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| 24 | 25 | 26 | 27 | 28 | 29 | 30 |
Take the linear combination \(1\cdot(2^{26}+1)-2\cdot(2^{25}+1) =2^{26}+1-2^{26}-2=-1\). The greatest common divisor must divide this value so the solution is 1.
Let \(a=\sqrt[3]{7+\sqrt{50}}\) and \(b=\sqrt[3]{7-\sqrt{50}}\). Then \[x^3=(a+b)^3=a^3+b^3+3ab(a+b)=(7+\sqrt{50})+(7-\sqrt{50}) +3\sqrt[3]{(7+\sqrt{50})(7-\sqrt{50})}x=14-3x\] so the values of \(x\) are solutions to \(x^3+3x-14=0\). By the rational root theorem, we can find \(x=2\) is a solution. Then it factors to \((x-2)(x^2+2x+7)=0\). The only real solution is \(x=2\). The other solutions are \(x=-1\pm\sqrt{-6}\).
Find the minimum vertical distance between the graphs of \(3+\sin(x)\) and \(\cos(x)-\sqrt{2}\).
Let \(f(x)=3+\sin(x)\) and \(g(x)=\cos(x)-\sqrt{2}\). The goal is to maximize \(|f(x)-g(x)|\). This will occur when \(f(x)-g(x)\) is at a maximum or minimum. The derivative is \({d\over dx}(f(x)-g(x))=\cos(x)+\sin(x)\). This is 0 when \(x=-\pi/4\) or \(x=3\pi/4\). Testing these to points, we find \(f(x)-g(x)\) to be \(3-2\sqrt{2}\) and 3. The absolute value is maximized at 3.
What is the maximum number of lattice points that can be contained within a circle of radius 1?
If the boundary is allowed, then we would have 5 as the answer by using the standard unit circle, so we assume the boundary is not allowed and the lattice points must be strictly inside the circle. It is possible to have 4 points, such as the circle \((x-1/2)^2+(y-1/2)^2=1^2\). In any other arrangement of 5 adjacent lattice points, if there are 3 in a straight line along a grid axis, then the length is 2 so it cannot fit on the circle diameter. If there are not 3 in such a straight line, then they form a "zig zag" which also cannot fit in the unit circle.
Consider the following right triangles in orange.
The smaller one has legs of lengths \(1/3\) and \(2/3\) and the right triangles are similar. Therefore, the larger one has legs of lengths \(1\) and \(2\). By the pythagorean theorem, the side length of the blue square is \(\sqrt{5}\) so \(x=5\).
It should be more clearly stated, but it asks us to find the \(x\). To find the intersection point, we first find \(u\) such that \(f(u)=g(u)\). \[u^2-2=(u-5)^2+3\Rightarrow u^2-2=u^2-10u+25+3\Rightarrow10u=30\Rightarrow u=3\] Then we can evaluate \(f(3)=g(3)=7\) so the intersection point is \((3,7)\) so the answer would be 7.
Find the sum of the angles of a decagon divided by the sum of the angles of a triangle.
For a polygon with \(n\) sides, its angles sum to \(180(n-2)\) degrees. A triangle has 3 sides and a decagon has 10 sides so the ratio is: \[{180(10-2)\over180(3-2)}={8\over1}=8\]
Find the volume of a regular tetrahedron with side length \(3\sqrt{2}\).
Let \(s=3\sqrt{2}\). Since this is a pyramid, we can find the area of the base and the height. The base has side length \(s\) and a medians divides it into 2 30-60-90 triangles. From here, its height is \(h=s\sqrt{3}/2\) and the area is \[B={1\over2}sh={1\over2}s{s\sqrt{3}\over2}={s^2\sqrt{3}\over4} ={9\sqrt{3}\over2}\] Next we need the height of the tetrahedron. From the base triangle, drawing 3 medians divides it into 6 30-60-90 triangles. From this, we can find that the centroid is distance \(d={s\over2\sqrt{3}}\) from the midpoint of a side. Now, we use the pythagorean theorem to find the height of the tetrahedron. \[H^2=h^2-d^2={3s^2\over4}-{s^2\over12}={2s^2\over3} \Rightarrow H={s\sqrt{2}\over\sqrt{3}}={6\over\sqrt{3}}=2\sqrt{3}\] Then we use \(B\) and \(H\) in the volume of a pyramid formula. \[V={1\over3}BH={1\over3}{9\sqrt{3}\over2}{2\sqrt{3}}=9\]
\[u=\int_2^{24}{d(t^2+t)\over t^2+t}=\ln|t^2+t|\Big|_2^{24} =\ln(600)-\ln(6)=\ln(100)\] \[x=e^{u/2}=e^{{1\over2}\ln(100)}=(e^{\ln(100)})^{1/2}=100^{1/2}=10\]
We can see that \(\lim_{x\to\infty}{4x+30\over x+14}=4\) and \({d\over dx}{4x+30\over x+14}={26\over(x+14)^2}>0\) for all positive \(x\). Therefore, as \(x\) increases, this fraction is only increasing, and is always less than 4. So this fraction must be equal to 1, 2, or 3. This gives us 3 equations to solve \[4x+30=1\cdot(x+14),\ 4x+30=2\cdot(x+14),\ 4x+30=3\cdot(x+14)\] which give solutions \(-16/3,-1,12\) respectively. Clearly 12 is the only valid solution since \(x\in\mathbb{N}\).
Find the number of digits in \(11^{12}\)
First, we can write this as \((10\cdot1.1)^{12}=10^{12}\cdot1.1^{12}\). So we can put bounds on \(1.1^{12}\). Clearly \(1.1^{12}\geq1\) so the answer is at least \(10^{12}\) containing 13 digits. For a crude upper bound: \[\begin{align}&(1+0.1)^{12}=1+12\cdot0.1+{12\cdot11\over2}\cdot0.1^2 +{12\cdot11\cdot10\over6}0.1^3+\ldots=1+1.2+0.66+0.22+\ldots\\& \leq3.08+2^{12}\cdot0.1^4=3.08+0.4096=3.4896\end{align}\] That last part comes from taking the remaining terms \[{12\choose4}0.1^4+{12\choose5}0.1^5+\ldots \leq0.1^4\left({12\choose4}+{12\choose5}+\ldots\right)\leq0.1^4\cdot2^{12}\] We have replaced all greater powers of 0.1 with \(0.1^4\), and the sum of the binomial coefficients \({12\choose k}\) for \(0\leq k\leq12\) is \(2^{12}\) so that also is an upper bound. This shows that \(10^{12}\leq11^{12}\leq3.4896\cdot10^{12}\). This guarantees that it has 13 digits without knowing the exact value. If we were to compute the exact value, we would get 3138428376721, which clearly is 13 digits.
We must have \(a\neq0\). Cross multiply to find \(10a+10x=ax\) and \(5x=2a\). Using the 2nd of these to substitute into the first, we have \(ax=10a+2\cdot2a=14a\Rightarrow x=14\).
How many twin primes are less than 100?
The simplest way to solve this is probably thet sieve of Eratosthenes. We only need to cross out multiples of \(2,3,5,7\) since those are the only primes up to \(\sqrt{100}\). The table below has the primes in green, created as if starting entirely in green and then marking composite numbers in red.
| 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 |
Next, count the primes \(p\) such that either \(p+2\) or \(p-2\) are prime. We find the primes \(3,5,7,11,13,17,19,29,31,41,43,59,61,71,73\). This list has 15 primes.
\(x=\) the number of sums of positive integers that add up to 5 (order matters).
We can solve this by breaking it into smaller problems. For 1, there is just one way with the number 1. For 2, our sum can start with 1 or 2. If it starts with 1, we have 1 way to finish it by summing to 1. If it starts with 2, it is done. Therefore there are 2 ways. For 3, it starts with 1, 2, or 3. By similar reasoning, there are \(2+1+1=4\) ways to make a sum to 3. We can repeat this 2 more times to show that the answer is 16, but there is a pattern to notice here. Let \(P(n)\) mean there are \(2^{n-1}\) ways to write \(n\) as a sum. Clearly \(P(1)\) is true.
Now assume \(P(1),P(2),\ldots,P(n)\) is true. Then to sum to \(n+1\), we can have the sum start with \(1,2,\ldots,n,n+1\). If it starts with \(n+1\), we are done so count 1 way. For all the other cases, we sum the numbers from \(P(1),P(2),\ldots,P(n)\) so we have \[1+2^{n-1}+2^{n-2}+\ldots+2^0=1+(2^n-1)=2^n\] so \(P(n+1)\) is true. Therefore, the answer is \(P(5)\), which is that there are \(2^{5-1}=16\) sums of positive integers with order mattering that sum to 5.
Find the product of the four digits of \(54\times58\).
Product is \(54\times58=3132\). Product of digits is \(3\cdot1\cdot3\cdot2=18\).
The product of two positive numbers is 266 and their difference is 5. Find the larger number.
Let the numbers be \(a>b>0\). Then \(ab=266\) and \(a-b=5\Rightarrow a=5+b\). Substituting this, \((5+b)b=266\Rightarrow b^2+5b-266=0\). Then the solutions are \(b={-5\pm\sqrt{25+4\cdot266}\over2}=14,-19\). Since \(b>0\), take \(b=14\). Then the larger number is \(a=19\). This also can be seen easily from the prime factorization \(266=2\cdot7\cdot19\).
The number of ways to arrange the leters of \(XOXOXO\).
This can be thought of a permutations and dividing the factorials for each type of letter, so \({6!\over3!3!}=20\). Since there are just 2, it can also be seen as combinations, choosing positions for \(X\) and the remaining for \(O\), so \({6\choose3}=20\).
Alec scored 61 points. Gwen had three times the points of Sierra and two more than Alec. How many points did Sierra have?
Turning this into equations, we have \(a=61,g=3s,g=a+2\). From this, \(g=61+2=63\) so \(63=3s\Rightarrow s=21\). Therefore, Sierra has 21 points.
How many times do the minute and hour hands of a standard 12-hour clock overlap in one day?
We can think of this in terms of how many times they go around the clock. They are at the same position at hour 0. Then going to hour 24, the hour hand goes around twice and the minute hand goes around 24 times. They stop in the same position. The minute hand spins faster and passes thet hour hand \(24-2=22\) times during the day.
I did not find a particularly satisfying solution for this problem so I'll present a simple way of seeing a solution, without showing that is it all the solutions. As \(y\) grows, \(x\) grows much faster. When \(y=0\), \(x^2=4\) is a perfect square, but we need \(x>5\) so keep trying more values of \(y\). When \(y=4\), we find \(x^2=529=23^2\) so \(x=23\). This is the solution, but I have not found a way to show whether or not more solutions exist.
The order of the symmetry group of the regular dodecagon.
The order of a polygon symmetry group is twice its number of sides, so for the dodecagon this is \(2\cdot12=24\). Each one describes one of 12 rotations either with or without a reflection.
Consider another sum that we can evaluate more easily. \[\begin{align}&\sum_{n=1}^{N^2-1}\lfloor\sqrt{n}\rfloor =\sum_{m=1}^{N-1}\sum_{n=m^2}^{(m+1)^2-1}\lfloor\sqrt{n}\rfloor =\sum_{m=1}^{N-1}\sum_{n=m^2}^{m^2+2m}m=\sum_{m=1}^{N-1}m(2m+1) =2\sum_{m=1}^{N-1}m^2+\sum_{m=1}^{N-1}m\\& ={(N-1)N(2N-1)\over3}+{(N-1)N\over2} ={N(N-1)\over6}\left({4N-2\over6}+{3\over6}\right)={N(N-1)(4N+1)\over6}\\ \end{align}\] Then the summation for our problem becomes \[\sum_{n=1}^{100}\lfloor\sqrt{n}\rfloor =\sum_{n=1}^{10^2-1}\lfloor\sqrt{n}\rfloor+\lfloor\sqrt{100}\rfloor ={10(10-1)(40+1)\over6}+10=625\] Then we take the square root so the answer is \(25\).
Let \(x=\cos(\pi/12)\) and \(y=\sin(\pi/12)\). Then we have \[{{x^3\over y^3}-{y^3\over x^3}\over2}={x^6-y^6\over2x^3y^3} ={(x^3+y^3)(x^3-y^3)\over2x^3y^3}\] Now we need to know what \(x,y\) are. The sum/difference trigonometric identities can be used. \[\begin{align}&x=\cos(\pi/12)=\cos(\pi/3-\pi/4) =\cos(\pi/3)\cos(\pi/4)+\sin(\pi/3)\sin(\pi/4)\\ &={1\over2}{\sqrt{2}\over2}+{\sqrt{3}\over2}{\sqrt{2}\over2} ={\sqrt{6}+\sqrt{2}\over4}\end{align}\] \[\begin{align}&y=\sin(\pi/12)=\sin(\pi/3-\pi/4) =\sin(\pi/3)\cos(\pi/4)-\cos(\pi/3)\sin(\pi/4)\\ &={\sqrt{3}\over2}{\sqrt{2}\over2}-{1\over2}{\sqrt{2}\over2} ={\sqrt{6}-\sqrt{2}\over4}\end{align}\] Next, using some algebra, we find \[x^3={3\sqrt{6}+5\sqrt{2}\over16},\quad y^3={3\sqrt{6}-5\sqrt{2}\over16}\] Then we substitute these. \[\begin{align}&{(x^3+y^3)(x^3-y^3)\over2x^3y^3} ={\left({3\sqrt{6}+5\sqrt{2}\over16}+{3\sqrt{6}-5\sqrt{2}\over16}\right) \left({3\sqrt{6}+5\sqrt{2}\over16}-{3\sqrt{6}-5\sqrt{2}\over16}\right) \over2{3\sqrt{6}+5\sqrt{2}\over16}{3\sqrt{6}-5\sqrt{2}\over16}}\\& ={{6\sqrt{6}\over16}{10\sqrt{2}\over16}\over2{9\cdot6-25\cdot2\over16^2}} ={60\sqrt{12}/256\over8/256}={120\sqrt{3}\over8}=15\sqrt{3}\end{align}\] Rounding this to the nearest integer, the answer is 26.
\(f(x)=x^3-9x^2+bx-c\) has three real roots \(>0\). Find the maximum value of \(c\).
This one can be solved by making use of Vieta's formulas. Let the roots be \(r_1,r_2,r_3>0\). Then using the 2 leading coefficients, \(r_1+r_2+r_3=9\). For the constant coefficient, we have \(r_1r_2r_3=c\). Now we need to find a way to relate these 2 expressions, which can be done with the property that arithmetic mean is always at least geometric mean. \[{r_1+r_2+r_3\over3}\geq\sqrt[3]{r_1r_2r_3}\Rightarrow3\geq\sqrt[3]{r_1r_2r_3} \Rightarrow27\geq r_1r_2r_3=c\] Therefore, the maximum value is \(c=27\). Given the constraint \(r_1+r_2+r_3=9\), the product \(r_1r_2r_3\) is a maximum when the values are equal to the average so \(r_1=r_2=r_3=3\). The polynomial corresponding to this solution is \(f(x)=(x-3)^3=x^3-9x^2+27x-27\).
Let \(x\) be the shorter square side length and \(y\) be the longer square side length. Consider the parts marked in orange below.
The desired quantity will be \(2x^2+2y^2\). To see how we can get there, we can write the area of the gray triangle by starting with the larger triangle of both the gray and orange parts, then subtracting the orange parts (2 triangles and 1 rectangle). The result is \[{1\over2}(2y-x)(2x+y)-{1\over2}xy-{1\over2}(y-x)(x+y)-x(y-x) ={1\over2}x^2+{1\over2}y^2\] This is just 1/4 of the quantity we desire, so the answer is 4 times the area of the gray triangle, 28.
The average of a finite set \(S\) is \(x\). The average of \(S\cup\{1\}\) is \(x-2\) and the average of \(S\cup\{99\}\) is \(x+5\).
Let \(n=|S|\) and \(t=\sum_{s\in S}s\). Then each statement corresponds to an equation: \[{t\over n}=x,\quad{t+1\over n}=x-2,\quad{t+99\over n}=x+5\] From the first equation, substitute \(t=nx\) into the others to get \(x-2n=3,x+5n=94\) which can be solved for \(n=13,x=29\). So the answer is \(x=29\).
But we made the assumption that \(1\not\in S\) and \(99\not\in S\). If we assume both \(1,99\in S\), then the last 2 equations become \({t\over n}=x-2=x+5\) which has no solution. If we have \(1\in S,99\not\in S\), then the system would gives us \(n=92/7,x=198/7\). If we have \(1\not\in S,99\in S\), then the solution is \(n=-8/7,x=5/7\). In both of these cases, \(S\) would not have a nonnegative integer size, so even if we are careful about our assumptions, we find that no matter what \(S\) is, \(1,99\not\in S\).
How many numbers from 1 to 100 are divisible by 4 or 5 but not 6?
This can be done by first drawing a venn diagram of 3 circles, one each for numbers divisible by 4, 5, or 6. Let's first start with numbers divisible by 4 or 5. That is \[\left\lfloor{100\over4}\right\rfloor+\left\lfloor{100\over5}\right\rfloor -\left\lfloor{100\over\text{lcm}(4,5)}\right\rfloor=25+20-5=40\] Next, we have to subtract numbers divisible by 4 and 6, as well as numbers divisible by 5 and 6. These are \(\lfloor100/12\rfloor=8\) and \(\lfloor100/30\rfloor=3\) respectively. So now we have \(40-8-3=29\). But the last thing we need to do is add numbers divisible by all of 4,5,6 because we just subtracted each twice. There is only 1 such number since \(\lfloor100/60\rfloor=1\) so the answer is 30.