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A relevant theorem to use for this problem is the limit root theorem. If \(\lim_{n\to\infty}{a_{n+1}\over a_n}=L\) exists for a sequence \(a_n\), then \[\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}{a_{n+1}\over a_n}\] For this problem, let \(a_n={n^n\over n!}\). Now evaluate the limit \[\lim_{n\to\infty}{a_{n+1}\over a_n} =\lim_{n\to\infty}{(n+1)^{n+1}\over(n+1)!}{n!\over n^n} =\lim_{n\to\infty}{(n+1)(n+1)^n n!\over(n+1)n!n^n}\] \[=\lim_{n\to\infty}{(n+1)^n\over n^n} =\lim_{n\to\infty}\left(1+{1\over n}\right)^n=e\] So this limit exists, which means \[\lim_{n\to\infty}\sqrt[n]{n^n\over n!}=e\] The problem asks for the log of this so the solution is \(1\).
Find the integer \(x\) where \(t^2-t+x\) divides \(t^{13}+t+90\).
Let \(f(t)=t^{13}+t+90\) and \(g(t)=t^2-t+x\). Then the quotient \(f(t)/g(t)\) will be a polynomial with integer coefficients because \(g(t)\) is a monic polynomial (leading coefficient is \(1\)). For some polynomial \(h(t)\) with integer coefficients, we would find \(f(t)=g(t)h(t)\). Now evaluate the polynomials at some easy values to find divisibility properties since \(h(t)\) will be an integer. \[\begin{align} t=0&:90=x\cdot h(0)\\ t=1&:92=x\cdot h(1)\\ t=-1&:88=(2+x)\cdot h(-1)\\ \end{align}\] Since \(x\mid90\) and \(x\mid92\), \(x\mid\gcd(90,92)\) so \(x\mid2\). This leaves possible values \(x=\pm1,\pm2\). Now check with the last condition that \((2+x)\mid88\) which narrows down the possibilities to \(x=-1\) and \(x=2\). We could choose more \(t\) values to find further divisibility properties or test the result with polynomial division, but here is another interesting thing we can do with properties of polynomial roots. Since \({d\over dt}f(t)=13t^{12}+1\), \(f(t)\) is strictly increasing so it must have exactly 1 real zero. But with \(x=-1\), the discriminant of \(g(t)\) is \(1-4(1)(-1)=5\) so there would be 2 real zeroes. Since \(g(t)\mid f(t)\), zeroes of \(g(t)\) are also zeroes of \(f(t)\), so we can eliminate \(x=-1\), leaving the only possibility of \(x=2\).
Finally, we can verify by polynomial division that \[\begin{align}t^{13}+t+90 =&(t^2-t+2)(t^{11}+t^{10}-t^9-3t^8-t^7+5t^6\\ &+7t^5-3t^4-17t^3-11t^2+23t+45)\\\end{align}\]
We have a square of area \(16\). The diagonal going from the lower left to upper right splits the square in half so the big upper left triangle has an area of \(8\). The smaller upper left triangle is similar and has a side of half the length so its area is \(1/4\) of \(8\), which is \(2\). This means the strip containing \(x\) (subtracting the area \(2\) triangle from the area \(8\) triangle) has area \(6\). Since \(x\) is half of this strip, \(x=3\).
Solve this by repeatedly squaring and subtracting. \[21+\sqrt{13+\sqrt{4+x}}=25\] \[\sqrt{13+\sqrt{4+x}}=4\] \[13+\sqrt{4+x}=16\] \[\sqrt{4+x}=3\] \[4+x=9\] \[x=5\]
What is the smallest perfect number?
A number \(n\) is a perfect number when its proper divisors sum to \(n\). We can try small numbers and find \(6=1+2+3\) is the smallest. They are also related to mersenne primes, the smallest which is \(2^2-1=3\). So the associated perfect number is \(2^1\cdot(2^2-1)=2\cdot3=6\).
On the top left there is a quarter circle of radius \(5+3\sqrt{2}\). On the bottom right, is a circle of radius \(x\). From the center of the full circle, we can draw radii to the right and down to form a square of length \(x\). From that circle center, we can also draw lines going left and up to the big square sides, forming a square of side length \(10+6\sqrt{2}-x\). A diagonal of that square has length \(5+3\sqrt{2}+x\) from adding the radii. So \[\left(10+6\sqrt{2}-x\right)\sqrt{2}=5+3\sqrt{2}+x\] \[10\sqrt{2}+12-x\sqrt{2}=5+3\sqrt{2}+x\] \[7\sqrt{2}+7=x+x\sqrt{2}\] \[7\left(1+\sqrt{2}\right)=x\left(1+\sqrt{2}\right)\] \[x=7\]
The total area of the square is \(4^2=16\). The quarter blue square is a \(1/2\) scaling so its area is \(16/4=4\). The blue triangle is formed by cutting the big square along the diagonal, and then cutting one of those triangle in half, so its area is also \(16/4=4\). The blue area is \(4+4=8\).
Find the net percentage loss of an investment that increases by 30% an then decreases by 30%.
This would be found from looking at the multiplications \[(1+0.3)(1-0.3)=1-0.3^2=1-0.09\] So we can see a 9% loss.
Notice how we have 2 right angles formed by triangle altitudes which share a hypotenuse. Right angles can subtend a circle diameter so if we make that hypotenuse the diameter, we can draw a circle. Also draw radii to the right angles.
Let \(AB\) be the line connecting the 2 right angles which is bisected. If we draw lines from \(A\) and \(B\) to another point on the line, both of the new segments must have the same length. So we can do this with circle radii. This shows us that \(x\) is another radius of the circle so \(x=10\).
First square both sides. \[(400-x^2)+(225-x^2)+2\sqrt{(400-x^2)(225-x^2)}=25^2\] Then since \(400+225=25^2\) we can rearrange and simplify to \[x^2=\sqrt{(400-x^2)(225-x^2)}\] Square both sides again and simplify \[x^4=(400-x^2)(225-x^2)=20^2\cdot15^2-25^2\cdot x^2+x^4\] \[5^4x^2=(5\cdot4)^2(5\cdot3)^2\Rightarrow x^2=4^2\cdot3^2\] We are told \(x>0\) so we take the positive solution \(x=12\).
If we notice a few things about the factors of the numbers given, \(27^2=729\), \(26\cdot27=702\), \(26^2=676\). So we can write this with the smaller numbers. \[27^{2x}-26^x\cdot27^x=26^{2x}\] Now divide by \(26^x\cdot27^x\) which gives us a way to turn the equation into 1 variable we can more easily solve for. \[{27^x\over26^x}-1={26^x\over27^x}\] Let \(y=(27/26)^x\). Then the equation becomes \[y-1={1\over y}\Rightarrow y^2-y-1=0\Rightarrow y={1\pm\sqrt{5}\over2}\] Clearly \(y>0\) so we take the positive root. Then we can find with some calculator help \[x={\ln(y)\over\ln(27/26)} ={\ln\left({1+\sqrt{5}\over2}\right)\over\ln(27)-\ln(26)}\approx12.7506\] Rounding this up gives us a solution of \(13\).
What is the most likely result for the total of the roll of four standard dice?
For one die, we have a uniform distribution across 1-6. The distribution remains symmetric and unimodal as we add more dice. So the most likely is the average, \(3.5\times4=14\). If we had an odd number of dice, the 2 closest to the middle would be equally likely. By the central limit theorem, the distribution looks normal as the number of dice approaches infinity.
Another way we can see this observation more rigorously is by counting outcomes with a generating function. For one die, the coefficients of this polynomial represent the number of outcomes where the exponents are the dice sums. \[p(x)=x^1+x^2+x^3+x^4+x^5+x^6=x\left(1+x+x^2+x^3+x^4+x^5\right)\] For four dice, we would have a generating function \[(p(x))^4=x^4(1+x+x^2+x^3+x^4+x^5)^4\] By a similar symmetry argument, we can see that the largest coefficient must come from the middle term of \((1+x+x^2+x^3+x^4+x^5)^4\), the \(x^{10}\) term. This forms the \(x^{14}\) coefficient so the most likely outcome is a sum of \(14\). We can prove this symmetry argument with induction by assuming unimodality of the distribution.
Form a coordinate system for the centroid calculation. Let \(C=(0,0)\). Then \(B\) is \(6\) units to the right so \(B=(6,0)\). We can see \(D\) is directly above \(C\) but we don't know by exactly how far so let \(D=(0,a)\). Finally, \(A\) is to the up and left from \(C\), so let \(A=(b,c)\). Now we will try to solve for these missing coordinates. From the centroid equation, we have \[D=(0,a)=\left({b+6+0\over3},{c+0+0\over3}\right)\] This allows us to find \(b=-6\) and \(c=3a\). Now focus on the square sharing an edge with the \(6\) square. Its side length is \(6+a\). We can also see that the square with \(AC\) has side length \(\sqrt{b^2+c^2}\). By applying the pythagorean theorem to the right triangle inside our \(6+a\) square, we have \[6^2+(6+a)^2=b^2+c^2=6^2+9a^2\] This simplifies to a quadratic equation we can factor and solve. \[36+12a+a^2=9a^2\Rightarrow2a^2-3a-9=0\Rightarrow(2a+3)(a-3)=0\] So \(a=-3/2\) or \(a=3\), but clearly we should take the positive solution \(a=3\) which also means \(c=9\). This tells us that the square next to the \(6\) square has side length \(9\) and the \(AC\) square has side length \[\sqrt{b^2+c^2}=\sqrt{(-6)^2+9^2}=12\] Next we use our coordinate system to find the length of \(AB\) which is the same as \(x\). \[x=\sqrt{(6-b)^2+(c-0)^2}=\sqrt{12^2+9^2}=15\]
Find the smallest whole number that has exactly five divisors.
If \(n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}\) is a prime factorization, then the number of divisors is \((a_1+1)(a_2+1)\ldots(a_k+1)\). So the only way we can factor \(5\) is just itself since it is prime. This means we must have a \(a_1+1=5\), so a 4th power in the prime factorization. We pick the smallest prime so we get the smallest possible \(n=2^4=16\).
Find the discriminant of \(x^2-3x-2\).
This is that quantity that tells us about the roots of a quadratic equation. Here, we have \(a=1,b=-3,c=-2\) so the discriminant is \[D=b^2-4ac=(-3)^2-4(1)(-2)=9+8=17\]
Find the maximum \(x\) such that \(20!\) is divisible by \(2^x\).
We have to count factors of \(2\) in \(20!\). We can do this by summing \(\lfloor20/2^k\rfloor\) for all positive integers \(k\). For \(k=1\), we count even numbers. Then multiples of \(4\) add another \(2\) to the factorization of \(20!\), so \(k=2\) counts those. \(k=3\) counts for multiples of \(8\), and so on. We stop once \(k\) is large enough to make that expression zero. \[\left\lfloor{20\over2}\right\rfloor+\left\lfloor{20\over4}\right\rfloor +\left\lfloor{20\over8}\right\rfloor+\left\lfloor{20\over16}\right\rfloor +\left\lfloor{20\over32}\right\rfloor=10+5+2+1+0=18\]
Notice how \(2t+1\) takes values \(-37,-35,\ldots,35,37\), so we have a symmetry. We can write the summation by pairing \(t\) and \(-t\) terms. \[\sum_{t=0}^{18}\left({1\over1+37^{2t+1}}+{1\over1+37^{-(2t+1)}}\right)\] Now multiply the 2nd fraction by \(37^{2t+1}/37^{2t+1}\) and simplify \[=\sum_{t=0}^{18}\left({1\over1+37^{2t+1}}+{37^{2t+1}\over37^{2t+1}+1}\right) =\sum_{t=0}^{18}{1+37^{2t+1}\over1+37^{2t+1}}=\sum_{t=0}^{18}1=19\]
We can find \(x\) with the pythagorean theorem. Let \(r=5\sqrt{6}-5\sqrt{2}\). One side of the rectangle is \(r\). The other can be found if we draw the following green right triangle.
We can see it has a hypotenuse of \(2r\) and leg of \(r\) so the remaining leg is \(r\sqrt{3}\) which makes the long side of the rectangle \(2r+r\sqrt{3}\). First compute \(r^2\) \[r^2=\left(5\sqrt{6}-5\sqrt{2}\right)^2 =25\cdot6+25\cdot2-2\cdot5\cdot5\sqrt{6}\sqrt{2} =200-100\sqrt{3}=100\left(2-\sqrt{3}\right)\] The square of the other side is \[\left(2r+r\sqrt{3}\right)^2=r^2\left(2+\sqrt{3}\right)^2 =100\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right) =100\left(2+\sqrt{3}\right)\] So now we can find \[x^2=r^2+\left(r\left(2+\sqrt{3}\right)\right)^2 =100\left(2-\sqrt{3}\right)+100\left(2+\sqrt{3}\right)=400\] And finally \(x=20\).
Find the number of eight digit binary numbers where no two consecutive digits are \(1\).
Let \(A_n\) be the number of \(n\) digit binary strings with no consecutive ones. We clearly have \(A_1=2\) and \(A_2=3\). For further terms \(A_n\), consider the cases of the first digit. If it is \(0\), then the remaining \(n-1\) digits need to have no consecutive ones so \(A_{n-1}\). If it is \(1\), then the next digit must be \(0\), and the remaining \(n-2\) digits must have no consecutive ones, so \(A_{n-2}\). This gives the recurrence \(A_n=A_{n-1}+A_{n-2}\) which is the fibonacci sequence. But there is a little offset so \(A_n=F_{n+2}\).
We have counted binary strings, but we were asked to count 8 digit binary numbers which must have a leading digit of \(1\). This means the second digit must be \(0\) and the remaining \(6\) must have no consecutive ones. So we actually look for term \(A_6=F_8=21\) which is the solution.
Let all modular congruences be modulo \(10\). Work from right to left. First, \(4T\equiv6\) implies \(T=4,9\). We will continue further for each of these.
If \(T=4\), then we carry \(1\) so \(3A+1\equiv0\) implies \(A=3\). This carries \(1\), so \(2E+1\equiv9\), which gives us \(E=4,9\). With \(E=4\), we carry \(0\) and \(N=7\). With \(E=9\), we carry \(1\) and \(N=6\).
If \(T=9\), then we carry \(3\) so \(3A+3\equiv0\) implies \(A=9\). This carries \(3\), so \(2E+3=9\), which gives us \(E=3,8\). With \(E=3\), we carry \(0\) so \(N=7\). With \(E=8\) we carry \(1\) so \(N=6\).
Here is the table showing all 4 possibilities we found. \[\begin{array}{c|c|c|c}N&E&A&T\\\hline 7&4&3&4\\6&9&3&4\\7&3&9&9\\6&8&9&9\\\end{array}\] However, typically a crypt arithmetic puzzle requires each letter to be a distinct digit. The only possibility we found is \(NEAT=6934\). We can verify \(6934+934+34+4=7906\). Finally \[x=N+E+A+T=6+9+3+4=22\]
Find the smallest number of people with greater than 50% probability that two will share a birthday.
We would look for the smallest \(n\) such that the following is below 50%, which is a way of calculating the probability that no birthdays are shared. \[\left({365-0\over365}\right)\left({365-1\over365}\right) \ldots\left({365-n+1\over365}\right)={365!\over(365-n)!\cdot365^n}\] We can find \(n=23\) is the solution, which is a well known result of the birthday problem. For \(n=23\), the probability of a shared birthday is about \(50.73\%\).
One approximation finding \(n\) from a probability \(p\) and number of days \(d\) is \[n(p,d)\approx\sqrt{2d\ln\left({1\over1-p}\right)}\] If we use \(p=0.5\) and \(d=365\), we find \(n(0.5,365)\approx22.494\) which we can round up. It is still an approximation so other methods are necessary to be fully certain.
The number of teaspoons in half a cup.
A teaspoon is \(5mL\). Half a cup is \(4\) ounces so \(4\times30mL=120mL\). This means we need \(24\) teasoons for half a cup.
Find the volume of the region bounded by \[\begin{align}&0\leq x\leq5\\&0\leq y\leq5\\&0\leq z\leq6\\ &6x+6y+5z\leq30\end{align}\]
Consider when all variables are nonnegative. The largest that \(x\) or \(y\) can be is \(5\) within the constraint of \(6x+6y+5z\leq30\). For \(z\), the largest is \(6\). This means the full triangular pyramid of region in the first octant is not trimmed at all by the rectangular bounds. The limit of \(z=6\) gives a height of \(6\) for the pyramid. To find the area of the base, let \(z=0\) and in the \(xy\) plane, \(6x+6y\leq30\Rightarrow x+y\leq5\). This describes a triangle with vertices \((0,0),(5,0),(0,5)\) which has area \({1\over2}(5)(5)={25\over2}\). So now that we have the base and height, the volume of this triangular pyramid is \[{1\over3}\left({25\over2}\right)(6)=25\] There is also a formula for the volume of a right angle tetrahedron we could have used which is \({1\over6}abc\) where \(a,b,c\) are the lengthts of the edges next to the right angle corner.
The side length of the square is \[13\left(\sqrt{5+4\sqrt{2}}-1\right)\]
Let \(a\) be the side length of the square. Let \(h\) be the other leg (height) of the right triangle outside the square. By similar triangles and the pythagorean theorem \[{a\over a+x}={h\over x},\quad x^2+h^2=a^2\] First multiply both sides by \(x\) to isolate \(h\), then square both sides so we can use the pythagorean theorem result. \[{ax\over a+x}=h\Rightarrow\left({ax\over a+x}\right)^2=h^2=a^2-x^2\] Now divide both sides by \(x^2\) \[\left({a\over a+x}\right)^2={a^2\over x^2}-1\] Next, divide the fraction numerator and denominator by \(x\) and substitute \(b=a/x\) \[\left({b\over b+1}\right)^2=b^2-1\] Multiply both sides by \((b+1)^2\) and expand to reveal a quartic polynomial \[b^4+2b^3-b^2-2b-1=0\] This one is challenging to solve, but there is a clever way to reduce it to a quadratic polynomial. Notice that the first 2 terms come from \((b^2+b)^2\). So we can use that \[(b^2+b)^2-b^2-b^2-2b-1=0\] Then the middle terms \(-2b^2-2b\) can use the same substitution. \[(b^2+b)^2-2(b^2+b)-1=0\] Now let \(c=b^2+b\). So this polynomial is \(c^2-2c-1=0\). Its solutions are \(c=1\pm\sqrt{2}\). But since \(b=a/x\) is a ratio of positive quantities, \(b>0\) so \(c>0\) as well, which means we should take the positive root. Now substitute \(b\) again, so we have this quadratic equation to solve and once again take the positive root since \(b>0\). \[b^2+b=1+\sqrt{2} \Rightarrow b={-1\pm\sqrt{1+4\left(1+\sqrt{2}\right)}\over2} ={\sqrt{5+4\sqrt{2}}-1\over2}\] Notice how \(b=a/26\). We have somehow found the nested roots given to us by going through 2 substitutions of transforming a problem into solving a quadratic equation. So going back to our earlier substitution, \(x=a/b=a/(a/26)=26\).
Let \(a=\sqrt[3]{36+x}\) and \(b=\sqrt[3]{36-x}\). Then \(a-b=2\). One way could be to cube both sides. \[(a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3+3ab(b-a)=8\] We would get \(b-a=-2\) and the difference of cubes simplifies to \(2x\). The problem is that we still have a cube root from \(ab\). We could cube things once again, but then we end up with really big numbers, although it does turn out to be a cubic polynomial with \(x=28\) as a solution.
To keep things more manageable, rearrange to \(a=b+2\) and then cube both sides. \[a^3=b^3+6b^2+12b+8\] We have a nice value for \(a^3+b^3=72\) so add \(b^3\) to both sides to use it. \[a^3+b^3=72=2b^3+6b^2+12b+8\Rightarrow b^3+3b^2+6b-32=0\] This cubic polynomial can be factored with the rational root theorem and it turns out to have one real solution \[(b-2)(b^2+5b+16)=0\] So \(b=2\). Then \(b^3=36-x=8\) and \(x=28\). We can verify that this solves the equation we are given.
Start with the first equation and expand. \[a^2+b+2a\sqrt{b}=11+6\sqrt{2}\] We have \(a^2+b=11\) and \(2a\sqrt{b}=6\sqrt{2}\). Square the 2nd equation so we find \(4a^2b=72\Rightarrow a^2b=18\). Multiply each side of the first equation by \(b\) so we can substitute it. \[a^2b+b^2=11b\Rightarrow18+b^2=11b\Rightarrow b^2-11b-18=0 \Rightarrow(b-2)(b-9)=0\] So \(b=9\) forces \(a=\pm\sqrt{2}\not\in\mathbb{Z}\) which means we must have \(b=2\). Then from \(2a\sqrt{b}=6\sqrt{2}\) we find \(a=3\). Next, expand the cube \[\left(3+\sqrt{2}\right)^3=27+27\sqrt{2}+18+2\sqrt{2}=45+29\sqrt{2}\] So we find \(x=29\) and \(y=45\).
Find half the perimeter of a rectangle where three of the side lengths add to \(56\) and three different side lengths add to \(37\).
Let \(a,b\) be the side lengths with \(a\geq b\). We add 3 sides for each given number which means one is excluded. We must exclude a longer side for \(56\) and a shorter side for \(37\). So in equations \[2a+b=56,\quad a+2b=37\] Adding these equations together, \[3a+3b=56+37=91=3\cdot31\] Half of the perimeter is just \(a+b\) so divide by \(3\) and we find that \(31\) is the solution.