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This may be over your head if you're in high school, but you want the maximum order of an element in the symmetric group on 12 letters. (In other words, you want Landau's function evaluated at 12.) This is 60, generated by the product of a 5-cycle, 4-cycle, and 3-cycle. (This is surely the way the people who wrote the test thought about this problem!) If ...

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You can't do better than $75\%$ because Sally's guess is now independent of all other events, and you can't change that by anything you do on your own. No matter what strategy you pursue, you have a $50\%$ chance of guessing Sally's coin correctly. The original strategy was able to do better than $75\%$ because Sally's guess was dependent on her coin. Now ...

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If $DH$ is the trapezium height, notice that $$EF=DH=\sqrt{DA^2-AH^2}=\sqrt{15^2-2.5^2}.$$ On the other hand, if $r$ is the circle radius, we have: $$EF=OF+OE=\sqrt{r^2-10^2}+\sqrt{r^2-7.5^2}$$ By comparing and solving for $r$ one gets $r^2=135$.

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From the data, we see that this cyclic quadrilateral is an isosceles trapezium. Splitting the trapezium into two triangles, sharing a common side $x$ which subtends angles $\theta$ and $180-\theta$ in each triangle respectively, we can apply the cosine rule and get $$x^2=625-600\cos\theta=450-450\cos(180-\theta)$$ from which we can deduce that $$\cos\theta=\... 2 The maximum sum of three square digits is 9^2+9^2+9^2=243, so first digit of any solution must be 0,1,2. But that means that the maximum sum of three square digits in any solution is 2^2+9^2+9^2=166, so first digit must be 0 or 1. If the number is in the range 100-166, then the maximum sum of its squared digits is either 1^2+6^2+6^2=73 if the 2nd ... 2 As p+q+r+s+t = 5r and q+r+s = 3r we have$$ 5r = m^2, \quad 3r = n^3 \tag1 $$for some integers m and n. Thus 5\mid r and 3^2\mid r, so r = 5\cdot3^2\cdot r_1 for som integer r_1. Substitute it to (1):$$ 5^2\cdot 3^2\cdot r_1 = m^2, \quad 3^3\cdot 5\cdot r_1 = n^3. \tag2 $$Thus r_1 is a perfect square (r_1 = m_1^2, where m_1 = m/15)... 1 Since \frac{3}{5}x^2=y^3 is a cube, the number of factors of 5 in x^2 must be even and one more than a multiple of 3. So choices are 4 and 10. Also the number of factors of 3 in x^2 must be even and one less than a multiple of 3, so choices are 2 and 8. Now, \frac{x^2}{5} \le 10000, so x must be a multiple of 3\cdot 5^2 = 75. And in ... 1 The problem is that of finding the value of \sqrt{5p+10} where p is an integer less than 9995 such that$$a^2=p+(p+1)+(p+2)+(p+3)+(p+4)=5p+10=5(p+2),b^3=(p+1)+(p+2)+(p+3)=3p+6=3(p+2),$$for some integers a and b. Then 5\mid a and 3\mid b, and hence 5\mid p+2 and 3^2\mid p+2, meaning that$$p=45n-2,$$for some integer n. It follows that ... 1 The sum of an arithmetic series is \frac n2(a+l), where n is the number of terms and a and l are the first and last terms respectively. You can think of it as the number of terms multiplied by the arithmetic means (simple average) of the bounding terms. Here, n=33 and l = a+(33-1)(2) = a + 64 so the sum is \frac{33}{2}(a + a + 64) = 33(a+32). ... 2 First we prove that 1+3+5+\ldots +(2n-1) = n^2 by induction: The base case 2\cdot 1-1=1=1^1 holds. Suppose the claim is true for n. Then 1+3+\ldots + 2(n+1)-1 = (1+\ldots +2n-1)+(2n+1) = n^2+2n+1 = (n+1)^2, as desired. Now, starting at number m, we have that (m+1)+(m+3)+\ldots+(m+2\cdot 33-1)=33^{33}. Hence 33m+33^2=33^{33}, and therefore m=... 18 The average of those 33 consecutive odd numbers is 33^{33}/33=33^{32}. And 16 of them are larger than the average, so the largest is... 4 33^{33}=33\times 33^{32}+(-32-30-\cdots -2-0+2+\cdots +30+32)= (33^{32}-32)+(33^{32}-30)+\cdots + (33^{32})+\cdots +(33^{32}+30)+(33^{32}+32) As 33^{32} is an odd number, then (33^{32}\pm 2k) is odd as well. So, the largest one is (33^{32}+32). 1 We have x-32,x-30,\ldots,x,\ldots,x+30,x+32 and thus 33x = 33^{33}. Thus, the largest number is 33^{32}+32. 0 This does not completely answer the question. This provides some more information than those given by previous answers and comments. Also, the method I provide here slightly differs from the following reference, but main ideas are from the paper. According to the reference http://www.fq.math.ca/Papers1/43-2/paper43-2-6.pdf We obtain common prime divisors ... 0 No one else seems to have suggested the following approach. Suppose we keep flipping a coin until we get five heads in a row. Define a "run" as either five consecutive heads or a tails flip plus the preceding streak of heads flips. (A run could be a single tails flip.) The number of coin flips is equal to the number of runs with at least four heads (R_{... 3 Let us assume (a^2 + b^2)/(ab + 1) = k \in \mathbb{N}. We then have a^2 - kab + b^2 = k. Let us assume that k is not an integer square, which implies k \ge 2. Now, we observe the minimal pair (a, b) such that a^2 - kab + b^2 = k holds. We may assume, without loss of generality, that a \ge b. For a = b, we get k = (2 - k)a^2 \le 0, so we ... 1 It holds that for any number n the exponent e_p(n) of a prime p in n! is exactly e_p(n) = \frac{n-d_p(n)}{p-1} where d_p(n) denotes the sum of digits in base p. This leads to the following reformulation of the problem: Find all positive integers k such that for all positive integers n \frac{(k-1)n-d_2(kn)+d_2(n)}{1}<(k-1)n+1 \\ d_2(n)-d_2(kn) ... 0 \text {let: } k = \frac {(b-a)}{2}\\ \text {then: } b = \frac {(b+a)}{2} + \frac {(b-a)}{2} = \frac 12 + k\\ a = \frac {(b+a)}{2} - \frac {(b-a)}{2} = \frac 12 - k a^4 + b^4\\ (\frac 12 - k)^4 + (\frac 12 + k)^2\\ [(\frac 12)^4 - 4(\frac 12)^3k + 6(\frac 12)^2k^2 -4(\frac 12)k^3 + k^4] + [(\frac 12)^4 + 4(\frac 12)^3k + 6(\frac 12)^2k^2 +4(\frac 12)k^3 +... 1$$\sqrt{\frac{x^2+y^2}{2}}\ge\frac{x+y}2 \Rightarrow x^2+y^2\ge\frac12(x+y)^2$$Then$$a^4+b^4\ge\frac12(a^2+b^2)^2\ge\frac12\cdot\left(\frac12(a+b)^2\right)^2=\frac18$$3 Hint: Note that  2(a^2 +b^2) \geq (a+b)^2, ~\forall a,b\in\mathbb{R} and use it twice to derive the desired inequality. 1 hint: use CS inequality twice: (ab + cd)^2 \leq (a^2+c^2)(b^2+d^2) 1 By Vieta's formulas,$$r+s+t=0,\quad rs+st+tr=-21,\quad rst=-35\tag1$$Now, we have$$r+s+t=(r^2+s^2+t^2)+a(r+s+t)+3b\implies b=-14\tag2$$By the way, as you noticed, noting that y=P(x) is a parabola whose axis of symmetry is x=-a/2, we know that$$\small\begin{align}&Q(P(x))\\&=(x^2+ax+b)^3-21(x^2+ax+b)+35\\&=x^6+3ax^5+(3a^2+3b)x^4+(a^3+...

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Hints: 1) $a>b$ denote $$a=b+x, x \geqslant 0$$ 2) Prove $$f= b+x+\frac{1}{bx} \geqslant 3$$

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kccu has basically answered the question in a comment. Here "these $k$ couples sit together" means that the spouses of each couple sit next to each other, not that all $k$ couples sit contiguously. And the missing factor $(2n-2k)!$ is indeed the number of ways of permuting the members of the remaining $n-k$ couples. It would have been simpler to just say ...

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You can use the properties of congruences: since $16\equiv 1\pmod{5}$, also $16a\equiv a\pmod{5}$. Thus you have $$9+a\equiv 12\pmod{5}$$ that becomes $$a\equiv 3\pmod{5}$$ Note that all steps are “reversible”, so you're sure that this is the only solution: \begin{gather} a\equiv3\pmod{5}\\[6px] a+9\equiv3+9\pmod{5}\\[6px] 15a+a+9\equiv12\pmod{5} \end{...

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Hint: $$\int_{(k-1)\pi}^{k\pi} \left| \frac{\sin x}{x} \right|\,dx \geq \int_{(k-1)\pi}^{k \pi} \frac{|\sin x|}{k \pi}$$

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We may start from: $$A_k = \int_{0}^{\pi}\frac{\sin x}{x+k\pi}\,dx = \int_{0}^{\pi/2}\sin(x)\left(\frac{1}{x+k\pi}+\frac{1}{\pi-x+k\pi}\right)\,dx$$ and notice that $f_k(x)=\frac{1}{x+k\pi}+\frac{1}{\pi-x+k\pi}$ is decreasing on $\left[0,\frac{\pi}{2}\right]$, hence: $$A_k \geq \int_{0}^{\pi/2} \sin(x)\,f_k\left(\frac{\pi}{2}\right)\,dx = \frac{4}{(2k+1)\... 3 We have$$\begin{align} \int_{0}^{\pi}\left|\frac{\sin\left(nx\right)}{x}\right|dx\stackrel{nx=v}{=} & \int_{0}^{n\pi}\left|\frac{\sin\left(v\right)}{v}\right|dv \\ = & \sum_{k=0}^{n-1}\int_{k\pi}^{\left(k+1\right)\pi}\left|\frac{\sin\left(v\right)}{v}\right|dv \\ = & \sum_{k=0}^{n-1}\left(-1\right)^{k}\int_{k\pi}^{\left(k+1\right)\pi}\frac{\...

1

Since $S$ has $1990$ elements there can be at most $1990 \times 1989 = 3958110$ possible ordered pairs. Now one $100$-ary sequence generates $\frac{100 \times 99}{2} = 4950$ ordered pairs. Now as each ordered pair appears in at most one element of $P$ it follows that the number of ordered pairs in $P$ is: $\mid P \mid \times 4950$. From all this we have: $$\... 1 Multiply by 8\sin\tfrac{\pi}{7} and simplify using 2\sin x \cos x = \sin 2x three times:$$8\sin\tfrac{\pi}{7} \cos \tfrac{\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7} = 4 \sin\tfrac{2\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7} = 2 \sin\tfrac{4\pi}{7} \cos \tfrac{4\pi}{7} = \sin\tfrac{8\pi}{7}$$But you also have:$$\sin\tfrac{8\pi}{7} = \sin\...

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We have $2^n=1,2,4,3\bmod5$ and $3^n=1,3,4,2\bmod5$ for $n=0,1,2,3\bmod4$. Hence $2^n-3,3^n-2$ are both divisible by 5 iff $n=3\bmod4$ and $\gcd(2^n-3,5)=5$ iff $n=3\bmod4$ (for other $n$ it is 1). The original question on this site asked for a proof that $\gcd(2^n-3,5)=\gcd(2^n-3,3^n-2)$ for all $n$. Given the observation above, that amounted to the ...

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By the sine duplication formula $\sin(2x)=2\sin(x)\cos(x)$ it follows that: $$\cos(2^n A) = \frac{\sin(2^{n+1} A)}{2\sin(2^n A)}\tag{1}$$ hence: $$\prod_{n=0}^{N-1}\cos(2^n A) = \frac{\sin(2^N A)}{2^N \sin(A)}\tag{2}$$ comes from a telescopic product.

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Turn the double-angle formula for sine "inside out". Put in \begin{align}\cos(A)& =\frac{\sin(2A)}{2·\sin(A)} \\[6pt] \cos(2A)&=\frac{\sin(4A)}{2·\sin(2A)} \end{align} etc., and use telescoping.

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For a fixed $n$, assume the all the prime numbers not greater than $n$ are $p_1,p_2,\dots.p_r$. Let $$\alpha_i=\text{ the largest integer j such that }[\frac{n}{p_i^j}]>0,$$ then $D=\text{lcm}(1,2,\dots,n)=\prod_{i=1}^rp_i^{\alpha_i}$. Now, let $m_i=[\frac{n}{p_i^{\alpha_i}}]$, then the largest $m=\min_{1\leqslant i\leqslant r}\{m_i\cdot p_i^{\... 0 You clearly need to have the range$[m,n]$include all primes$p_i>n/2$, since those will not occur as$2p_i$in the modified range. Similarly you will need any highest prime power that occurs between$n/2$and the least$p_i$. All other numbers in this interval that are not primes or prime powers will be disposable due to their occurrence in other ... 0 Let$L_n$be the left-hand side, and let$R_{m,n}$be the right-hand side. Note that$L_n$is divisible by all of the greatest prime powers$p^r \leq n$. In fact, $$L_n= \prod_{p \leq n} p^r$$ where$p$is prime and$r$is the greatest exponent such that$p^r \leq n$. Clearly$R_{m,n} \mid L_n$. Therefore, $$L_n=R_{m,n} \iff p^r \mid R_{m,n}$$ for each ... 3 It works for all positive$n$. The proof involves 10-adic idempotents (https://divisbyzero.com/2008/12/29/a-10-adic-number-that-is-a-zero-divisor/). These are 10-adic numbers that are their own squares so they are automatically their own cubes, their own fourth powers, etc. In ordinary numbers only$0$and$1$satisfy this condition. But in 10-adics ... 2 Another way to think of this is "does there exist an$n$-digit number$a$such that$a^2 \equiv a \pmod{10^n}$?" Since$10^n=2^n\cdot 5^n$, we can conclude that$a^2 \equiv a \pmod{2^n}$and$a^2 \equiv a \pmod{5^n}$. However, according to this question, we know that the only way this can be true is if: $$a \equiv 0,1 \pmod{2^n}$$ and likewise: $$a \equiv 0,... 5 The number of factors 2 in a number a! is:$$ \left\lfloor \frac a 2 \right\rfloor + \left\lfloor \frac a 4 \right\rfloor + \left\lfloor \frac a 8 \right\rfloor + \cdots $$So we have to find all k such that for every n, we have$$ kn - n + 1 > \sum_{x=1} \left\lfloor \frac{kn}{2^x} \right\rfloor - \sum_{x=1} \left\lfloor \frac{n}{2^x} \right\... 1 I know that this question has been sufficiently answered already, but just in case you're curious, I made this python program that finds these numbers: x = input("Input Amount of Digits") y = 1 z = 9 for a in range(1, int(x)): y = y*10 #Lower Bound a = a+1 for b in range(1, int(x)): z = ((z*10)+9) b = b+1 print("Puzzle: ... 1 Here is the solution from the American Mathematical Monthly: 3 Note that the integral$I$as given by $$I=\int_0^{\pi/2}\int_0^{\pi/2} \frac{\log(2-\sin(\theta)\cos(\phi))}{2-2\sin(\theta)\cos(\phi)+\sin^2(\theta)\cos^2(\phi)}\,\sin(\theta)\,d\theta\,d\phi$$ is the surface integral of$f(x)=\frac{\log(2-x)}{2-2x+x^2}over a sphere in the first octant. Therefore, we can write \begin{align} I&=\int_0^1 \int_0^{... 3 We may exploit the following fact: \forall a\in\mathbb{N},\qquad \int_{0}^{\pi/2}\int_{0}^{\pi/2}\sin(x)^{a+1}\sin(y)^a\,dx\,dy = \frac{\pi}{2+2a}.\tag{1}$$It follows that, assuming:$$ \frac{\log(2-z)}{2-2z+z^2}=\sum_{n\geq 0} c_n z^n \tag{2}$$our integral equals:$$ \frac{\pi}{2}\sum_{n\geq 0}\frac{c_n}{n+1} = \frac{\pi}{2}\int_{0}^{1}\frac{\log(2-z)}{... 5 For positivesa$,$b$and$c$by AM-GM we obtain:$2\sum\limits_{cyc}\frac{1+ab}{(a+b)^2}\geq\sum\limits_{cyc}\frac{a^2+b^2+c^2+ab+ac+bc+2ab}{(a+b)^2}=3+\sum\limits_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq3+3=6$0 Hint:$A(0\mid 1)B(-\cot{30°}\mid 0)C(\cot{15°}\mid 0)O(0\mid 0)$What is the midpoint$M\,$of$BC\,$? What sort of triangle is$AOM\,$? 1 If we let$\alpha$be the angle$BAM,$then after strenuous calculation, we have that $$\alpha = \sin^{-1}\left(\sqrt{\frac{2 - \sqrt{3}}{2}}\right),$$ which we observe to be equal to$15$degrees. With this, we see that the desired angle is$180 - 30 - 15 = \boxed{135}$degrees. And no, this is not worth a bounty. 5 I'm not sure that my solution is elegant. So let$g(x) = f(x) - x$. We may rewrite the first equation from task as $$f(f^2(x)) - f^2(x) + 3f^2(x)-3f(x) + 4f(x) - 4x = 0,$$ or using$g(x)$$$g(f^2(x)) + 3g(f(x)) + 4g(x) = 0.$$ Let$a_n = g(f^n(x))$. Then$a_n + 3a_{n-1} + 4a_{n-2} = 0$. We may make this recurrent relation simpler: let$b_n = a_n + \frac{...

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I'm not confident about this answer but i'll post it anyways, since there's no answers. Look at the equation $f^{3} + 2f^{2} + f = 4x$. This tell us that the left hand terms are a linear combination of $x$ summing up to $4x$, unless any of them cancel eachother out. Suppose none of the left hand terms cancel out. Then $f^3 + 2f^2 + f = ax + bx + f$ \$f = ...

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