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We can multiply the first equation by \(2\) so both have \(2y^2\). \[2y^2+4\sqrt{x}=12\] Next we can find 2 different expressions for \(2y^2\), which tells us \[3+5\sqrt{x}=12-4\sqrt{x}\Rightarrow9\sqrt{x}=9\Rightarrow x=1\] so the solution is \(x=1\). We can also use this to show \(y=\pm2\).
First expand the 2 equations. \[\begin{align}&abc-ab-ac-bc+a+b+c-1=abc-1\\ &abc-2ab-2ac-2bc+4a+4b+4c-8=abc-2\end{align}\] Then substitute \(x=a+b+c\) and \(y=ab+ac+bc\). \[\begin{align}&abc-y+x-1=abc-1\\&abc-2y+4x-8=abc-2\end{align}\] Then simplify a little so we have a system of equations \(x-y=0\) and \(4x-2y=6\). By solving this we find \(x=y=3\).
Since \(f(0)=4\), the maximum must be at least 4. By using the triangle inequality, we have \[|f(x)|\leq|\cos(x)|+|\cos(2x)|+|\cos(3x)|+|\cos(4x)| \leq1+1+1+1=4\] so \(-4\leq f(x)\leq4\). Since \(f(x)\) is bounded above by 4 and we found that it achieves this value at \(x=0\), the maximum is 4.
First complete the rectangle by drawing 2 edges.
This is a 3-4-5 right triangle so \(x=5\).
The 100th digit of \({4\over13}\) in decimal.
By long division, we can evaluate the approximation and see where it repeats. The value is \(0.\overline{307692}\). We need the 100th digit which will be the same as the 4th because \(100\equiv4\mod6\) so the answer is 6.
Expanding the base representation gives us \(1\cdot(2x)^2+2\cdot(2x)+9 =4\cdot x^2+5\cdot x+1\). This is a quadratic equation to solve with 1 solution. \(4x^2+4x+9=4x^2+5x+1\Rightarrow x=8\).
If you memorize cubes, you would recognize \(9^3=729\) which shows you that \(x\) and \(x+1\) should be 9 and 10. Without that, we can expand this polynomial equation to \(2x^3+3x^2+3x-1728=0\). Using various means, this can be factored to \((x-9)(2x^2+21x+192)=0\). This shows \(x=9\) is a solution and the other 2 are imaginary because \(21^2-4\cdot2\cdot192<0\).
Three of the roots of \(f(y)=y^4+ay^2+by+c\) are 2, 3, and -4. Find \(b\).
Since there are 4 roots, let \(z\) be the last root so we have \[\begin{align}&f(y)=(y-2)(y-3)(y+4)(y-z)=(y^2-5y+6)(y^2+(4-z)y-4z)\\ &=y^4+y^3(-5+4-z)+y^2(6-5(4-z)-4z)+y(6(4-z)+20z)-24z\end{align}\] Then by matching terms, we have \(-5+4-z=0\Rightarrow z=-1\), \(6-5(4-z)-4z=a\Rightarrow a=-15\), \(6(4-z)+20z=b\Rightarrow b=10\), and \(-24z=c\Rightarrow c=24\). The solution is \(b=10\).
The right side can be simplified to \(a+12\). On the left, notice how the coefficient of \(x\) is 1 and the difference between \(a\) and \(a+1\) is 1. The numerator can be factored to \((x-a)(x+a+1)\), which shows we can divide out the \(x-a\). We are left with \(x+a+1=a+12\) so the solution is \(x=11\).
Square both sides to find \[{x^2\over36}={12+4\sqrt{5}\over3+\sqrt{5}}={4(3+\sqrt{5})\over3+\sqrt{5}}=4\] It only makes sense to take the positive value here so \({x\over6}=\sqrt{4}=2\) and we find \(x=12\).
On the left is the formula for the geometric sum \(1+3+3^2+\ldots+3^{998}\). By observing that \(3^3\equiv1\mod27\), we can reduce it to \((1+3+9)+(1+3+9)+\ldots+(1+3+9)\equiv13\cdot333\mod26\). This value is odd and divisible by 13 so it is \(\equiv1\mod2\) and \(\equiv0\mod13\). Since \(\gcd(2,13)=1\) we can apply the Chinese remainder theorem to find that this number is \(\equiv13\mod26\).
We can solve this by making use of the angle sum identity for tangent, which is \[\tan(x+y)={\tan(x)+\tan(y)\over1-\tan(x)\tan(y)}\] First use \(x=y=\pi/8\) to find \(\tan(\pi/8)\). \[\begin{align}&1=\tan(\pi/4)=\tan(\pi/8+\pi/8) ={2\tan(\pi/8)\over1-\tan^2(\pi/8)}\\ &1-\tan^2(\pi/8)=2\tan(\pi/8)\end{align}\] Then let \(u=\tan(\pi/8)\) and solve this quadratic equation \(u^2+2u-1=0\Rightarrow u=-1\pm\sqrt{2}\). Since \(\tan(\pi/8)\) is positive, \(\tan(\pi/8)=\sqrt{2}-1\). Next we use the identity again. \[\begin{align}&\tan(3\pi/8)=\tan(\pi/4+\pi/8) ={\tan(\pi/4)+\tan(\pi/8)\over1-\tan(\pi/4)\tan(\pi/8)}\\ &={1+(\sqrt{2}-1)\over1-(\sqrt{2}-1)}={\sqrt{2}\over2-\sqrt{2}} {2+\sqrt{2}\over2+\sqrt{2}}\end{align}=1+\sqrt{2}\] Then the solution is \[\begin{align}&\tan^3(3\pi/8)-\tan^3(\pi/8)=(1+\sqrt{2})^3-(\sqrt{2}-1)^3\\ &(1+3\sqrt{2}+6+2\sqrt{2})-(2\sqrt{2}-6+3\sqrt{2}-1)=14\end{align}\]
This means \(x=100\cdot{3\over20}=15\), since a percent is 100 times the ratio it represents.
100 books are sold for \(x\) dollars (a whole number) or for half price. The total sale is $963.
Let \(a\) be how many are sold for full price and \(b\) for half price. Then \(a+b=100\) and \(ax+bx/2=963\). Starting from the 2nd equation we have \(1926=2ax+bx=ax+100x=x(a+100)\). Since \(x\) and \(a\) are integers, \(x\) and \(a+100\) must be factors of \(1926=2\cdot3\cdot3\cdot107\). Since \(0\leq a\leq100\), \(100\leq a+100\leq200\). The only divisor of 1926 satisfying this is 107 so \(a=7\). Then \(x=1926/107=18\). 7 books are sold for $18 and 93 books are sold for $9.
Find the smallest two-digit number that is equal to the sum of its digits plus the product of its digits.
Let the 2 digit number be \(10a+b\) with \(a\neq0\). Then \(10a+b=(a+b)+ab \Rightarrow 9a=ab\Rightarrow b=9\). So this works for any 2 digit number with \(b=9\). The smallest is with \(a=1\) so the answer is 19.
How many numbers under 66 are relatively prime to 66?
This can be solved with Euler's totient function once we know the prime factorization. Since \(66=2\cdot3\cdot11\), \(\phi(66)=66\cdot{1\over2} \cdot{2\over3}\cdot{10\over11}=20\).
Split take out the \(n=0\) term. Then combine terms \(i\) and \(-i\). Next is some algebra. \[\begin{align}&1+\sum_{n=1}^{10}\left({2\over1+10^n}+{2\over1+10^{-n}}\right) =1+\sum_{n=1}^{10}{2(1+10^{-n})+2(1+10^n)\over(1+10^n)(1+10^{-n})}\\ &=1+\sum_{n=1}^{10}{4+2\cdot10^n+2\cdot10^{-n}\over1+10^n+10^{-n}+1} =1+\sum_{n=1}^{10}2=21\end{align}\]
First we can split up the fraction into 2 parts. \[{n^3+5\over n^2+7}={(n^3+7n)+(5-7n)\over n^2+7}=n+{5-7n\over n^2+7}\] Since \(n\) is an integer, focus on the new fraction, which cannot be 0. We must have \[|5-7n|\geq n^2+7\] otherwise the answer would be between -1 and 1, and it cannot be 0. First suppose \(5-7n\geq0\). Then we have \[5-7n\geq n^2+7 \Rightarrow n^2+7n+2\leq0\] which is always false. Alternatively we can see that \(5-7n<0\) for all \(n\in\mathbb{N}\) so this case was not necessary. Now suppose \(5-7n<0\). Then \[7n-5\geq n^2+7\Rightarrow n^2-7n+12\leq0\Rightarrow(n-3)(n-4)\leq0\] which is clearly true when \(n=3,4\). If \(n<3\) or \(n>5\) then \((n-3)(n-4)>0\), so the only \(n\) that work are \(n=3,4\). We can verify \[{3^3+5\over3^2+7}={27+5\over9+7}={32\over16}=2\quad\text{and}\quad {4^3+5\over4^2+7}={64+5\over16+7}={69\over23}=3\] are both integers. Then \[x=10+7\cdot3-3^2=10+21-9=22=10+7\cdot4-4^2=10+28-16=22\] so we have the same value \(x=22\) for each \(n=3,4\).
Find the number of proper factors of \(4004\).
The number of factors can be computed from the prime factorization. With trial division or other means, we find \(4004=2^2\cdot7\cdot11\cdot13\). The number of factors can be computed from the exponents on the prime factors so it would be \((2+1)(1+1)(1+1)(1+1)=3\cdot2^3=24\). But since we are only interested in counting proper factors, we exclude \(4004\) itself, so the total is 23.
Begin by drawing the following lines.
This forms a right triangle so we have \(10^2+x^2=26^2\Rightarrow x=24\).
From the 2nd equation, \(\sqrt{y}=16-\sqrt{x}\Rightarrow y=256-32\sqrt{x}+x\). Substitute this into the first equation and we get \(256-32\sqrt{x}=96\Rightarrow-32\sqrt{x}=-160 \Rightarrow\sqrt{x}=5\Rightarrow x=25\). We can also find \(y=121\).
Find the number of trailing zeroes in \(111!\).
This would be counting how many times 10 divides \(111!\). We can do this by counting how many times its factors 2 and 5 divide \(111!\). For 2, it is \[\left\lfloor{111\over2}\right\rfloor+\left\lfloor{111\over4}\right\rfloor +\left\lfloor{111\over8}\right\rfloor+\left\lfloor{111\over16}\right\rfloor +\left\lfloor{111\over32}\right\rfloor+\left\lfloor{111\over64}\right\rfloor =55+27+13+6+3+1=105\] and for 5, it is \[\left\lfloor{111\over5}\right\rfloor+\left\lfloor{111\over25}\right\rfloor =22+4=26\] Therefore 10 can divide \(111!\) up to 26 times because each divide by 10 takes away a 2 and a 5, and we run out of 5's first. Therefore, there are 26 trailing zeroes.
\(f\) is a function such that \(\forall x,y\in\mathbb{R}: f(xy)=f(x)/y\). If \(f(10)=81\), find \(f(30)\).
The easiest way is to use the given information. Let \(x=10,y=3\). Then we have \[f(xy)=f(10\cdot3)=f(10)/3=81/3=27\] Alternatively, we can find out what the function \(f(x)\) is. First see what happens when differentiating each side by \(x\) and then by \(y\). \[f'(xy)y=f'(x)/y,\quad f'(xy)x=-f(x)/y^2\] Then from this, we find \[f'(xy)y^2=f'(x)=-f(x)/x\Rightarrow{f'(x)\over f(x)}={-1\over x}\] Then by integrating and solving, \[\ln|f(x)|=\int{-dx\over x}=-\ln|x|+C\Rightarrow f(x)=e^Ce^{-\ln|x|} =C'x^{-1}\] then with the given \(f(10)=81\), we find \(C'=810\). So \(f(x)=810/x\) and \(f(30)=810/30=27\).
A cone with volume \(9687\) has radius \(x\) (rounded) and height \(11\).
The volume of the cone is \({11\pi x^2\over3}=9687\) so \(x\approx28.999\approx29\).
Find the coefficient of \(x^5\) in \((1+x+x^3)^5\).
We can use the multinomial theorem for this. The possible ways to select 5 terms that multiply to \(x^5\) are: \[x^3\cdot x\cdot x\cdot1\cdot1,\ \text{and}\ x\cdot x\cdot x\cdot x\cdot x\] The coefficients are \[{5\choose1,2,2}={5!\over1!2!2!}=30\ \text{and}\ {5\choose5}=1\] so the coefficient of \(x^5\) is \(30+1=31\).