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This problem has a mistake. Either change the 99 to 100 or change the 101 to 100. The same answer results. Here I will change the 101 to 100. Consider the pattern of partial sums of \[{1\over2!}+{2\over3!}+{3\over4!}+{4\over5!}+\ldots\] which are \[{1\over2}=1-{1\over2},{5\over6}=1-{1\over6},{23\over24}=1-{1\over24},\ldots\] This suggests the formula \[\sum_{n=1}^x{n\over(n+1)!}=1-{1\over(x+1)!}\] We can easily see it is true when \(x=1\), and can be proven by induction. \[\sum_{n=1}^{x+1}=1-{1\over(x+1)!}+{x+1\over(x+2)!} =1-{x+2\over(x+2)!}+{x+1\over(x+2)!}=1-{1\over(x+2)!}\] So from this, we have \[y=1-{1\over100!},\quad x=1\]
This is an exponential equation that can be solved as a quadratic since \(4^x=(2^x)^2\). Let \(y=2^x\). Then \[y^2-32-4y=0\Rightarrow y={4\pm\sqrt{16+4\cdot32}\over2} ={2\pm\sqrt{4+32}}=2\pm\sqrt{36}=2\pm6=-4,8\] So then solving for \(x\), we have \(x=3\) since there is no solution for \(y=-4\).
Consider the right triangle on the right with side lengths 4 and 12. The remaining leg has length \(\sqrt{12^2-4^2}=\sqrt{128}\). Then let \(a\) be the altitude of this triangle and we can form the ratio \[{12\over\sqrt{128}}={4\over a}\Rightarrow a={\sqrt{128}\over3}\] Now let \(h\) be the height of the shaded triangle. This is also a leg of one of the right triangles the altitude formed so \[h=\sqrt{4^2-a^2}=\sqrt{16-{128\over9}}=\sqrt{144-128\over9}={4\over3}\] Then the area of the shaded triangle is \[{1\over2}\cdot6\cdot{4\over3}=4\]
Another exponential equation that can be solved as a polynomial. Let \(y=2^x\). \[y^3+y=32y^2+32\Rightarrow y(y^2+1)=32(y^2+1)\Rightarrow y=32\] Then \(x=\log_2(y)=5\)
To solve for \(x\) here, we do not actually need to use all the information. This is a quadrilateral with interior angle sum \(360^\circ\). The 2 angles on the bottom sum to \(180^\circ\) and only involve \(x\) so we can find it from there. \[(10x+25x-30)=180\Rightarrow35x=210\Rightarrow x=6\]
Jamal Brown is Kate's brother. Jamal has one brother, and Kate has twice as many sisters as brothers. How many Brown siblings are there?
Jamal is a brother and we have to assume Kate is a female name here so she is a sister. If Jamal has 1 brother, then there are 2 males. This means Kate has 2 brothers, and twice as many sisters, so 4 sisters. There are 2 males and 5 females in the family so 7 siblings total.
The 2nd equation tells us \(a=7c\) so combining with the 3rd gives us \(7c-c=18\Rightarrow c=3\). Then we can use the first equation for the other variables (which we do not need to know the individual values of) so \[x=a+b+c=5+c=5+3=8\]
Find the number of line segments in the complex plane that connect two of the sixth roots of 1 where at least one root is real.
The roots form a hexagon around the origin and the only real roots are \(1,-1\). There is a line connecting these 2 real roots. Then for each of these real roots we can connect a line to 4 imaginary roots. In total, \(1+4+4=9\) lines.
There is not a whole lot that can be done to simplify the algebra, but one observation is \[x=t^3(t^2+3t+1)+2t+10=t^3((t+1)^2+t)+2t+10\] Then we can evaluate \((t+1)^2=2\) and \(t^3=5\sqrt{2}-7\) and substitute \[x=\left(5\sqrt{2}-7\right)\left(\sqrt{2}+1\right)+2\sqrt{2}+8 =10-7\sqrt{2}+5\sqrt{2}-7+2\sqrt{2}+8=11\]
Let \(r,s\) be the solutions of \(y+{1\over\sqrt{y}}=2\). Then let \(z=r+s+\sqrt{r}+\sqrt{s}\) and find \(x=z^2+z\). First we can find solutions for \(y\). \[y\sqrt{y}+1=2\sqrt{y}\] Let \(u=\sqrt{y}\), then we have a cubic \[u^3-2u+1=0\Rightarrow(u-1)(u^2+u-1)=0\Rightarrow u=1,{-1\pm\sqrt{5}\over2}\] Since \(u\geq0\), the solutions are \(u=1,u={-1+\sqrt{5}\over2}\) which can both be verified. We square these to find solutions \(y=1,{3-\sqrt{5}\over2}\). Let \(r=1,s={3-\sqrt{5}\over2}\). Then after a bit of algebra \[z={7\over2}-{\sqrt{5}\over2}+t,\quad t=\sqrt{{3\over2}-{\sqrt{5}\over2}}\] \[z^2+z=z+{49\over2}+{5\over4}+t^2 -2{7\over2}{\sqrt{5}\over2}+2{7\over2}t-2{\sqrt{5}\over2}t ={37\over2}-{9\sqrt{5}\over2}+t\left(8-\sqrt{5}\right)\] Now supposing we can find an integer solution, we should be able to find some \(a,b\) such that \[a+b\sqrt{5}=t\left(8-\sqrt{5}\right)\] Looking at the terms we have, suppose \(b=9/2\). Then by squaring, \[a^2+{81\cdot5\over4}+9a\sqrt{5} =\left({3\over2}-{\sqrt{5}\over2}\right)\left(69-16\sqrt{5}\right) ={287\over2}-{117\sqrt{5}\over2}\] Trying to get the parts with \(\sqrt{5}\) to match, we have \[9a=-{117\over2}\Rightarrow a={-13\over2}\] We can verify that this choice of \(a\) solves the equation. So now we have \[z^2+z={37\over2}-{9\sqrt{5}\over2}-{13\over2}+{9\sqrt{5}\over2}=12\]
Using the definition of \(f(t)\), \[f(g(t))=2g(t)-1=4t+3\Rightarrow g(t)=2t+2\] So then \(x=g(6)=2\cdot6+2=14\)
First simplify \[y=\sqrt{{4\over9}+{9\over25}-{4\over5}}=\sqrt{{4\cdot25\over9\cdot25} +{9\cdot9\over25\cdot9}-{4\cdot5\cdot9\over5\cdot5\cdot9}} ={1\over3\cdot5}\sqrt{100+81-180}={1\over15}\] So then \(x=1/y=15\)
For the line through \((-4,0)\) and \((2,6)\), we can find the slope and then use point slope form \[{6-0\over2-(-4)}=1\Rightarrow y-6=1(x-2)\Rightarrow y=x+4\] This gives us a y-intercept of 4 so \(a=4\). Doing the same for the other line \[{0-6\over4-2}=-3\Rightarrow y-6=-3(x-2)\Rightarrow y=-3x+12\] So the y-intercept is 12 and we have \(b=12\). Then \(a+b=16\)
Let \(\sigma_0(n)\) be defined as equal to the number of divisors of \(n\). Find \({\sigma_0(11!)\over\sigma_0(6!)}\)
Consider the prime factorizations \[6!=1\cdot2\cdot3\cdot4\cdot5\cdot6 =2\cdot3\cdot2\cdot2\cdot5\cdot2\cdot3 =2^4\cdot3^2\cdot5\] So \(6!\) has \((4+1)(2+1)(1+1)=30\) divisors. \[11!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11 =2^8\cdot3^4\cdot5^2\cdot7\cdot11\] So \(11!\) has \((8+1)(4+1)(2+1)(1+1)(1+1)=540\) divisors. The solution is \(540/30=18\).
Find the difference between the sum of the four element subset of \(1,2,\ldots,9\) with the largest sum and the sum of the four element subset with the least sum.
The largest subset of 4 has the 4 largest numbers and sum \(9+8+7+6=30\). The smallest has the 4 smallest and sum \(1+2+3+4=10\). The difference is \(30-10=20\)
How many integers from 80 to 130 contain the digit 1 exactly once?
Below 100, we have tens digit 8 or 9 so the only such numbers to count are 81 and 91. After 100, we need numbers without a 1 in the tens or ones digit. The only tens digits we can have are 0 or 2. For each, there are 9 digits other than 1. Finally we count 130. The total is \(2+9+9+1=21\)
Let \(a\) be the side length of one of the double tick marked edges. Then by observing the 45-45-90 right triangle, we can form the equation \[a^2+(2a)^2=\left(2\sqrt{55}\right)^2\Rightarrow a=2\sqrt{11}\] To find the shaded area, we can subtract from the area of the whole 45-45-90 triangle. \[{1\over2}(2a)^2-{1\over2}a^2-{1\over2}a(2a)={1\over2}a^2 ={1\over2}4\cdot11=22\]
Juan is given 40 cookies on a Monday and eats 2 cookies at lunch every school day and one cookie each weekend day. How many days will the cookies last?
In a whole week, he eats 12 cookies (10 on weekdays, 2 on the weekend). So after going through the week cycle 3 times, he has \(40-36\) cookies left. Then he has 4 cookies on a Monday, enough for 2 weekdays. In total it is \(21+2=23\) days.
Start with \(y=\log_{12}(18)\). Then we evaluate that fraction exponent \[{y-5\over y-2}={\log_{12}(18)-5\over\log_{12}(18)-2} ={\log_{12}(18)-\log_{12}(12^5)\over\log_{12}(18)-\log_{12}(12^2)} ={\log_{12}\left(18\over12^5\right)\over\log_{12}\left(18\over12^2\right)}\] \[={\ln(18/12^5)\over\ln(18/12^2)}={\ln(18)-5\ln(12)\over\ln(18)-2\ln(12)} ={\ln(2)+2\ln(3)-10\ln(2)-5\ln(3)\over\ln(2)+2\ln(3)-4\ln(2)-2\ln(3)}\] \[={-9\ln(2)-3\ln(3)\over-3\ln(2)}={3\ln(2)+\ln(3)\over\ln(2)} ={\ln(8)+\ln(3)\over\ln(2)}={\ln(24)\over\ln(2)}=\log_2(24)\] So then \[x=2^{y-5\over y-2}=2^{\log_2(24)}=24\]
Squaring the first equation and then adding the 2nd to cancel the \(y\) \[x+y+2\sqrt{xy}=81\Rightarrow2x+2\sqrt{xy}=90\Rightarrow \sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)=45\Rightarrow\sqrt{x}=45/9=5 \Rightarrow x=25\] We can also find \(y=16\).
An 8' by 3' rectangular pool is surrounded on all sides by 1' of grass. How many square feet of grass are there?
Adding 1' on each side increases the dimensions to 10' by 5'. The grass has increased the area from the original \(8\cdot3=24\) to \(10\cdot5=50\). The new area of \(50-24=26\) square feet is the grass added.
Find the number of two digit numbers larger than 20 whose digits sum to a number divisible by 3.
There is a theorem about divisibility by 3 which is that every number divisible by 3 has a base 10 digit sum divisible by 3. So this is the same as counting how many numbers divisible by 3 there are between 20 and 100. The largest is 99 and the smallest is 21. This gives us \({99-21\over3}=78/3=26\) steps of 3 so a total of 27 such numbers.
Square the middle equation and things will simplify \[4^r+4^s+2\cdot2^r\cdot2^s=36\Rightarrow4^r+4^s+2\cdot2^{r+s}=36\] \[\Rightarrow4^r+4^s+2\cdot2^2=36\Rightarrow4^r+4^s=36-8=28\]
The average monthly salary at a company is $8000. If 10 employees have an average salary of $12000 and the rest average $6000, how many employees are there?
We can express this as an average equation in thousands of dollars to make it simpler. \[{10\cdot12+x\cdot6\over10+x}=8\Rightarrow120+6x=80+8x \Rightarrow2x=40\Rightarrow x=20\] So there are \(x=20\) employees with an average salary of $6000. In total there are \(10+20=30\) employees.
One idea is to see \(x\) as a geometeric series and write \[x={(z^3)^{31}-1\over z^3-1}\] But we are going to see that this does not work since it requires \(z^3\neq1\). We can similarly see the first equation as a geometric series and write is as \[{z^3-1\over z-1}=0\Rightarrow z^3=1\] And clearly this works since \(z\neq1\). So \(z\) must be another cube root of unity. We could also find this with the quadratic formula. The exact values are \[z=-{1\over2}\pm i{\sqrt{3}\over2}\] But we do not actually need these. We know \(z^3=1\) so all the other powers in that formula for \(x\) are also 1. There are 31 terms so \(x=31\).