| August | ||||||
|---|---|---|---|---|---|---|
| Mo | Tu | We | Th | Fr | Sa | Su |
| 01 | 02 | 03 | ||||
| 04 | 05 | 06 | 07 | 08 | 09 | 10 |
| 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| 18 | 19 | 20 | 21 | 22 | 23 | 24 |
| 25 | 26 | 27 | 28 | 29 | 30 | 31 |
We could recognize that both of these are Mersenne primes and from there, they must be coprime so the solution is \(1\). Alternatively, we can approach the problem with the following theorem for integers \(a>1\) and \(m,n>0\): \[\gcd(a^m-1,a^n-1)=a^{\gcd(m,n)}-1\] Assume than \(m\geq n\). Then subtract a multiple of \(a^n-1\) from \(a^m-1\). \[(a^m-1)-a^{m-n}(a^n-1)=a^m-1-a^m+a^{m-n}=a^{m-n}-1\] So now we have \[\gcd(a^m-1,a^n-1)=\gcd(a^n-1,a^{m-n}-1)\] Notice that the exponents behave like a \(\gcd\) algorithm: \(\gcd(m,n)=\gcd(n,m-n)\). This can be repeated until we end up with \(a^0-1=0\) and find that the other exponent is \(\gcd(m,n)\). Therefore, \[\gcd(a^m-1,a^n-1)=a^{\gcd(m,n)}-1\] For this problem specifically, both \(107\) and \(127\) are prime so \(\gcd(107,127)=1\).
Another problem with roots and conjugates. Let \(x\) be equal to this expression and then square both sides: \[x^2=\left(6+\sqrt{20}\right)+\left(6-\sqrt{20}\right) -2\sqrt{\left(6+\sqrt{20}\right)\left(6-\sqrt{20}\right)}\] \[=12-2\sqrt{6^2-20}=12-2\sqrt{16}=4\] So then \(x=\sqrt{16}=2\). Another way is to find things that square to \(6+\sqrt{20}\) and \(6-\sqrt{20}\). Focusing on \(6+\sqrt{20}=6+2\sqrt{5}\), we expect it to look something like \[\left(a+b\sqrt{5}\right)^2=a^2+5b^2+2ab\sqrt{5}\] So we would solve for \(a^2+5b^2=6\) and \(2ab=2\). In general, this can be done similar to other systems with quadratic equations, but we can just spot that \(a=b=1\) works. A similar result can be found for the other root. So \[\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}} =\left(1+\sqrt{5}\right)-\left(\sqrt{5}-1\right)=2\]
We can compute the value \(7!+1=5041\). This is not so easy to factor manually but it is \(71^2\) so its divisors are \(1,71,71^2\), 3 in total.
Find the radius of the circle inscribed inside a right triangle with shorter leg \(10\) and hypotenuse \(26\).
We can draw a few radii of the circle which create some more right triangles sharing a hypotenuse. Focus on the 2 new right triangles on the right. They each share a side and have a leg which is a radii. By equality to the hypotenuse, \[(26-s)^2+r^2=(24-r)^2+r^2\] Now on the left we can use \(r+s=10\) and substitute. \[(26-10+r)^2+r^2=(24-r)^2+r^2 \Rightarrow (16+r)^2=(24-r)^2\] Taking square roots we have 2 cases. First \[16+r=24-r \Rightarrow 2r=8 \Rightarrow r=4\] This is the solution. We should also check the other case, which is invalid. \[16+r=-(24-r)=r-24 \Rightarrow 16=-24\] Another way that is actually easier is to recognize that we can construct the hypotenuse from \(s=10-r\) and similarly a \(24-r\) segment. So \[26=(10-r)+(24-r)=34-2r\Rightarrow r=4\]
We should try partial fractions. This denominator looks complicated but it can be factored by grouping. \[4n^4+8n^3-n^2-2n=4n^3(n+2)-n(n+2)=(4n^3-n)(n+2)\] \[=n(n+2)(4n^2-1)=n(n+2)(2n-1)(2n+1)\] Now we evaluate a partial fraction decomposition. \[{20n^2+8n-4\over n(n+2)(2n-1)(2n+1)} ={a\over n}+{b\over n+2}+{c\over2n-1}+{d\over2n+1}\] Each factor is linear and has multiplicity 1 so we can use the cover up method for all of them. It works by multiplying both sides by one of the linear factors and then evaluating at its root, which isolates one of the constants. \[n:{20n^2+8n-4\over(n+2)(2n-1)(2n+1)}\Big|_{n=0}={-4\over2(-1)(1)}=2\] \[n+2:{20n^2+8n-4\over n(2n-1)(2n+1)}\Big|_{n=-2}={80-16-4\over(-2)(-5)(-3)} ={60\over-30}=-2\] \[2n-1:{20n^2+8n-4\over n(n+2)(2n+1)}\Big|_{n=1/2}={5+4-4\over(1/2)(5/2)(2)} ={5(2)\over5}=2\] \[2n+1:{20n^2+8n-4\over n(n+2)(2n-1)}\Big|_{n=-1/2}={5-4-4\over(-1/2)(3/2)(-2)} ={-3(-2)\over-3}=-2\] So our summation factors to \[\sum_{n=1}^\infty{20n^2+8n-4\over4n^4+8n^3-n^2-2n} =\sum_{n=1}^\infty\left({2\over n}-{2\over n+2} +{2\over2n-1}-{2\over2n+1}\right)\] We get some telescoping summations. Use the first 2 terms, but split them into odd and even terms so the cancellation is clear. \[2\sum_{n=1}^\infty\left({1\over n}-{1\over n+2}\right) =2\sum_{n=1}^\infty\left({1\over2n-1}-{1\over2n+1}\right) +2\sum_{n=1}^\infty\left({1\over2n}-{1\over2n+2}\right)\] We can see only the first term remains in both of these summations so \[2\left({1\over1}\right)+2\left({1\over2}\right)=3\] Then the other 2 terms is exactly the same as one of these 2 summations: \[\sum_{n=1}^\infty\left({2\over2n-1}-{2\over2n+1}\right)=2\] Adding all results, the solution is \(5\).
For the sequence \(a_n\) with \(a_1=2,a_n=2a_{n-1}+1\), \(a_x\) is the first composite number of the sequence.
We can evaluate terms until one is composite: \[\begin{align} &a_1=2\\ &a_2=2a_1+1=2(2)+1=5\\ &a_3=2a_2+1=2(5)+1=11\\ &a_4=2a_3+1=2(11)+1=23\\ &a_5=2a_4+1=2(23)+1=47\\ &a_6=2a_5+1=2(47)+1=95=5\cdot19\\ \end{align}\] We have found the first composite term so \(x=6\).
Find the length of the long diagonal of a box with side lengths 2, 3, 6.
This is a euclidean distance. \(\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7\).
A set of test scores are normally distributed with mean 100 and standard deviation 15. What percentile is a score of 79?
We can find that the z score of 79 is \({79-100\over15}={-21\over15}=-1.4\). In a table, this corresponds to a cumulative value of \(\approx0.0808\) which puts the score in the 8th percentile since it is between \(0.08\) and \(0.09\).
Instead of table lookup, it can be computed with tedious work or on a calculator with an expression such as \[\Phi(x) ={1\over2}\left[1+\text{erf}\left({x-\mu\over\sigma\sqrt{2}}\right)\right]\] or \[\Phi(z)={1\over2}\left[1+\text{erf}\left({z\over\sqrt{2}}\right)\right]\] This would still be a pain to do on paper but the taylor series is \[\text{erf}(x)={2\over\sqrt{\pi}} \sum_{n=0}^\infty{(-1)^n x^{2n+1}\over n!(2n+1)}\]
Find the count of two-digit primes where the number with digits reversed is still prime.
List the two-digit primes and identify which ones have a reverse pair. \[\boxed{11},\boxed{13},\boxed{17},19,23,29,\boxed{31},\boxed{37},41, 43,47,53,59,61,67,\boxed{71},\boxed{73},\boxed{79},83,89,\boxed{97}\] In total there are 9.
We can first use properties of complementary angles to identify similar triangles.
Both are 3-4-5 right triangles. We can find that the smaller one is \(4.5,6,7.5\) and the larger one with \(6,8,10\). The hypotenuse of the larger one is \(x=10\).
What is the maximum number of regions in the plane four lines can form?
This can be maximized in general by creating lines such that no pair is parallel and there are never more than 2 lines intersecting at a point. The first line divides the plane into 2 regions. Beyond that, when drawing a line, it starts in a region and switches to a new region every time it crosses a line, so it splits a number of regions equal to 1 more than how many lines it crosses. So the 2nd line crosses 1 line, creating 2 more regions. The 3rd line crosses 2 lines and creates 3 more regions. The 4th line crosses 3 regions and creates 4 more regions. In total, there are 11 regions. The general formula for \(n\) lines is \[1+{n(n+1)\over2}\]
Find the smallest whole number \(x\) such that \(2^7+4+x\) is a perfect square.
This is equal to \(132+x\). It is larger than \(11^2=121\) but smaller than \(12^2=144\). So the solution is \(x=12\) and \(2^7+4+x=12^2\).
Since we only need \(x\) for the calendar problem, we can use Cramer's rule which gives us just 1 value. If the system is \(Ax=b\) then \(x_i={\det(A_i)\over\det(A)}\) where \(A_i\) is \(A\) but with the \(i\)th column replaced by \(b\). \[x={\begin{vmatrix}24&-3&1\\32&-4&6\\22&-5&-1\\\end{vmatrix}\over \begin{vmatrix}2&-3&1\\1&-4&6\\3&-5&-1\\\end{vmatrix}} ={96-396-160+88+720-96\over8-54-5+12+60-3}={252\over18}=14\] If we also wanted the other variables, we could use other methods and find \[y=3,\quad z=5\]
Solve this by considering 2 cases for the sign of what is inside absolute value.
For positive, \(|4-x|+11=22\Rightarrow|4-x|=11\). This has solutions \(x=15,-7\) so \(x=15\) is the only one relevant here.
For negative, \(-|4-x|-11=22\Rightarrow|4-x|=-33\). This is impossible so there are no more solutions. We can actually see that \(|4-x|+11\geq11\) so this case is not even necessary.
Let \(y=\log_{16}(x)\). Then we can solve a simpler looking equation \[\sqrt{y+3}+\sqrt{4y+12}=6=\sqrt{y+3}+2\sqrt{y+3}=3\sqrt{y+3}\] \[\Rightarrow\sqrt{y+3}=2\Rightarrow y+3=4\Rightarrow y=1\] Then \(x=16^y=16\).
Multiply each side by 10 to make the numbers look nicer, then simplify. \[4(x+3)=10-5(3-x)\Rightarrow4x+12=10-15+5x\Rightarrow x=17\]
Find the sum of the digits of \(1000002^3\).
\[(10^6+2)^3=10^{18}+3\cdot2\cdot10^{12}+3\cdot2^2\cdot10^6+2^3\] The numbers are small and put in parts of a large number with zeroes in between. The numbers we get are \(1,6,12,8\) so the digit sum is \(1+6+1+2+8=18\).
Ten numbers average to 10. One of those numbers is \(x\), and when it is removed, the average of the remaining nine numbers is 9.
Let \(y\) be the sum of the other 9 numbers. Then \[x+y=10\times10=100\] But the average of \(y\) is 9, so \[y=9\times9=81\] Then \(x=100-y=100-81=19\).
How many four digit numbers \(abcd\) satisfy:
One detail missing is that no other digit can be equal to \(d\), it must be the unique smallest otherwise there are 29 solutions. We only have \(b\) in the first constraint, and \(a=b\) means it does not add any additional solutions so we can ignore \(b\).
Since \(d\) is smallest, \(a>d\). Since \(c=(a+d)/2\), \(a+d\) needs to be even. We can also verify \(2c=a+d>d+d=2d\) so \(c>d\) as required. Now we make a table with values \(a\) with allowed values \(d\) such that \(a+d\) is even.
| \(a\) | \(d\) values |
|---|---|
| 0 | (none) |
| 1 | (none) |
| 2 | 0 |
| 3 | 1 |
| 4 | 0,2 |
| 5 | 1,3 |
| 6 | 0,2,4 |
| 7 | 1,3,5 |
| 8 | 0,2,4,6 |
| 9 | 1,3,5,7 |
In total, there are 20.
Each path must go up 2 times and right 5 times. So out of the 7 steps, we can pick any 2 to go up, and the other 5 go right. \[{7\choose2}={7\cdot6\over2}=21\]
Find the positive root of \(x^2-2x-440\).
By looking at the factorization \(440=2\cdot2\cdot2\cdot5\cdot11\) we can search and find that \(20,22\) differ by 2. So we can factor it \[x^2-2x-440=(x+20)(x-22)=0\] The roots are \(x=-20,22\) so the positive one is \(x=22\).
Separate the square root and square both sides. \[\sqrt{6x-134}=25-x\Rightarrow6x-134=625-50x+x^2\] \[\Rightarrow x^2-56x+759=0=(x-23)(x-33)\] From here, we can see that \(x=23\) is the solution and that \(x=33\) is extraneous because \(33\) plus a positive would be greater than \(25\).
First we can simplify the function \(f\) and get a nicer expression for it. \[f(3y+5)=3y(3y+10)=(3y+5-5)(3y+5+5)\] Now let \(z=3y+5\) so \(f(z)=(z-5)(z+5)=z^2-25\). Then \(f(7)=7^2-25=49-25=24\).
To determine the constants \(f(1)\) and \(f(0)\), we pick a value for \(t\) to substitute which will give us a linear system: \[t=0:\quad f(1)=2(0)f(1)+f(0)+2=f(0)+2\] \[t=-1/3:\quad f(0)={-2\over3}f(1)+f(0)+2\] Substitute the first into the second \[f(0)={-2\over3}(f(0)+2)+f(0)+2\Rightarrow{2\over3}f(0)={-4\over3}+2 \Rightarrow f(0)=1\] Then \(f(1)=f(0)+2=1+2=3\). So the function is \(f(3t+1)=6t+3=2(3t+1)+1\) which means we can also write it as \(f(u)=2u+1\). Then \(f(12)=25\).
If we factor a bit, we find \(\sqrt{88556}=26\sqrt{131}\). Then we would like to find if the quantities inside square roots are perfect squares of something. \[\left(a+b\sqrt{131}\right)^2=a^2+131b^2+2ab\sqrt{131}\] We need \(a^2+131b^2=300\) in both cases, so \(a=\pm13,b=\pm1\) works just by recognizing that \(300=169+131\) gives us a perfect square. Then we need \(ab=13\). For the first root, \(a=13,b=1\) (note that it is important we pick the positive quantity that works). So we find \[\left(13+\sqrt{131}\right)^2=300+26\sqrt{131}\] For the other root, \(a=13,b=-1\) works and leaves us a positive quantity. \[\left(13-\sqrt{131}\right)^2=300-26\sqrt{131}\] So now we just add the roots \[\left(13+\sqrt{131}\right)+\left(13-\sqrt{131}\right)=26\]
A 54 liter solution that is 99% water is dehydrated to an \(x\) liter solution that is 98% water.
We can find that the amount of other content is \(54\times0.01=0.54\). Now if it becomes 2% of the volume, then \(0.02\times x=0.54\) (2% of the new volume is equal to the same non water content). The solution is \(x=27\).
Another way to look at this is the 1% other content increases to 2%. This means we have half as much total volume so the same amount of content takes up twice as much. \(x=54/2=27\).
We have 2 squares of area 16, but they overlap and share 1/4 of the area, so avoid double counting. \[16+16-4=28\]
First compute logs of the part with \(x\). \[\ln(x)=r\ln(2)=s\ln(3)=t\ln(5)\] Then we can use these to substitute \(r,s,t\) for something with \(x\). \[1={\ln(2)\over\ln(x)}+{\ln(3)\over\ln(x)}+{\ln(5)\over\ln(x)}\] Multiply by \(\ln(x)\). \[\ln(x)=\ln(2)+\ln(3)+\ln(5)=\ln(30)\] Then the solution is \(x=30\).