Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$

Solution: Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$ . We only need to find $\tan x \tan y$
We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$ . Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25$
Similarly, we have $30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}$
Dividing:
$\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}$ . Plugging in to the earlier formula, we have $\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}$

Answer:  150
Problem: If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$

Solution: Let $a=\tan x$ and $b=\tan y$ . This simplifies the equations to:
\[a + b = 25\]
\[\frac{1}{a} + \frac{1}{b} = 30\]
Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$
We can use substitution to solve the system of equations from above: $b = -a + 25$ , so $\frac{1}{a} + \frac{1}{-a + 25} = 30$
Multiplying by $-a(a-25)$ ,  we get $a + (-a + 25) = -30a(a-25)$ , which is $-30a^2 + 750a = 25$ . Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$
Using the quadratic formula gives $a = \frac{150 \pm \sqrt {22380}}{12}$ . Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$ , which turns out to also be $b = \frac{150 \pm \sqrt {22380}}{12}$ , provided that the sign of the radical in $a$ is opposite the one in $b$
WLOG, assume $a = \frac{150 + \sqrt{22380}}{12}$ and $b = \frac{150 - \sqrt{22380}}{12}$ . Multiplying them gives $ab = \frac{22500 - 22380}{144}$ which simplifies to $\frac{5}{6}$
THe denominator of $\frac{25}{1 - ab}$ ends up being $\frac{1}{6}$ , so multiplying both numerator and denominator by 6 gives $\boxed{150}$

Answer:  150
Problem: Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below.

Solution: Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ .  Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$ , so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$

Answer:  181
Problem: Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below.

Solution: Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, \[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}\]

Answer:  181
Problem: The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice?

Solution: Denote the page number as $x$ , with $x < n$ . The sum formula shows that $\frac{n(n + 1)}{2} + x = 1986$ . Since $x$ cannot be very large, disregard it for now and solve $\frac{n(n+1)}{2} = 1986$ . The positive root for $n \approx \sqrt{3972} \approx 63$ . Quickly testing, we find that $63$ is too large, but if we plug in $62$ we find that our answer is $\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}$

Answer:  033
Problem: The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice?

Solution: Use the same method as above where you represent the sum of integers from $1$ to $n$ expressed as $\frac{n(n + 1)}{2}$ , plus the additional page number $k$ . We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)
$\frac{n(n + 1)}{2} = 1986$ is the quadratic you must solve to obtain the upper bound of $n$ and $\frac{n(n + 1)}{2} + n = 1986$ is the quadratic you must solve to obtain the lower bound of $n$
Solving the two equations gives values that are respectively around $62.5$ and $61.5$ with the quadratic formula, and the only integer between the two is $62$
This implies that we can plug in $62$ and come to the same conclusion as the above solution where $x = \boxed{033}$

Answer:  033
Problem: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$ , so in binary form we get $1100100$ . However, we must change it back to base 10 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed{981}$

Answer:  981
Problem: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: Notice that the first term of the sequence is $1$ , the second is $3$ , the fourth is $9$ , and so on. Thus the $64th$ term of the sequence is $729$ . Now out of $64$ terms which are of the form $729$ $'''S'''$ $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$ , i.e. $972$ , is greater than the largest term which does not, or $854$ . So the $96$ th term will be $972$ , then $973$ , then $975$ , then $976$ , and finally $\boxed{981}$

Answer:  981
Problem: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: After the $n$ th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$ th power of 3 is equal to the number of terms before the $n$ th power, because those terms after the $n$ th power are just the $n$ th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of $3$ and the terms that come after them, we see that the $100$ th term is after $729$ , which is the $64$ th term. Also, note that the $k$ th term after the $n$ th power of 3 is equal to the power plus the $k$ th term in the entire sequence. Thus, the $100$ th term is $729$ plus the $36$ th term. Using the same logic, the $36$ th term is $243$ plus the $4$ th term, $9$ . We now have $729+243+9=\boxed{981}$

Answer:  981
Problem: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$ . From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ $729$ and the 128th term is $3^7$ $2187$ . Writing out more terms of the sequence until the next power of 3 again (81) we can see that the ( $2^n$ $2^{n+1}$ )/2 term is equal to $3^n$ $3^{n-1}$ . From here, we know that the 96th term is $3^6$ $3^5$ $972$ . From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is $972 + 1 = 973$ , the 98th term is $972 + 3 = 975$ , the 99th term is $972 + 3 + 1 = 976$ , and finally the 100th term is $972 + 9 = \boxed{981}$

Answer:  981
Problem: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: The number of terms $3^n$ produces includes each power of 3 ( $1, 3^1, ..., 3^n$ ), the sums of two power of 3s(ex. $3^1 + 1$ ), three power of 3s (ex. $3^1 + 1 + 3^n$ ), all the way to the sum of them all. Since there are $n+1$ powers of 3, the one number sum gives us ${n+1\choose 1}$ terms, the two number ${n+1\choose 2}$ terms, all the way to the sum of all the powers which gives us ${n+1\choose n+1}$ terms. Summing all these up gives us $2^{n+1} - 1 ^ {*}$ according to the theorem
Since $2^6$ is the greatest power $<100$ , then $n=5$ and the sequence would look like { $3^0, ..., 3^5$ }, where $3^5$ or $243$ would be the $2^5 - 1 = 63$ rd number. The next largest power $729$ would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces. 
Noting that $729$ plus any combination of lower powers ${1, 3^1 . . .3^4}$ is < $729 + 243$ , so we can add all those terms( $2^5 - 1 = 31$ ) into our sequence:
Our sequence now has $63 + 1 + 31 = 95$ terms. The remaining $5$ would just be the smallest sums starting with $729 + 243$ or $972$
Hence the 100th term would be $972 + 9 = \boxed{981}$ . ~SoilMilk

Answer:  981
Problem: Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$

Solution: The prime factorization of $1000000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) = 49$ divisors, of which $48$ are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.
Writing out the first few terms, we see that the answer is equal to \[\log 1 + \log 2 + \log 4 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).\] Each power of $2$ appears $7$ times; and the same goes for $5$ . So the overall power of $2$ and $5$ is $7(1+2+3+4+5+6) = 7 \cdot 21 = 147$ . However, since the question asks for proper divisors, we exclude $2^65^6$ , so each power is actually $141$ times. The answer is thus $S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}$

Answer:  141
Problem: Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$

Solution: The formula for the product of the divisors of $n$ is $n^{(d(n))/2}$ , where $d(n)$ is the number of divisor of $n$ .
We know that $\log_{10} a + \log_{10} b + \log_{10} c + \log_{10} d...$ and so on equals $\log_{10} (abcd...)$ by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of the product of the proper divisors of $10^6$ . The product of the divisors, by the earlier formula, is ${(10^6)}^{49/2} = 10^{49*3}$ , and since we need the product of only the proper divisors, which means the divisors NOT including the number, $10^6$ , itself, we divide $10^{(49*3)}$ by $10^6$ to get $10^{(49*3-6)} = 10^{(141)}$ . The base-10 logarithm of this value, in base 10, is clearly $\boxed{141}$

Answer:  141
Problem: Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$

Solution: Since the prime factorization of $10^6$ is $2^6 \cdot 5^6$ , the number of factors in $10^6$ is $7 \cdot 7=49$ . You can pair them up into groups of two so each group multiplies to $10^6$ . Note that $\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6$ . Thus, the sum of the logs of the divisors is half the number of divisors of $10^6 \cdot 6 -6$ (since they are asking only for proper divisors), and the answer is $(49/2)\cdot 6-6=\boxed{141}$

Answer:  141
Problem: Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$

Solution: Note that we can just pair terms up such that the product is $10^{6}.$ Now, however, note that $10^{3}$ is not included. Therefore we first exclude. We have $\displaystyle\frac{49-1}{2} = 24$ pairs that all multiply to $10^{6}.$ Now we include $10^{3}$ so our current product is $24 \cdot 6 - 3.$ However we dont want to include $10^6$ since we are considering proper factors only so the final answer is $144 + 3 - 6 = \boxed{141}.$

Answer:  141
Problem: In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$

Solution: Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are parallelograms
By similar triangles, $BE'=\frac{d}{510}\cdot450=\frac{15}{17}d$ and $EC=\frac{d}{425}\cdot450=\frac{18}{17}d$ . Since $FD'=BC-EE'$ , we have $900-\frac{33}{17}d=d$ , so $d=\boxed{306}$

Answer:  306
Problem: In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$

Solution: Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are parallelograms
Since $PDAF'$ is a parallelogram, we find $PF' = AD$ , and similarly $PE = BD'$ . So $d = PF' + PE = AD + BD' = 425 - DD'$ . Thus $DD' = 425 - d$ . By the same logic, $EE' = 450 - d$
Since $\triangle DPD' \sim \triangle ABC$ , we have the proportion
Doing the same with $\triangle PEE'$ , we find that $PE' =510 - \frac{17}{15}d$ . Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}$

Answer:  306
Problem: In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$

Solution: Refer to the diagram in solution 2; let $a^2=[E'EP]$ $b^2=[D'DP]$ , and $c^2=[F'FP]$ . Now, note that $[E'BD]$ $[D'DP]$ , and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$ . Similarly, we find that $[D'AF]=(b+c)^2$ and $[F'CE]=(c+a)^2$ , so $[ABC]=(a+b+c)^2$ . Now, again from similarity, it follows that $\frac{d}{510}=\frac{a+b}{a+b+c}$ $\frac{d}{450}=\frac{b+c}{a+b+c}$ , and $\frac{d}{425}=\frac{c+a}{a+b+c}$ , so adding these together, simplifying, and solving gives $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$

Answer:  306
Problem: In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$

Solution: Refer to the diagram above. Notice that because $CE'PF$ $AF'PD$ , and $BD'PE$ are parallelograms, $\overline{DD'} = 425-d$ $\overline{EE'} = 450-d$ , and $\overline{FF'} = 510-d$
Let $F'P = x$ . Then, because $\triangle ABC \sim \triangle F'PF$ $\frac{AB}{AC}=\frac{F'P}{F'F}$ , so $\frac{425}{510}=\frac{x}{510-d}$ . Simplifying the LHS and cross-multiplying, we have $6x=2550-5d$ . From the same triangles, we can find that $FP=\frac{18}{17}x$
$\triangle PEE'$ is also similar to $\triangle F'PF$ . Since $EF'=d$ $EP=d-x$ . We now have $\frac{PE}{EE'}=\frac{F'P}{FP}$ , and $\frac{d-x}{450-d}=\frac{17}{18}$ . Cross multiplying, we have $18d-18x=450 \cdot 17-17d$ . Using the previous equation to substitute for $x$ , we have: \[18d-3\cdot2550+15d=450\cdot17-17d\] This is a linear equation in one variable, and we can solve to get $d=\boxed{306}$

Answer:  306
Problem: In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$

Solution: Let $m$ be the number $100a+10b+c$ . Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$ . But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$ . 
Recall that $a, b, c$ refer to the digits the three digit number $(abc)$ , so of the four options, only $m = \boxed{358}$ satisfies this inequality.

Answer:  358
Problem: In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$

Solution: As in Solution 1, $3194 + m \equiv 222(a+b+c) \pmod{222}$ , and so as above we get $m \equiv 136 \pmod{222}$ .
We can also take this equation modulo $9$ ; note that $m \equiv a+b+c \pmod{9}$ , so
\[3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.\]
Therefore $m$ is $7$ mod $9$ and $136$ mod $222$ . There is a shared factor in $3$ in both, but the Chinese Remainder Theorem still tells us the value of $m$ mod $666$ , namely $m \equiv 358$ mod $666$ . We see that there are no other 3-digit integers that are $358$ mod $666$ , so $m = \boxed{358}$

Answer:  358
Problem: In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$

Solution: Let $n=abc$ then \[N=222(a+b+c)-n\] \[N=222(a+b+c)-100a-10b-c=3194\] Since $0<100a+10b+c<1000$ , we get the inequality \[N<222(a+b+c)<N+1000\] \[3194<222(a+b+c)<4194\] \[14<a+b+c<19\] Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$ , we quickly find $n=\boxed{358}$

Answer:  358
Problem: In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$

Solution: The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$ ） and $122 \equiv 5$ （mod $9$ ） so we need to make sure that $a+b+c \equiv 7$ （mod $9$ ） by some testing. So we let $a+b+c=9k+7$
Then, we know that $1\leq a+b+c \leq 27$ so only $7,16,25$ lie in the interval
When we test $a+b+c=25, 10b+11c=16$ , impossible
When we test $a+b+c=16, 10b+11c=138, b=5,c=8,a=3$
When we test $a+b+c=7, 10b+11c=260$ , well, it's impossible
The answer is $\boxed{358}$ then

Answer:  358
Problem: The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the Binomial Theorem , this is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$

Answer:  816
Problem: The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Again, notice $x = y - 1$ . So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.
We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$ . The Hockey Stick Identity tells us that this quantity is equal to ${18\choose 3} = \boxed{816}$

Answer:  816
Problem: The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}$ . This is actually the sum of the first 16 triangular numbers, which evaluates to $\frac{(16)(17)(18)}{6} = \boxed{816}$

Answer:  816
Problem: The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$
Then, since $f(x)=g(y)$ \[\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}\] $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.
Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})$
Now, notice that if $x = -1$ , then $y = 0$ , so $f^{''}(-1) = g^{''}(0)$
$g^{''}(0)= 2a_2$ , and $f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17$
Now, we can use the hockey stick theorem to see that $2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}$
Thus, $2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}$

Answer:  816