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Find the area of the two regions bounded by \(y=x^2,y=x,x=2\)
Above is the plot of the given information. There are 2 regions bounded with the 2 graphs crossing at \((1,1)\). The first area is \[\int_0^1(x-x^2)dx=\left[{x^2\over2}-{x^3\over3}\right]_0^1 ={1\over2}-{1\over3}={1\over6}\] The second area is \[\int_1^2(x^2-x)dx=\left[{x^3\over3}-{x^2\over2}\right]_1^2 ={8\over3}-{4\over2}-{1\over3}+{1\over2}={5\over6}\] So the total area is \(1/6+5/6=1\).
The left side is strictly increasing so there is exactly 1 solution. We can conveniently factor \(270=10\cdot27\) and factor the left side \[3^{2x-1}(1+9)=27\cdot10\Rightarrow3^{2x-1}=27 \Rightarrow2x-1=3\Rightarrow x=2\]
Find the ones digit of \(7^{343}\)
We can find the cycle for powers of 7 modulo 10 \[7^1\equiv7,\qquad7^2\equiv9,\qquad7^3\equiv3,\qquad7^4\equiv1\] So we can factor out some power of \(7^4\equiv1\) to simplify it modulo 10 \[7^{343}\equiv7^{340+3}\equiv(7^4)^{85}\cdot7^3\equiv1^{85}\cdot7^3\equiv7^3 \equiv3\]
First square both sides \[16+8\sqrt{3}+16-8\sqrt{3} +2\sqrt{\left(16+8\sqrt{3}\right)\left(16-8\sqrt{3}\right)}=3x^2\] Then simplify \[32+2\sqrt{16^2-64\cdot3}=32+2\sqrt{256-192}=32+2\sqrt{64}=32+2\cdot8=48\] So the equation is \(3x^2=48\Rightarrow x=\pm4\). But the quantity we started with must be positive so \(x=4\).
Find the number of Mersenne primes less than 10000.
The Mersenne primes are well known with many lists available on the internet. The first 5 are \[2^2-1=3,\qquad2^3-1=7,\qquad2^5-1=31,\qquad2^7-1=127,\qquad2^{13}-1=8191\] The next one is \(2^{17}-1=131071\) which exceeds the limit.
We need to find what some multiple compositions of \(f\) are. \[f^2(t)={{t-1\over t+1}-1\over{t-1\over t+1}+1}={t-1-(t+1)\over t-1+(t+1)} ={-2\over2t}={-1\over t}\] \[f^3(t)={{-1\over t}-1\over{-1\over t}+1}={-1-t\over-1+t}\] \[f^4(t)={{-1-t\over-1+t}-1\over{-1-t\over-1+t}+1} ={-1-t-(-1+t)\over-1-t+(-1+t)}={-2t\over-2}=t\] \[f^5(t)={t-1\over t+1}=f^1(t)\] From here we can see a cycle of 4 forms so \(f^9=f^5\) and \(f^8=f^4\). \[x={-1\over{1\over2}}+{9-1\over9+1}\cdot10=-2+{4\over5}\cdot10=-2+8=6\]
The cubic function \(x^3+x-350\) has derivative \(3x^2+1\) which is always positive so there is exactly 1 real solution. The constant \(350=2\cdot5\cdot5\cdot7\) has quite a few factors that could be tried with the rational root test to find \(x=7\) which is the solution. We could guess \(x=7\) because \(x^3=343\) is close to \(350\) and see that this happens to work. We could factor \(x^3+x=x(x^2+1)\) and see that this matches the factorization \(350=7\cdot50=7\cdot(7^2+1)\).
For solving with more general methods, we could observe that this is a depressed cubic. Cardano's method applies because the \(x\) coefficient and constant satisfy the required condition: \[t^3+pt+q=0,\qquad{q^2\over4}+{p^3\over27}>0\] The value of this positive value is \[D={(-350)^2\over4}+{1\over27}={3307500+4\over108}={826876\over27}\] Then Cardano's formula says there is a single real root of \[\sqrt[3]{-{q\over2}+\sqrt{D}}+\sqrt[3]{-{q\over2}-\sqrt{D}} =\sqrt[3]{175+{2\over3}\sqrt{206719\over3}} +\sqrt[3]{175-{2\over3}\sqrt{206719\over3}}\] Approximate calculation shows this is \(7\). There should be some gross algebra to prove it is exactly equal to \(7\).
This can be solved by trying various \(a^b<100\) and checking if adding \(b^a\) gets to \(100\). Due to the symmetry we can use \(a\leq b\) without loss of generality. Starting with \(a=2\) we have \[2^2,2^3,2^4,2^5,2^6\] From here we find \(2^6+6^2=64+36=100\) which is the solution. Therefore \(x=2+6=8\). To show there are no other solutions, we continue with \(3^3\) and \(3^4\) and find that neither of these create a solution. We can stop here because \(4^4>100\).
\(a,b,c,\ \text{and}\ x\) are distinct digits such that \(ac\times bc=xxx\)
The following equation describes what we are looking for. \[(10a+c)(10b+c)=100ab+10bc+10ac+c^2=100x+10x+x=111x=3\cdot37\cdot x\] From this, \(37\) must divide \(10a+c\) (without loss of generality due to symmetry) so \(10a+c=37\) or \(10a+c=74\) since it must be 2 digits.
If \(10a+c=37\) then \(a=3,c=7\) and \(370b+259=111x\). Since \(x\) would be a single digit, we must have \(b=1\) or \(b=2\). It turns out to work only for \(b=2\) which gives us the product \(37\times27=999\) so \(x=9\).
If \(10a+c=74\) then \(a=7,c=4\) and \(740b+296\) which does not work for any \(b\) because \(b=1\) makes this larger than 1000.
How many minutes after 2pm do the hour and minute hands meet? (Round)
Starting at 2pm, the minute hand is at 0 degrees and the hour hand is at \((2/12)\cdot360=60\) degrees. From here, every minute that passes move the minute and hour hands to \(6x\) degrees and \(60+6x/60\) degrees since the minute hand moves 6 degrees per minute and the hour hand moves 60 times slower. Solving this equation \[6x=60+0.1x\Rightarrow5.9x=60\Rightarrow x=60/5.9\approx10.17\] So it is about 10 degrees when rounded.
How many ways can one partition 6?
These can be counted by brute force, there are few enough of them for this to be feasible. This can be counted more intelligently by using a recursive formula for how many partitions of \(n\) are there with largest partition \(k\). Below are the 11 partitions of 6. \[\begin{align}&6\\&5+1\\&4+2\\&4+1+1\\&3+3\\&3+2+1\\&3+1+1+1\\& 2+2+2\\&2+2+1+1\\&2+1+1+1+1\\&1+1+1+1+1+1\end{align}\]
First use the triangle on the bottom which is 30-60-90. The ratio of the hypotenuse to the longer leg is \(2/\sqrt{3}\) so the hypotenuse is \[3\sqrt{6}\cdot{2\over\sqrt{3}}=6\sqrt{2}\] This is a leg of the larger triangle which is 45-45-90. So the length of its hypotenuse (which is \(x\)) is \[6\sqrt{2}\cdot\sqrt{2}=12\]
First summation using the sum of squares formula with \(n=6\). \[\sum_{i=1}^n{i^2}={n(n+1)(2n+1)\over6}={6\cdot7\cdot13\over6}=91\] Second summation using sum of integers formula with \(n=12\). \[\sum_{i=1}^n{i}={n(n+1)\over2}={12\cdot13\over2}=78\] Therefore the expression evaluates to \(91-78=13\).
The 2 semicircles add to a single circle. The diameter of this circle is the side length of the square which is \(\sqrt{56/\pi}\) so the circle area is \[\pi\left({1\over2}\sqrt{56\over\pi}\right)^2={\pi\over4}{56\over\pi} ={56\over4}=14\]
We can evaluate \(y\) with the substitution \(u=t^2+t,du=(2t+1)dt\) \[y=\int_2^9{d(t^2+t)\over t^2+t}=\int_{2^2+2}^{9^2+9}{du\over u} =\int_6^{90}{du\over u}=\ln(90)-\ln(6)=\ln\left({90\over6}\right)=\ln(15)\] Then \(x=e^y=e^{\ln(15)}=15\)
60 pennies are divided into four separate piles of distinct sizes. At least how many are in the largest pile?
If every pile were smaller than 15, then there would be fewer than \(4\cdot15=60\) pennies so the answer is at least 15. If every pile were at most 15, then they would all need to be 15 to add to 60 pennies so the answer must be bigger than 15. If the largest pile is 16, then the largest sum of pennies we could make with distinct pile sizes is \(16+15+14+13=58\), too small. With a pile of size 17, we could have \(17+16+15+14=62\). So only once a pile of size 17 exists is it possible to have 4 distinct pile sizes summing to 60. One such example is \(17+16+15+12=60\)
Since the bottom triangle is isosceles, the bottom left angle is also \(2y^\circ\). Let \(z\) be the missing angle in the upper triangle. Then we form the equations \[2x+2y+2y=180,\quad2x+y+z=180,\quad2y+z=180\] Subtract the middle equation from the one on the left to get \[3y-z=0\Rightarrow3y=z\] Then substituting into the equation on the right we have \[2y+3y=180\Rightarrow y=36\] Then \(z=3\cdot36=108\) and we can use these in the middle equation to find \[2x=180-y-z=180-36-108=36\Rightarrow x=18\]
How many times does the digit 4 appear among all two digit numbers?
In the ones place, 9 times, once for each of the tens digits 1-9. In the tens place, 10 times for the numbers 40-49. In total \(9+10=19\).
The average of seven numbers, including \(x\), is \(12\). When \(x\) is removed, the average of the remaining numbers is \(10.5\).
Let \(S\) be the sum of the other \(6\) numbers. Then \[{S+x\over7}=12,\quad{S\over6}=10.5,\quad\Rightarrow\quad S+x=84,\quad S=63\] From here we can find \(x=84-S=84-63=21\)
The individual squares can easily be counted: \(1+2+3+4+5=15\). The 2x2 squares are formed from 4 of the small squares and they are positioned in a similar stairs structure with length 3 instead of 5, so there are \(1+2+3=6\) of them. Finally, there is a single 3x3 square positioned at the bottom right. In total, there are \(15+6+1=22\) squares.
Two cars begin at the same location and travel in opposite directions at 40mph and 60mph respectively. How many minutes does it take for them to be 40 miles apart?
The cars move apart at \(40\text{mph}+60\text{mph}=100\text{mph}\). To reach \(40\) miles apart, it takes \[{40\text{mi}\over100\text{mph}}={2\over5}\text{hour} ={2\over5}\cdot60\text{min}=24\text{min}\]
How many primes are less than 100?
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | ||
| 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 |
| 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 |
| 40 | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 |
| 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 |
| 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 |
| 70 | 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 |
| 80 | 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 |
| 90 | 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 |
The easiest way to solve this would be to count them with the sieve of Eratosthenes or similar. We pick the first prime \(2\) and cross off multiples starting at \(2^2=4\). Then we do the same for \(3,5,7\) and can stop because the next prime is \(11\) and \(11^2>100\). The result would be the table above with the primes in green. We can count 25 of them.
Notice how \(t/4\) and \(4/t\) are multiplicative inverses. If we make \(t/4\) both \(7\) and \(1/7\), then we will have 2 equations. First, set \(t=28\) \[f(7)=3f\left({1\over7}\right)-196\] Then set \(t=4/7\) \[f\left({1\over7}\right)=3f(7)-4\] Now we have a system of equations in \(f(7)\) and \(f(1/7)\). Solving this system, we will find \(f(1/7)=74\) and \(f(7)=26\)
We can go a step further and solve for what the function \(f(t)\) is. Let \(u=t/4\) (this is a bijective mapping of the domain). Then substitute it \[f(u)=3f(1/u)-28u\] Similarly substitute \(u=4/t\) (also a bijective mapping of the domain) \[f(1/u)=3f(u)-28/u\] From here, we can find 2 equations for \(f(1/u)\) \[f(1/u)=3f(u)-28/u={1\over3}(f(u)+28u)\] Then solving this equation in \(f(u)\) \[9f(u)-84/u=f(u)+28u\Rightarrow8f(u)=84/u+28u\Rightarrow f(u)=10.5u^{-1}+3.5u\] Therefore the function is \[f(t)={21\over2t}+{7t\over2}\]
How many digits are in \(20^{20}\)?
One way to work this out with manageable numbers is \[20^{20}=2^{20}\cdot10^{20}\approx1.05\cdot10^6\cdot10^{20} \approx1.05\cdot10^{26}\] This uses the knowledge that \(2^{20}\) is a little larger than \(10^6\). The result is a little larger than \(10^{26}\), a 27 digit number.
We could also use logarithms. \[\log_{10}(20^{20})=20\log_{10}(20)\approx26.02\] This tells us it is between \(10^{26}\) and \(10^{27}\) so it must be 27 digits.