Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)

Solution: Please understand the machinery of an escalator before proceeding to read this solution.
Let the number of steps that disappear at the top of the escalator equal $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him \[\frac{150}{3}=50\] seconds to traverse the distance of the escalator moving downwards. Since Bob counts $75$ steps, it takes him \[\frac{75}{1}=75\] seconds to traverse the distance of the escalator moving downwards.
For the sake of this solution, we activate the emergency stop button on the escalator.
Now, the escalator is not moving, or is simply a staircase. Imagine that Al is taking $3$ steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards $x$ steps. This is equivalent to Al taking $3-x$ steps downwards every second. Since we discovered that it takes him $50$ seconds to get from the top to the bottom of the escalator, and we are forcing Al to imitate the movement of the escalator, it also takes him $50$ seconds to move from the top to the bottom of the staircase. Thus, Al takes a total of \[(3-x) \cdot 50=150-50x \qquad (\heartsuit)\] steps.
The explanation for Bob is similar except now we pick him up and place him forward $x$ steps immediately after he takes his usual step per second, and since we discovered he does this for $75$ seconds, it takes him \[(1+x) \cdot 75=75+75x \qquad (\clubsuit)\] steps to get from the bottom to the top.
Note that because the escalator is broken and is now a staircase, Al and Bob must have had to take an equal amount of steps to get from the bottom to the top or from the top to the bottom. (Clearly, there are an equal amount of steps from the bottom to the top, and from the top to the bottom.) Therefore, we may equate $\heartsuit$ and $\clubsuit$ to get \[150-50x=75+75x\] \[x=\frac{3}{5}.\] Therefore, substituting $x$ in the expression we discovered in $\heartsuit,$ Al takes a total of \[150-50x=150-50\left(\frac{3}{5}\right)=\boxed{120}\] steps, and we are done.

Answer:  120
Problem: Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)

Solution: WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\frac{150}{15}\ = 10$ seconds and Bob was on the escalator for $\frac{75}{5}\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r = 75 + 15r$ . Al is getting $10$ seconds of resistance at rate $r$ from the escalator, while Bob is getting $15$ seconds of help at rate $r$ . Solving for $r$ , we have $r = 3$ steps per second. Then, we can plug $r$ into the previous equation or subtract/add it to Al/Bob's rate (respectively) then multiply by their respective time. Either way, we get $\boxed{120}$ and we are done.

Answer:  120
Problem: Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers

Solution: Let us write down one such sum, with $m$ terms and first term $n + 1$
$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$
Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$ .  However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3^{11}} < 3^6$ .  Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$ .  The largest such factor is clearly $2\cdot 3^5 = 486$ ; for this value of $m$ we do indeed have the valid expression $3^{11} = 122 + 123 + \ldots + 607$ , for which $k=\boxed{486}$

Answer:  486
Problem: Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers

Solution: Proceed as in Solution 1 until it is noted that $m$ is a divisor of $2\cdot 3^{11}$ . The divisors of $2\cdot 3^{11}$ are $3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}$ . Note that the factors of $m(2n + m + 1)$ are of opposite parity (if $m$ is odd, then $(2n + m + 1)$ is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. $(2n + m + 1)$ will represent the greater factor while $m$ will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of $m$ . The factor pair which maximizes the lesser factor is $2\cdot 3^{5}$ and $3^{6}$ . It follows that $m$ $2\cdot 3^{5}$ $\boxed{486}$

Answer:  486
Problem: Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$

Solution: In order to keep $m$ as small as possible, we need to make $n$ as small as possible.
$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$ .  Since $r < \frac{1}{1000}$ and $m - n^3 = r(3n^2 + 3nr + r^2)$ is an integer, we must have that $3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000$ .  This means that the smallest possible $n$ should be less than 1000. In particular, $3nr + r^2$ should be less than 1, so $3n^2 > 999$ and $n > \sqrt{333}$ $18^2 = 324 < 333 < 361 = 19^2$ , so we must have $n \geq 19$ .  Since we want to minimize $n$ , we take $n = 19$ .  Then for any positive value of $r$ $3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000$ , so it is possible for $r$ to be less than $\frac{1}{1000}$ .  However, we still have to make sure a sufficiently small $r$ exists.
In light of the equation $m - n^3 = r(3n^2 + 3nr + r^2)$ , we need to choose $m - n^3$ as small as possible to ensure a small enough $r$ .  The smallest possible value for $m - n^3$ is 1, when $m = 19^3 + 1$ .  Then for this value of $m$ $r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}$ , and we're set.  The answer is $\boxed{019}$

Answer:  019
Problem: Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$

Solution: To minimize $m$ , we should minimize $n$ . We have that $(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}$ . For a given value of $n$ , if $(n + \frac{1}{1000})^3 - n^3 > 1$ , there exists an integer between $(n + \frac{1}{1000})^3$ and $n^3$ , and the cube root of this integer would be between $n$ and $n + \frac{1}{1000}$ as desired. We seek the smallest $n$ such that $(n + \frac{1}{1000})^3 - n^3 > 1$
\[(n + \frac{1}{1000})^3 - n^3 > 1\] \[\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1\] \[3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3\]
Trying values of $n$ , we see that the smallest value of $n$ that works is $\boxed{019}$

Answer:  019
Problem: Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$

Solution: Since $r$ is less than $1/1000$ , we have $\sqrt[3]{m} < n + \frac{1}{1000}$ . Notice that since we want $m$ minimized, $n$ should also be minimized. Also, $n^3$ should be as close as possible, but not exceeding $m$ . This means $m$ should be set to $n^3+1$ . Substituting and simplifying, we get \[\sqrt[3]{n^3+1} < n + \frac{1}{1000}\] \[n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}\] The last two terms in the right side can be ignored in the calculation because they are too small. This results in $1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}$ . The minimum positive integer $n$ that satisfies this is $\boxed{019}$ . ~ Hb10

Answer:  019
Problem: A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes".  A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$ , with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass.  The numbers compared at each step are underlined.
Suppose that $n = 40$ , and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order.  Let $p/q$ , in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place.  Find $p + q$

Solution: If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$ , one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position.  Thus, $r_{20}$ must be the largest of the first 20 terms.  In addition, $r_{20}$ must be larger than $r_{21}, r_{22}, \ldots, r_{30}$ but smaller than $r_{31}$ in order that it move right to the 30th position but then not continue moving right to the 31st.
Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?
This is much easier to solve: there are $31!$ ways to order the first thirty-one numbers and $29!$ ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of $\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}$ , so the answer is $\boxed{931}$

Answer:  931
Problem: A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes".  A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$ , with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass.  The numbers compared at each step are underlined.
Suppose that $n = 40$ , and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order.  Let $p/q$ , in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place.  Find $p + q$

Solution: Note that you can solve the restated problem differently:
We see that the numerator is $1$ because there is only $1$ way to make the $31^{\text{st}}$ term the largest and the $20^{\text{th}}$ the second-largest. To calculate the denominator, we note that there are $31\times30$ ways to pick the largest term and the second-largest term. Therefore, our answer is $\boxed{931}$ .
~Yiyj1

Answer:  931
Problem: Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\]

Solution: The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\ &= (x(x-6) + 18)(x(x+6)+18), \end{align*} so the original expression becomes \[\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]},\] which simplifies to \[\frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.\] Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}.$

Answer:  373
Problem: Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\]

Solution: In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$
We factor $N^4+18^2$ by completing the square, then applying the difference of squares: \begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\ &=\left((N-3)^2+9\right)\left((N+3)^2+9\right). \end{align*} The original expression now becomes \[\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.\] ~MRENTHUSIASM

Answer:  373
Problem: Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\]

Solution: We rewrite $N$ to the polar form \[N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , we have \[N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),\] from which
By the Factor Theorem , we get \begin{align*} N^4+18^2&=\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr) \\ &=\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggr] \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

Answer:  373
Problem: Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\]

Solution: We rewrite $N$ to the rectangular form \[N=a+bi\] for some real numbers $a$ and $b.$
Note that $N^2=\pm18i,$ so there are two cases:
By the Factor Theorem , we get \begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

Answer:  373
Problem: Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png

Solution: 1987 AIME-15a.png
Because all the triangles in the figure are similar to triangle $ABC$ , it's a good idea to use area ratios . In the diagram above, $\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.$ Hence, $T_3 = \frac {440}{441}T_1$ and $T_4 = \frac {440}{441}T_2$ . Additionally, the area of triangle $ABC$ is equal to both $T_1 + T_2 + 441$ and $T_3 + T_4 + T_5 + 440.$
Setting the equations equal and solving for $T_5$ $T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}$ . Therefore, $441T_5 = 441 + T_1 + T_2$ . However, $441 + T_1 + T_2$ is equal to the area of triangle $ABC$ ! This means that the ratio between the areas $T_5$ and $ABC$ is $441$ , and the ratio between the sides is $\sqrt {441} = 21$ . As a result, $AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}$ . We now need $(AC)(BC)$ to find the value of $AC + BC$ , because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2$
Let $h$ denote the height to the hypotenuse of triangle $ABC$ . Notice that $h - \frac {1}{21}h = \sqrt {440}$ . (The height of $ABC$ decreased by the corresponding height of $T_5$ ) Thus, $(AB)(h) = (AC)(BC) = 22\cdot 21^2$ . Because $AC^2 + BC^2 +  2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2$ $AC + BC = (21)(22) = \boxed{462}$

Answer:  462
Problem: Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png

Solution: Let $\tan\angle ABC = x$ . Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$ . Using the second square, $AB=\sqrt{440}(1+x+x^{-1})$ . We have $AC^2+CB^2=AB^2$ , or \[441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).\] Rearranging and letting $u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+2u-440=0.$ We take the positive root, so $u=20$ , which means $AC+CB=21(2+x+x^{-1})=21(2+u)=\boxed{462}$

Answer:  462
Problem: Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png

Solution: Let $\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\sqrt{441}(2+\tan{\theta}+\cot{\theta})$ . We observe that the short leg has length $\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})$ . Grouping and squaring, we get $\sqrt{\frac{440}{441}} = \frac{\sin{\theta}+\cos{\theta}}{1+\sin{\theta}\cos{\theta}}$ . Squaring and using the double angle identity for sine, we get, $110(\sin{2\theta})^2 + \sin{2\theta} - 1 = 0$ . Solving, we get $\sin{2\theta} = \frac{1}{10}$ . Now to find $\tan{\theta}$ , we find $\cos{2\theta}$ using the Pythagorean
Identity, and then use the tangent double angle identity. Thus, $\tan{\theta} = 10-3\sqrt{11}$ . Substituting into the original sum,
we get $\boxed{462}$

Answer:  462
Problem: Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png

Solution: [asy] size(200); pair A, B, C, D, E, F; A = (0, 5); B = (12, 0); C = (0, 0); D = (0, 60/17); E = (60/17, 60/17); F = (60/17, 0); draw(A--B--C--cycle); draw(D--E--F); label("$A$",A,N); label("$B$",B,dir(0)); label("$C$",C,SW); label("$D$",D,W); label("$E$",E,NE); label("$F$",F,S); label("$S_1$",(30/17,30/17)); [/asy]
[asy] size(200); pair A, B, C, W, X, Y, Z; A = (0, 5); B = (12, 0); C = (0, 0); real m = 1.31004366812; real n = 3.1441048035; W = (0,m); X = (m,m+n); Y = (m+n,n); Z = (n,0); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle); label("$A$",A,N); label("$B$",B,dir(0)); label("$C$",C,SW); label("$W$",W,dir(180)); label("$X$",X,NE); label("$Y$",Y,NE); label("$Z$",Z,S); label("$S_2$",(2.22707423581,2.22707423581)); [/asy]
Label points as above. Let $x=AC$ $y=BC$ $s_1 = 21$ be the side length of $S_1$ , and $s_2 = \sqrt{440}$ be the side length of $S_2$
Since $\triangle ABC\sim\triangle AED$ , we have $\frac{x}{y} = \frac{x-s_1}{s_1}$
$\implies xs_1=xy-ys_1$
$\implies xy=s_1(x+y)$
$\implies xy=21(x+y) \qquad \qquad (*)$
Since $\triangle ABC\sim\triangle AWX\sim\triangle ZBY$ , we have $s_2 + \frac{s_2x}{y} + \frac{s_2y}{x}=\sqrt{x^2+y^2}$
$\implies s_2(x^2+xy+y^2)=xy\sqrt{x^2+y^2}$
$\implies s_2^2(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$
$\implies 440(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$
Let $t=x+y$ . Repeatedly applying $(*)$ , we get \[440(t^2-21t)^2 = 441t^2(t^2 - 42t)\] \[440(t-21)^2 = 441(t^2-42t)\] \[440t^2 - 42\cdot 440t + 440\cdot 441 = 441t^2 - 441\cdot 42t\] \[t^2-42t-440\cdot 441=0\] \[(t-21)^2 = 441^2\] \[t-21=441\] \[t=\boxed{462}\]

Answer:  462
Problem: What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$

Solution: Let $y = \sqrt[4]{x}$ . Then we have $y(7 - y) = 12$ , or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]
This means that $\sqrt[4]{x} = y = 3$ or $4$
Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \boxed{337}$

Answer:  337
Problem: Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]

Solution: More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$
We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\ &= \boxed{104} Remark

Answer:  104
Problem: Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]

Solution: We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104} ~Azjps (Solution)

Answer:  104
Problem: Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]

Solution: Notice that in a triangle with side-lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7,$ by Heron's Formula, the area is the square root of the original expression.
Let $\theta$ be the measure of the angle opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(2\sqrt7\right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},\] so $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.$
The area of the triangle is then \[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104},\] so our answer is $\left(\sqrt{104}\right)^2=\boxed{104}.$

Answer:  104
Problem: If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$

Solution: Since $\cot$ is the reciprocal function of $\tan$
$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$
Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$
Using the tangent addition formula:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}$

Answer:  150