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Working left to right, we compute \(123456789\equiv5\mod11\). Then by modular exponentiation, \(5^2\equiv3\mod11\), \(5^4\equiv-2\mod11\), \(5^8\equiv4\mod11\), then (modulo \(11\)) \[123456789^{10}\equiv5^{10}\equiv5^2\cdot5^8\equiv3\cdot4\equiv12\equiv1\]
Find the number of connected components of the Lie Group \(GL(2,\mathbb{R})\)
This is the general linear group of degree 2, defined with multiplication on \[\left\{A=\begin{pmatrix}a&b\\c&d\\\end{pmatrix}:(A)=ad-bc\neq0\right\}\] This group has 2 components. One for the positive determinants and one for the negative determinants. This is because the determinant on this set is continuous so the inverse of an open set is open. The 2 components correspond to the inverse images of \((0,\infty)\) and \((-\infty,0)\).
How many times do \(y=x\) and \(y=x^{11}\) intersect?
For this, we solve \(x=x^{11}\Rightarrow x^{11}-x=0\Rightarrow x(x^{10}-1)=0 \Rightarrow x(x^5+1)(x^5-1)=0\). From here, we can see the only real solutions are \(x=0,-1,1\) so these graphs intersect 3 times.
What is the largest odd number that cannot be represented as the sum of three primes?
We can show 5 is not the sum of 3 primes since the smallest such sum is \(2+2+2\). Goldbach's weak conjecture states that every odd number greater than 5 is the sum of some 3 primes. This was proven by Harald Helfgott in 2013.
This very not to scale graph is a linear equation. Using the 2 points \((-2,0),(-1,3)\), we can find the slope is \({3-0\over-1-(-2)}=3\). This can be extended 1 more step to the right from \((-1,3)\) to find y-intercept \(6\). Alternatively, we can use point-slope form to find the line equation \(y-3=3(x-(-1))\Rightarrow y=3x+6\) which shows the y-intercept is \(6\).
Find the period of \(12\sin\left(2\pi x\over7\right)\)
The 12 does not change the period, only the amplitude. The period of \(\sin(x)\) is \(2\pi\). Multiplying the argument \(x\) by something divides the period by that amount. In this case, we would divide the period by \({2\pi\over7}\) and find the period is \(7\).
Find the number of irreducible \({a\over b}<1\) such that \(a,b\in\mathbb{N}\) and \(a+b=40\).
More generally, this is counting the number of pairs \((a,b)\) with \(0<a<b\), \(\gcd(a,b)=1\), and \(a+b=N\). Suppose \(\gcd(a,b)=\gcd(a,N-a)=1\). Then \(\gcd(a,N)=1\) so this is equivalent to counting \((a,b)\) pairs with \(\gcd(a,N)=1\). There are \(\phi(N)\) such pairs which is always even for \(N>2\). Half of these pairs satisfy \(a<b\) as required so the solution is \(\phi(N)/2\). For this problem, \(N=40\) has distinct prime factors \(2,5\) so the solution is \({1\over2}\cdot40\cdot{1\over2}\cdot{4\over5}=8\)
This is a simple pythagorean theorem problem with \(x^2=41^2-40^2=1681-1600=81\) so \(x=\sqrt{81}=9\).
Find the Frobenius number of the numerical semigroup generated by 4, 7, and 9.
These integers under addition generate the elements \(\{4a+7b+9c\mid a,b,c\in\mathbb{N}\}\). The Frobenius number is the largest value \(n\) such that \(4a+7b+9c=n\) has no solution. We can check numbers individually to find that \(n=10\) has no solution. Then \(11=4+7,12=4+4+4,13=4+9,14=7+7\) and any larger number number is 4 plus a smaller number that has a solution.
How many rectangles can be formed from 1400 identical squares?
This is how many ways we can factor 1400 into integers. We can factor it into \(1400=2^3\cdot5^2\cdot7\) and see that it has \(4\cdot3\cdot2=24\) divisors. Since 1400 is not a perfect square, each divisor below \(\sqrt{1400}\) corresponds to a divisor above \(\sqrt{1400}\) so half of the number of divisors is how many rectangles we can make. The solution is 12.
How many numbers can be formed using the digits 1, 2, and 3 such that two digits are 1 or 2 and two wdigits are 2 or 3?
This can be solved with a few cases. If there are no 2's, then there must be 2 1's and 2 3's so there are 6 permutations of 1133. If there is one 2, there must also be a 1 and 3 so there are 6 permutations of 123. If there are two 2's, then no other digits are included so we count the single number 22. The solution is \(6+6+1=13\).
An urn has 21 black balls and \(x\) white balls with \(x<21\). The chance that two balls chosen (with replacement) are the same mcolor is \(52\%\).
There are \(21+x\) balls in total. The probability of 2 black is \({21^2\over(21+x)^2}\). The probability of 2 white is \({x^2\over(21+x)^2}\). The equation to solve is \[{21^2+x^2\over(21+x)^2}={52\over100}={13\over25}\] Multiplying stuff gets \(11025+25x^2=13x^2+546x+5733\). This can be combined and common factors removed to get \(2x^2-91x+882=0\) which factors to \((2x-63)(x-14)=0\). Therefore, the only integer solution is \(x=14\).
Find the sum of the x-intercept and the y-intercept of \(y=-{2\over3}x+6\)
The x-intercept is found by \(y=0\Rightarrow x=9\). The y-intercept is the 6. The solution is \(9+6=15\).
This problem can be solved by using the geometric summation \[f(x)=1+x+x^2+\ldots={1\over1-x}\] By integration, \[\int(1+x+x^2+\ldots)=\int{dx\over1-x}\] \[x+{x^2\over2}+{x^3\over3}+\ldots=-\ln|1-x|+C\] Clearly \(C=0\) since the left is 0 when \(x=0\). So we have found \[\sum_{n=1}^\infty{x^n\over n}=-\ln|1-x|\] Then going back to the givens, \[{t\over4}=\sum_{n=1}^\infty{1\over n}\left(1\over2\right)^n\] which matches our equation with \(x={1\over2}\). So we get \[t=-4\ln\left|1-{1\over2}\right|=-4\ln\left({1\over2}\right)=4\ln(2)\] Then using the original \(x\) instead of our \(x\) from the derivation, \(x=e^t=e^{4\ln(2)}=2^4=16\)
How many ways can \(1225\) be written as the sum of at least two consecutive numbers.
Let \(x\) be the first number in a sum of \(k\) numbers. Then the sum is \[x+(x+1)+(x+2)+\ldots+(x+k-1)=kx+{(k-1)k\over2}\] Then we need this sum to be equal to \(1225\) for some starting integer \(x\) and integer \(k\geq2\). \[1225=kx+{(k-1)k\over2}\Rightarrow2450=2kx+(k-1)k\Rightarrow k(2x+k-1)=2\cdot5^2\cdot7^2\] Since \(k\) is an integer, it is a divisor of \(2450\). Then \(2x+k-1\) has opposite parity of \(k\) and we can solve for \(x\). So we must factor \(2450\) into an odd and even number. This will always be the case since there is only a single \(2\), it will onlyn be part of 1 factor. So since there are \(18\) divisors of \(2450\), we can find a \(k\) and \(2x+k-1\) in \(18\) different ways. We only have to exclude \(k=1\) since it must be at least 2 consecutive numbers. Therefore, the answer is 17. Each possible way corresponds to \(k\) numbers where \(k>1\) and \(k\mid2450\). The starting number is \(x={1\over2}\left({2450\over k}+1-k\right)\)
A sphere of radius \(x\) is inscribed in the tetrahedron with vertices \((162,0,0)\), \((0,108,0)\), \((0,0,54)\), and \((0,0,0)\).
3 of the faces are on the main axis planes for the first octant (\(x,y,z\geq0\)). Since the sphere is tangent to each of these, its center must be \((r,r,r)\) with radius \(r\). From this center point, we can make a line of length \(r\) to the point of tangency with the last plane, which will be normal to that plane.
The non-axis plane is described by the 3 points other than the origin. The plane can be described by \(ax+by+cz=1\) for some \(a,b,c\). We can use those 3 points to form the equations \(162a=1,108b=1,54c=1\) which gives us the plane equation: \[{1\over162}x+{1\over108}y+{1\over54}z=1\Rightarrow2x+3y+6z=324\] From \((r,r,r)\), we add a length \(r\) vector in the direction of the plane's normal vector \((2,3,6)\) to find a point on the plane. This gives us the point: \[(r,r,r)+r{(2,3,6)\over\Vert(2,3,6)\Vert}=(r,r,r) +r\left({2\over7},{3\over7},{6\over7}\right) =r\left({9\over7},{10\over7},{13\over7}\right)\] Next we put these 3 points into the equation for the plane and solve for \(r\) \[2\cdot{9\over7}r+3\cdot{10\over7}r+6{13\over7}r=324\Rightarrow {18+30+78\over7}r=324\Rightarrow{126\over7}r=324\Rightarrow18r=324\Rightarrow r=18\]
3 kg of 24 carat gold and 5 kg of 16 carat gold are melted to form a bar of \(x\) carat gold.
This is a weighted average \({3\over8}\cdot24+{5\over8}16=9+10=19\)
Multiplying the 2nd equation by \(xy\), we have \(x-y={xy\over20}\). So by the first equation \(xy=200\). Then substitute for \(y\) and we solve the quadratic \[x(x-10)=200\Rightarrow x^2-10x-200=0\Rightarrow (x+10)(x-20)=0\] So the only positive solution as required is \(x=20,y=10\).
How many seven digit binary numbers are there such that no two consecutive digits are 0?
Seven digit binary numbers start with a 1 and are followed by a binary string of length 6. So the answer is the number of binary strings of length 6 with no consecutive 0. Let \(C(n)\) be the number of length \(n\) binary strings with no consecutive 0s. Then \(C(0)=1,C(1)=2\) (for length 0 it is the empty string). Then \(C(n)\) is computed recursively. If the first bit is 0, the 2nd bit must be 1 and there are \(C(n-2)\) possible ways for the next bits. If the first bit is 1, then there are \(C(n-1)\) possible ways for the next bits. This is exactly the fibonacci sequence and we can find \(C(6)=21\) is the solution.
Find the average queue size (rounded) of an M/M/1 queue with arrival rate 22 and service rate 23.
There is a lot of theory behind these queue models so this will probably not be a very complete answer.
The M/M/1 queue is a stochastic model with a state space describing the number of jobs in the queue. Arrivals are poisson distributed with mean \(\lambda=22\) and service times are exponentially distributed with parameter \(\mu=23\) so they have mean \(1/\mu=1/23\). This says that every time unit, we expect about 22 processes to arrive and 23 to be completed.
Based on the Wikipedia article, the value \(p=\lambda/\mu=22/23\) describes the queue utilization. The model is stable when \(p<1\), when jobs are completed on average faster than they arrive. The probability of the system being in state \(i\) is \((1-p)p^i\). This is a geometric distribution which has well-known results, but for the purpose of adding more math to this solution, we can compute the average state index with the following summation \[\sum_{i=0}^\infty i(1-p)p^i=(1-p)\sum_{i=1}^\infty ip^i\] Now visualize a grid of the values summed in the summmation \[\begin{array}{ccc}p&&&\\p^2&p^2&&\\p^3&p^3&p^3&\\\vdots&&&\ddots\\\end{array}\] The summation sums by rows so instead we can write a summation by columns \[(1-p)\sum_{i=1}^\infty p^i\left(1+p+p^2+\ldots\right) =(1-p)\sum_{i=1}^\infty p^i{1\over1-p}=\sum_{i=1}^\infty p^i={p\over1-p}\] So now substitute \(p\) to find the average queue size \[{p\over1-p}={22/23\over1-22/23}={22/23\over1/23}=22\]
Find the number of weeks in 231840 minutes.
By factoring, we find \(2^5\cdot3^2\cdot5\cdot7\cdot23\). We divide out \(60\cdot24=2^5\cdot3^2\cdot5\) for minutes in a day and \(7\) for days in a week. What remains in the factor \(23\)
The 3 equations have a nice factorization if 1 is added to each side. We get \[\begin{align}&(x+1)(y+1)=525=3\cdot5\cdot5\cdot7\\ &(y+1)(z+1)=147=3\cdot7\cdot7\\&(z+1)(w+1)=105=3\cdot5\cdot7\end{align}\] From the first 2 of these, \(y+1\) divides both \(525\) and \(147\) so it divides their GCD which is \(21\). Similarly, \(z+1\) divides \(21\). Since \(y>0\), the 3 cases are \(y+1=3,7,21\). If \(y+1=3\), then \(z+1=49\) which does not divide \(21\) so it can't be that. If \(y+1=7\), then \(x+1=75\) so we find \(37\) is a factor of \(x\) but it is not a factor of \(8!\) so this case does not work either. Finally, if \(y+1=21\), then \(z+1=7,x+1=25,w+1=15\). So by multiplying the variables \(x=24,y=20,z=6,w=14\), we can confirm \(xyzw=8!\). Therefore the solution is \(x=24\).
Alec lost \(20\%\) of his money. He then gained \(x\%\) of what he had remaining, just breaking even.
By losing \(20\%\), Alec has \(0.8M\) where \(M\) is the amount he started with. From here, we solve the equation \(0.8M(1+x/100)=M\) to find what \(\%\) he has to gain to get back to \(M\). \[0.8M(1+x/100)=M\Rightarrow1+x/100=1.25\Rightarrow x=25\]
Find the smallest positive integer \(x\) such that the sum of \(x\) and the next \(50\) consecutive numbers is a square number.
The sum is \[x+(x+1)+(x+2)+\ldots+(x+50)=51x+{50\cdot51\over2}=51(x+25)\] The factors are \(51=3\cdot17\) so \(x+25\) must have these same factors to make this quantity a square. Therefore, \(x+25=51\Rightarrow x=26\).
It turns out there is actually no point in 3D space that simultaneously satisfies all 3 of these equations. We can find a solution for \(x\) which may correspond to imaginary solutions. Start with the 2nd equation, multiply \(x\), then substitute the 3rd \[x^2+xy+xz=72x\Rightarrow yz+xy+xz=72x\] Now square both sides of the 2nd equation and use some substitutions \[\begin{align}&x^2+y^2+z^2+2xy+2xz+2yz=72^2\\ &1296+2(xy+xz+yz)=72^2\\ &36^2+2(72x)=72^2\\ &144x=3\cdot36^2\Rightarrow x=27\\\end{align}\] Now if we substitute this into the equations, we get a new system in 2 variables \[\begin{align}&y^2+z^2=567\\&y+z=45\\&yz=729\\\end{align}\] But by using the first 2 equations \[y^2+(45-y)^2=567\Rightarrow y^2-45y+729=0\Rightarrow y={45\pm9\sqrt{-11}\over2}\] Then we can use this to show that there are 2 solutions in complex 3D space that satisfy all 3 equations simultaneously \[\begin{align}&x=24,y={45+9\sqrt{-11}\over2},z={45-9\sqrt{-11}\over2}\\ &x=24,y={45-9\sqrt{-11}\over2},z={45+9\sqrt{-11}\over2}\end{align}\]
The 3 given angles \(66^\circ,114^\circ,152^\circ\) are supplementary to angles of measures \(114^\circ,66^\circ,28^\circ\) respectively. These 3 angles are interior to a convex quadrilateral so the remaining angle (which is supplementary to \(x^\circ\)) is \(360^\circ-114^\circ-66^\circ-28^\circ=152^\circ\). From here we find \(x^\circ=180^\circ-152^\circ=28^\circ\)
A set of 5 positive integers has a median of 17 and mean of 13. Find the largest possible value in the set.
For clarification, the word used is "set" is used but equal elements are allowed. Describe the set by the 5 integers \[0<a\leq b\leq c\leq d\leq e\] where \(c=17\). From the mean, we know \[a+b+c+d+e=13\cdot5=65\] Subtract \(c=17\) so we have \[a+b+d+e=48\] Next subtract 2 since \(a,b\geq1\) which shows \[d+e\leq46\] And finally since \(d\geq c=17\), we find the largest element is \[e\leq46-17=29\] So the largest possible value is 29. The solution set is 1,1,17,17,29.
Consider the circumscribed circle around this regular hexagon. The angle \(x^\circ\) subtends an arc of \(60^\circ\) since it is \(1/6\) of the circle. Therefore, \(x=30\), half of the arc measure.