta)$$where \alpha and \beta are the two roots of the equation x^2 + 4x + 1 = 0. 0 Here is an idea I received from a friend. Use \displaystyle 1/2\int_{0}^{2\pi}\left[log(a-e^{ix})+log(a-e^{-ix})\right]^{2}dx Then, use Gauss's Mean Value Theorem: \displaystyle 2\pi f(a)=\int_{0}^{2\pi}f(a-e^{\pm ix})dx with f(z)=log^{2}(z). This means that \displaystyle 1/2\int_{0}^{2\pi}log(a-e^{ix})dx+1/2\int_{0}^{2\pi}log(a-e^{-ix})dx=2\pi ... 3 It seems that http://oeis.org/A094190 doesn't provide complete list of such numbers. Complete list (without leading zeros is here: http://oeis.org/A003226). If not consider trivial 0 and 1, then for each n there are 2 n-digital such numbers, that last n digits of their square are the same that number (sometimes solution has leading zeroes):$$ ...

4

It suffices to find the solutions to $x^2 \equiv x \mod 16$ and $x^2 \equiv x \mod 625$. As these polynomials are degree two and we're looking at solutions mod prime powers, there are at most two solutions to each by Hensel's lemma. Each has only the solutions $x \equiv 0, 1$. These extend via CRT to solutions $0, 1, 625, 9376$. The OEIS lists a sequence of ...

0

Since $S=\frac{abc}{4R}$ and $2m_a^2=b^2+c^2-\frac{a^2}{2}$, we can put the given inequality in the following form: $$\sum_{cyc} a\cdot(m_a-R)^2 \leq \sum_{cyc}a\cdot\left(R^2-\frac{a^2}{4}\right).\tag{1}$$ Let $O$ be the circumcenter of $ABC$ and $\Gamma_A$ the circle having center $A$ and radius $R$: obviously $O\in\Gamma_A$. Let $M_A$ be the midpoint of ...

0

Perhaps a simpler way to see the solution: Let there be an optimal motorway connecting all the cities. It is clear that the road from A to C must lie within the square, similarly the road from D to B. Hence these roads must intersect in possibly multiple points. Let the first such point be P and the last Q. (A diagram may help at this point). Now, it ...

3

For this to be true, we need to specify that $p$ has integer coefficients: without this assumption, $p(x) = \frac16x(x-1)(x+1)$ is a counterexample, with roots at $-1,0,$ and $1$, but $p(2)=1$. Suppose a polynomial $p(x)$ with integer coefficients has three or more distinct integral roots. This means that $p(x) = (x-a_0)(x-a_1)(x-a_2)q(x)$, and $q(x)$ also ...

4

I think that the original question (where coefficients need not be integral) is false. A counter example would be $$f(x)=(x -(a-1))(x-(a-2))(x-(a-3))\left(x-\left(a-\frac{1}{2}\right)\right)\left(x-\left(a-\frac{1}{3}\right)\right)$$ It has three distinct integral roots $a-1,a-2,a-3$, and it also satisfies $f(a)=1$.

0

Given @Henning Makholm's input, I have to revise my answer to an X connecting the four cities, with an intersection in the middle. This problem is reminiscent of the famous traveling salesman problem. If I now actually understand this question ;p then it's interesting how miserably greedy algorithms would fail. As for adding more details, I think ...

1

I will give a brief review to Yiyuan's answer, and then complete the proof. We have: $$3^x-5^y=z^2$$ Working $\pmod4$, we have $(-1)^x-1\equiv 0,1 \pmod4$, so $x$ has to be even. Substituting $x=2k$ and $y=a+b$: $$(3^k-z)(3^k+z)=5^y\implies 3^k-z=5^a, 3^k+z=5^b\implies 2\times3^k=5^a+5^b$$ If $a,b\ge1$, then the RHS is a multiple of $5$ but the LHS is not, ...

0

Hint  The odd terms show that for $\,c = b/a\,$ we have $\, b c^{2n+1}\!\in \Bbb Z\,$ for all $\,n\in \Bbb N,\,$ hence $\,c\in\Bbb Z\,$ (else, by Euclid's Lemma, $b$ is divisible by unbounded powers of $c$'s reduced denominator $> 1).\,$ Hence $\,c\in\Bbb Z\,\Rightarrow\,a\mid b.\,$ Similarly $\,b\mid a\,$ from the even terms. Remark $\$ Domains are ...

0

One way to go about this is to use the identity: $$\int_{0}^{\frac{\pi}{2}}\cos^{p-1}(x)\cos(ax)dx=\frac{\pi}{2^{p}}\cdot \frac{\Gamma(p)}{\Gamma\left(\frac{a+p+1}{2}\right)\Gamma\left(\frac{p-a+1}{2}\right)}....(1)$$ Then, diff this twice w.r.t 'a', and let a=0. Then, diff twice w.r.t p and let p=1. The diffing on the right side may be a little tedious, ...

2

Hint: Each tile must be moved an odd number of times. Hint: Each tile must be moved more than once. Hence, each tile must be moved at least 3 times. Hence, the number of moves is at least $\frac{4 \times 3 } { 2} = 6$. We still need to establish that 6 is possible. Trial and error suffices.

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1

Hint: Let $I(p)=\displaystyle\int_0^\infty\frac{dx}{x^2+p}$ , and then try to express $\displaystyle\int_0^\infty\frac{dx}{(x^2+p)^{n+1}}$ in terms of $I^{(n)}(p)=$ $=(-1)^n\dfrac{(2n-1)!!}{2^{n+1}p^n\sqrt p}\cdot\pi$.

2

Well, as an alternative (and I promise, none of the dreaded complex analysis stuff), we could use Parseval's theorem for Fourier transforms: For example, the FT of $(\sin{x}/x)^2$ is $$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$ ...

2

Consider $$I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx$$ Differentiate it twice. Since $$\int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k}$$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$I(a)=\frac{\pi}{4}(-1+2a+e^{-2a})$$ It is remains to substitute $a=1$.

0

Clearly, the net velocity vector will be larger when the hands' velocity vectors are perpendicular than when they are aligned, hence 12:15.

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1

This can certainly be proven using graph theory. In fact, this is a classical result related to digraphs called (appropriately) tournaments. Your problem is equivalent to that of finding a Hamiltonian path in the tournament. This statement is proven right in the Wikipedia article on tournaments which I linked to above. For completeness, let me reproduce the ...

0

How is similarity for degenerate triangles determined? One may assume that a deg. triangle is a line segment, so they are all similar. Or are they different, in which case the side ratios have to be considered? How will MAA deal with this problem? Degenerate Triangles are triangles, after all, and if commonsense was sufficient for #13, the writers would not ...

1

After a straightforward application of the formula $\cos{2A = 2\cos^2{A} - 1}$ to the right side and some algebraic simplification, we obtain $$\cos{(2x)}\cos{\left( \frac{2014\pi^2}{x^2} \right)} = 1$$ Note that we cannot have $\cos 2x \in (0, 1)$ for that would mean that the other factor was greater than 1, clearly impossible. Similarly, we cannot have ...

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