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Minimal number of sequential evaluations of \(p(x)\in\mathbb{N}[x]\) needed to determine the coefficients of \(p(x)\).
This is worded a bit confusingly. It is asking how many evaluations are needed to determine coefficients of a polynomial with nonnegative integer coefficients. Suppose \(p(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n\) where each \(a_i\in\mathbb{N} =\{0,1,2,\ldots\}\). First evaluate \(p(1)=\sum a_i\) and let this value be \(q\). Next evaluate \(p(q)=a_0+a_1q+a_2q^2+\ldots+a_nq^n\). If only one of the \(a_i>0\), then we would find \(p(x)=a_nq^n\) where \(a_n=q\) and \(n=\log_q(p(q))-1\). If multiple \(a_i>0\), then we will have \(q>a_i\) for each \(a_i\), so \(p(q)\) is represented such that each \(a_i\) are the digits of the base \(q\) representation, which allows us to determine them.
If \(\log_ab+\log_bc+\log_ca=0\), find \((\log_ab)^3+(\log_bc)^3+(\log_ca)^3\).
To make this simpler, let \(x=\log_ab,y=\log_bc,z=\log_ca\). Then we have \(x+y+z=0\) and need to find \(x^3+y^3+z^3\). Now to get the cubes we need, cube each side of \(x+y=-z\), since this avoids cubing a side with 3 terms. \[x^3+y^3+3x^2y+3xy^2=-z^3 \Rightarrow x^3+y^3+z^3=-3xy(x+y) =-3xy(-z)=3xyz\] Now we just use the change base formula: \[3xyz=3{\ln(b)\over\ln(a)}{\ln(c)\over\ln(b)}{\ln(a)\over\ln(c)} =3{\ln(a)\ln(b)\ln(c)\over\ln(a)\ln(b)\ln(c)}=3\]
For the Fibonacci sequence \(F_i\), \[\begin{align}&a=F_0+F_1+\ldots+F_{59}\\&b=F_2+F_4+\ldots+F_{60}\\ &c=F_2+F_5+\ldots+F_{59}\\&x={a+b+c\over c}\end{align}\]
We will make use of the identity \(F_n=F_{n-1}+F_{n-2}\) (for \(n\geq2\)). This allows us to show that \(b=(F_0+F_1)+(F_2+F_3)+\ldots+(F_{58}+F_{59})=a\). Doing the same again, \(c=(F_0+F_1)+(F_3+F_4)+\ldots+(F_{57}+F_{58})\). This can be added to the original \(c\) to get a sum of all the terms \(F_0\) to \(F_{59}\) so \(c+c=2c=a=b\). Then \[x={a+b+c\over c}={2c+2c+c\over c}={5c\over c}=5\]
This is another good problem for square each side and see what happens. \[\begin{align}&(9+6\sqrt{2})+(9-6\sqrt{2})+2\sqrt{(9+6\sqrt{2})(9-6\sqrt{2})} =4x\\&18+2\sqrt{81-72}=4x \Rightarrow 24=4x \Rightarrow x=6\\\end{align}\]
Start by evaluating \(x^3\). \[x^3=y^3+y^{-3}+3(y+y^{-1})=322+3x \Rightarrow x^3-3x-322=0\] This gives us a polynomial to factor. By using the factorization \(322=2\cdot7\cdot23\), we can find the rational root \(x=7\). Then dividing that out results in the polynomial \(x^2+7x+46\) which has imaginary roots. Therefore, the solution is \(x=7\).
We can start with the smaller numbers to make this easier. \[\text{gcf}(504,120)=\text{gcf}(4\cdot120+24,120)=\text{gcf}(24,120) =\text{gcf}(24,5\cdot24)=\text{gcf}(24)\] Next compute \[\text{gcf}(7000,24)=\text{gcf}(291\cdot24+16,24)=\text{gcf}(16,24) =\text{gcf}(16,16+8)=\text{gcf}(16,8)=\text{gcf}(2\cdot8,8)=8\]
Multiply each side by 10 and then combine similar terms. \[4x+15=5(x+1)\Rightarrow4x+15=5x+5\Rightarrow15-5=5x-4x\Rightarrow x=10\]
The sum of the roots of \(x^{45}+(0.5-x)^{45}\).
There is a Vieta formula that can be used to solve this problem. Suppose \(p(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0\) is a polynomial. Then if we know its roots \(z_1,z_2,\ldots,z_n\), \[p(z)=a_n(z-z_1)(z-z_2)\ldots(z-z_n)\] Then expand it partially. \[p(z)=a_n\left(z^n-(z_1+z_2+\ldots+z_n)z^{n-1} +(\text{other complex terms not needed})+(-1)^nz_1z_2\ldots z_n\right)\] Because the other complex terms not needed are lower powers of \(z\), we can compare the coefficients of \(z^{n-1}\) in both of these to find \[a_{n-1}=-a_n(z_1+z_2+\ldots+z_n)\] This gives us the general formula \(-{a_{n-1}\over a_n}\). Now we need to find the 2 relevant coefficients of the given polynomial. \[x^{45}+(0.5-x)^{45}=x^{45}-x^{45}+{45\choose1}0.5\cdot x^{44} -{45\choose2}0.5^2\cdot x^{43}+\ldots\] The \(x^{45}\) terms go away so this is a degree 44 polynomial. The coefficients we need tell us the sum of roots is \[-{-{45\choose2}0.5^2\over{45\choose1}0.5}={{45\cdot44\over2}0.5\over45} ={44\cdot0.5\over2}=11\]
Find the difference between the roots of \(x^2+6x-27\).
In general, with roots \(x={-b\pm\sqrt{b^2-4ac}\over2a}\), we can find that the difference is \({1\over a}\sqrt{b^2-4ac}\). For this problem, with \(a=1,b=6,c=-27\) we get \({1\over1}\sqrt{6^2-4\cdot1(-27)}=\sqrt{36+108} =\sqrt{144}=12\). Alternatively, this one is easy to factor to \((x+9)(x-3)\) so we can see the roots are \(x=-9,x=3\) and \(|-9-3|=12\).
\[\begin{pmatrix}1&1&0&38\\0&1&1&81\\1&0&1&69\\\end{pmatrix}\sim \begin{pmatrix}1&0&0&13\\0&1&0&25\\0&0&1&56\\\end{pmatrix}\] Therefore by linear algebra, \(x=13,y=25,z=56\).
How many positive two-digit numbers are divisible by each of their digits?
The digits cannot be 0, so there are \(9\cdot9=81\) possible numbers. The easiest to start with are those with same digit: 11, 22, 33, 44, 55, 66, 77, 88, 99. These are 11 times their digit. Next, for the tens, we try and find that 12 and 15 also work. For the twenties, we start at 22, going up in multiples of 2, and only find 24. For thirties, we find 36. For forties, we find 48. Past this, there are no others besides the same digit number. In total, we found 14 such two digit numbers.
Find the number of cards that make either a flush or straight with \({\color{red}2\diamondsuit},{\color{red}3\diamondsuit}, {\color{red}4\diamondsuit},\text{ and }{\color{red}5\diamondsuit}\).
Any other \(\color{red}\diamondsuit\) makes it a flush, so that is 9 possible cards. To make a straight using the other 3 suits, we select either an ace or 6 so there are 6 other possible cards. \(9+6=15\).
\(x=\text{the natural number less than 50 that has exactly 5 factors}\)
Consider a factorization \(x=p_1^{a_1}p_2^{a_2}\ldots\). We need \((a_1+1)(a_2+1)\ldots=5\) for 5 factors. The only way to form a product equal to 5 is with 1s and 5s, so the number we are looking for is a prime 4th power. The only such number below 50 is \(2^4=16\) and its factors are 1, 2, 4, 8, 16.
The sum of the first four primes.
\(2+3+5+7=17\)
\[\sqrt[3]{8}\cdot27^{2\over3}=2\cdot3^{3\cdot{2\over3}}=2\cdot3^2=18\]
Find the larger solution to \(7x^2+10x+8\equiv0\mod25\).
Let \(f(x)=7x^2+10x+8\). Since \(25=5^2\), we start mod \(5\) and lift solutions. Reducing this quadratic polynomial modulo 5, we have \(2x^2+3\equiv0\mod5\). The solutions are \(x\equiv1,4\). If we start with \(x\equiv1\), then lift it modulo 25, we get \[1-f(1)f'(1)^{-1}\equiv1-(7+10+8)f'(1)^{-1}\equiv1-25f'(1)^{-1}\equiv1\] For \(x\equiv4\), lifting modulo 25 gets us \[4-f(4)f'(4)^{-1}\equiv4-160\cdot66^{-1}\equiv4-10\cdot16^{-1} \equiv4-10\cdot11\equiv-106\equiv19\] where we have found that \(16^{-1}\equiv11\) modulo 25 so the solution is 19.
On the right side, \((3y+20)+(2y+10)=5y+30=180\Rightarrow y=30\). Then on the left side, \(x=180-90-(2y+10)=90-70=20\).
We can evaluate \(0.\overline{45}\) as a geometric series. \[45\sum_{i=1}^{\infty}100^{-i}=0.45\sum_{i=0}^{\infty}100^{-i} ={9\over20}{1\over1-{1\over100}}={9\over20}{100\over99}={5\over11}\] Then the answer is: \[{1\over{1\over2}-{5\over11}}={1\over{11\over22}-{10\over22}} ={1\over{1\over22}}=22\]
Find the sum of all natural numbers, \(n<17\) where \(17\) divided by \(n\) gives a remainder of \(2\).
To have a remainder of 2, we need \(n>2\) and \(17\equiv2(\mod n)\) or \(n\mid(17-2)\Rightarrow n\mid15\). If \(n<17\), then \(n\leq15\) or \(n=16\). Since \(16\nmid15\), we sum divisors of 15 and subtract those which are smaller than 2. Since \(15=3\cdot5\), its divisors sum to \((1+3)(1+5)=24\). 1 divides 15 but 2 does not so we subtract 1 so the answer is \(24-1=23\).
A boat travels downriver at 30mph, then goes back up along the same path at 20mph. What is the boat's average speed?
If the distance is \(d\), then the time taken is \(d/30+d/20=d/12\). The average speed is distance divided by time so \(2d/(d/12)=24\). The \(d\) divides out so the answer does not depend on distance.
Find the final two digits of \(5^{1000}\).
We compute \(5^{1000}\) modulo 100. Clearly \(5^2\equiv25\) mod 100. From this, \(5^3\equiv5\cdot25\equiv125\equiv25\) mod 100. So by induction, when \(n\geq3\) then \(5^n\equiv5\cdot5^{n-1}\equiv5\cdot25\equiv125\equiv25\). Therefore, \(5^{1000}\equiv25\mod100\) so the final 2 digits are 25.
By evaluating a few terms, we can see \(a_1-a_0=1,a_2-a_1=4,a_3-a_2=16,\ldots\) so the clear conjecture is \(a_{n+1}-a_n=4^n\). It is true when \(n=0\), so when \(n>0\), \[a_{n+1}-a_n=5a_n-4a_{n-1}-a_n=4a_n-4a_{n-1}=4(a_n-a_{n-1}) =4\cdot4^{n-1}=4^n\] Therefore, \(\log_2(a_{14}-a_{13})=\log_2(4^{13})=\log_2(2^{26})=26\)
Number of divisors of \(4900\).
This one is particularly easy to factor to \(4900=2^2\cdot5^2\cdot7^2\). The number of divisors comes from choosing a power of each prime factor, so it is \((2+1)(2+1)(2+1)=27\).
All of these shapes are squares. We know that square in the middle has side length 7, so label the next smallest square with side length \(y\) and start finding the side lengths of the other squares.
From here, we can see the bottom edge is \(y+14+y+21=2y+35\). Now look at the other squares toward the top to subtract from this the part that is not included in \(x\). We would subtract \(y+7+7\) and that little bit between thet line \(x\) and the middle square which is \(y-7\). So in total, we subtract \(2y+7\) which gives us an answer of \((2y+35)-(2y+7)=28\).
The sum of two numbers is \(37\) and their product is \(132\). What is their difference?
We can start with the equations \[a+b=37,\quad ab=132\] Clearly both are nonzero so \(b=132/a\) which can be substituted: \(a+132/a=37\). Multiply each side by \(a\) and we have a quadratic, \(a^2-37a+132=0\), which is not too hard to factor to \((a-33)(a-4)=0\). This gives us \(a=33\) corresponding to \(b=4\) and \(a=4\) corresponding to \(b=33\). The difference is \(|4-33|=29\).
What angle do clock hands make at 1PM?
The minute hand is pointing straight up and the hour hand has rotated \(1/12\) of the way around so \(360^\circ/12=30^\circ\).
Find the largest prime factor of \(10!+11!+12!+13!\).
\[\begin{align}&10!+11!+12!+13!=10!(1+11+11\cdot12+11\cdot12\cdot13)\\ &=10!(12+12(12-1)+12(12-1)(12+1))=10!\cdot12(1+12-1+(12-1)(12+1))\\ &=10!\cdot12(12+12^2-1)=10!\cdot12\cdot155=10!\cdot2^2\cdot3\cdot5\cdot31 \end{align}\] From this factorization, clearly \(31\) is the largest prime factor.