Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Solution: A less technical approach that still gets the job done:
Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers above that can be achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score. If this is already confusing, I suggest not looking further.)
For example, the number $"21"$ can be achieved with only $1$ method $(4+4+4+4+5).$ However, $25$ , which is a larger number than $21$ , can be achieved with multiple methods (e.g. $5 \cdot 5$ or $4 \cdot 5 + 5$ ), hence $21$ is not the number we are trying to find.
If we make a table of adding $4$ or adding $5$ , we will see we get $4, 8, 12, 16, 20,$ etc. if we add only $4$ s and if we add $5$ to those numbers then we will get $9, 13, 17, 21, 25,$ etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to $20$ , then there will be multiple methods (because $20 = 4 \cdot 5 = 5 \cdot 4$ ).
Hence, the number we are looking for cannot be $20$ plus one of those base numbers. Instead, it must be $10$ plus that base number, because that results in the same last digit while maintaining only one method to solve. For example, if we start with $4$ , the number $14$ would have only $1$ method to solve, but the number $24$ would have multiple (because $4 + 20 = 24$ and we are trying to avoid adding $20$ ). The largest number we see that is in our base numbers is $21.$ Hence, our maximum number is $21 + 10 = 31.$
Note that if we have the number $25$ , that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding $20$ to a previous base number, which we don't want.
And since the maximum number of this game is $31$ , that is the number we subtract from the maximum score of $150$ , so we get $150 - 31 = \boxed{119}.$

Answer:  119
Problem: Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Solution: Based on the value of $c,$ we construct the following table: \[\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}    &\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&&&&&&&&&&&&& \\ [-2.5ex] \boldsymbol{c} &\boldsymbol{\cdots}&\boldsymbol{12}&\boldsymbol{13}&\boldsymbol{14}&\boldsymbol{15}&\boldsymbol{16}&\boldsymbol{17}&\boldsymbol{18}&\boldsymbol{19}&\boldsymbol{20}&\boldsymbol{21}&\boldsymbol{22}&\boldsymbol{23}&\boldsymbol{24}&\boldsymbol{25}&\boldsymbol{26}&\boldsymbol{27}&\boldsymbol{28}&\boldsymbol{29}&\boldsymbol{30} \\  \hline \hline &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\min}} &\cdots&60&65&70&75&80&85&90&95&100&105&110&115&120&125&130&135&140&145&150 \\ \hline   &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\max}} &\cdots&78&82&86&90&94&98&102&106&110&114&118&122&126&130&134&138&142&146&150 \end{array}\] For a fixed value of $c,$ note that $s_{\min}$ occurs at $w=30-c,$ and $s_{\max}$ occurs at $w=0.$ Moreover, all integers from $s_{\min}$ through $s_{\max}$ are attainable. To find Mary's score, we look for the lowest score $\boldsymbol{s}$ such that $\boldsymbol{s\geq80}$ and $\boldsymbol{s}$ is contained in exactly one interval.
Let $S(c)$ denote the interval of all possible scores $s$ with $c$ correct answers. We need:
It follows that the least such value of $c$ is $23,$ from which the lowest such score $s$ is $\boxed{119}.$

Answer:  119
Problem: Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Solution: Given that Mary's score is $30+4c-w$ , two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$ .
Since it is clear that $c>1$ , we must have $w<4$ . In order to minimize the score, assume that $w=3$ .
The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$ , making the number of correct problems $23$ .
Substituting, we get that $s=30+23{\,\cdot\,}4-3$ , so $s=\boxed{119}$

Answer:  119
Problem: A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Solution: First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.)
The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\choose5} = 56$ different ways to arrange this.
There are ${12 \choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\frac{56}{792} = \frac{7}{99}$
The answer is $7 + 99 = \boxed{106}$

Answer:  106
Problem: A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Solution: Let $b$ $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$ s to separate the $5$ $b$ s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$ s, we are placing the extra $3$ $n$ s into the $6$ intervals beside the $b$ s.
Now doing simple casework.
If all $3$ $n$ s are in the same interval, there are $6$ ways.
If $2$ of the $3$ $n$ s are in the same interval, there are $6\cdot5=30$ ways.
If the $n$ s are in $3$ different intervals, there are ${6 \choose 3} =20$ ways.
In total there are $6+30+20=56$ ways.
There are ${12\choose5}=792$ ways to distribute the birch trees among all $12$ trees.
Thus the probability equals $\frac{56}{792}=\frac{7}{99}\Longrightarrow m+n=7+99=\boxed{106}$

Answer:  106
Problem: A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Solution: Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion.
The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees.
The total number of configurations is given by $\frac{12!}{3! \cdot 4! \cdot 5!}$ . To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE.
$\#$ (configurations with at least one pair of adjacent Birch trees) $=$ $\#$ (configurations with one pair) $-$ $\#$ (configurations with two pairs) $+$ $\#$ (configurations with three pairs) $-$ $\#$ (configurations with four pairs).
To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\frac{11!}{3! \cdot 3! \cdot 4!}$ configurations.
For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\frac{10!}{2! \cdot 3! \cdot 4!}$ cases. So our second term is $\frac{2 \cdot 10!}{2! \cdot 3! \cdot 4!}$
The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\frac{2 \cdot 9!}{3! \cdot 4!}$ arrangements.
The final term can happen in one way (BBBBB). This gives $\frac{8!}{3! \cdot 4!}$ arrangements.
Substituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\frac{12!}{3! \cdot 4! \cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees.
Thus, the probability of a given configuration having no two adjacent Birch trees is given by $\frac{1960}{\frac{12!}{3! \cdot 4! \cdot 5!}} = \frac{7}{99}$
Therefore, the desired result is given by $7+99 = \boxed{106}$

Answer:  106
Problem: A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Solution: Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based.
Arrange first the $3$ maple and $4$ oaks as $MMMOOOO$ . We then notice that for none of the $5$ birch trees to be adjacent, they must be put in between these $M$ 's and $O$ 's. We then see that there are $8$ spots to put these $5$ birch trees in. So we can select $5$ spots for these birch trees in $\binom{8}{5}$ . But then, we can rearrange the $M$ 's and $O$ 's in $7!/(3!4!)=\binom{7}{3}$ ways. So then there are $\binom{8}{5}\binom{7}{3}$ valid arrangements with no given consecutive birch trees.
There are then a total of $\frac{12!}{3!4!5!}$ different total arrangements. Therefore the probability is given as $\frac{\binom{8}{5}\binom{7}{3}}{\frac{12!}{3!4!5!}}=\frac{7}{99}$ , so the answer is $7+99=\boxed{106}$

Answer:  106
Problem: function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Solution: If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\]
Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\] satisfies the conditions and has no other roots.
In the interval $-1000\leq x\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$ , therefore the minimum number of roots is $\boxed{401}$

Answer:  401
Problem: function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Solution: We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$ . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$ $QED \blacksquare$

Answer:  401
Problem: function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Solution: Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

Answer:  401
Problem: function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Solution: Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$ , respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$

Answer:  401
Problem: Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

Solution: We know that $\tan(\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ . Let $a = \cot^{-1}(3)$ $b=\cot^{-1}(7)$ $c=\cot^{-1}(13)$ , and $d=\cot^{-1}(21)$ . We have
so
and
so
Thus our answer is $10\cdot\frac{3}{2}=\boxed{015}$

Answer:  015
Problem: Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

Solution: On the coordinate plane, let $O=(0,0)$ $A_1=(3,0)$ $A_2=(3,1)$ $B_1=(21,7)$ $B_2=(20,10)$ $C_1=(260,130)$ $C_2=(250,150)$ $D_1=(5250,3150)$ $D_2=(5100,3400)$ , and $H=(5100,0)$ . We see that $\cot^{-1}(\angle A_2OA_1)=3$ $\cot^{-1}(\angle B_2OB_1)=7$ $\cot^{-1}(\angle C_2OC_1)=13$ , and $\cot^{-1}(\angle D_2OD_1)=21$ . The sum of these four angles forms the angle of triangle $OD_2H$ , which has a cotangent of $\frac{5100}{3400}=\frac{3}{2}$ , which must mean that $\cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}$ . So the answer is $10\cdot\left(\frac{3}{2}\right)=\boxed{015}.$

Answer:  015
Problem: Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

Solution: Recall that $\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta$ and that $\arg(a + bi) = \tan^{-1}\frac{b}{a}$ . Then letting $w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,$ and $z = 1 + 21i$ , we are left with
\[10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)\] \[= -10\cot(\arg wxyz).\]
Expanding $wxyz$ , we are left with \[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\] \[= (2+i)(13+i)(21+i)\] \[= (25+15i)(21+i)\] \[= (5+3i)(21+i)\] \[= (102+68i)\] \[= (3+2i)\] \[= 10\cot \tan^{-1}\frac{2}{3}\] \[= 10 \cdot \frac{3}{2} = \boxed{015}\]

Answer:  015
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: Take an even positive integer $x$ $x$ is either $0 \bmod{6}$ $2 \bmod{6}$ , or $4 \bmod{6}$ . Notice that the numbers $9$ $15$ $21$ , ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:
If $x \ge 18$ and is $0 \bmod{6}$ $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$ . Note that $9$ and $9+6n$ are both odd composites.
If $x\ge 44$ and is $2 \bmod{6}$ $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$ . Note that $35$ and $9+6n$ are both odd composites.
If $x\ge 34$ and is $4 \bmod{6}$ $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$ . Note that $25$ and $9+6n$ are both odd composites.
Clearly, if $x \ge 44$ , it can be expressed as a sum of 2 odd composites. However, if $x = 42$ , it can also be expressed using case 1, and if $x = 40$ , using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$ . Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

Answer:  038
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$ , then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$ , which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$ , yielding a maximal answer of 38. Since $38-25=13$ , which is prime, the answer is $\boxed{038}$

Answer:  038
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...
For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14.
If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\boxed{38}$

Answer:  38
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: Claim: The answer is $\boxed{038}$

Answer:  038
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: All numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \cdot 25 - 9 - 25 = 225-34 = 191$ . We then bash from top to bottom:
1. $178 = 89 \cdot 2 => 87 + 91$ - refuted
2. $166 = 83 \cdot 2 => 81 + 85$ - refuted
3. $158 = 79 \cdot 2 => 77 + 81$ - refuted
4. $146 = 73 \cdot 2 => 69 + 77$ - refuted
5. $142 = 71 \cdot 2 => 65 + 77$ - refuted
6. $134 = 67 \cdot 2 => 65 + 69$ - refuted
7. $122 = 61 \cdot 2 => 57 + 65$ - refuted
8. $118 = 59 \cdot 2 => 55 + 63$ - refuted
9. $106 = 53 \cdot 2 => 51 + 55$ - refuted
10. $94 = 47 \cdot 2 => 45 + 49$ - refuted
11. $86 = 43 \cdot 2 => 35 + 51$ - refuted
12. $82 = 41 \cdot 2 => 33 + 49$ - refuted
13. $74 = 37 \cdot 2 => 35 + 39$ - refuted
14. $62 = 31 \cdot 2 => 27 + 35$ - refuted
15. $58 = 29 \cdot 2 => 25 + 33$ - refuted
16. $46 = 23 \cdot 2 => 21 + 25$ - refuted
17. $38 = 19 \cdot 2 => = 19 + 19$ - it works!
Because we did a very systematic bash as shown, we are confident the answer is $\boxed{038}$

Answer:  038
Problem: What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution: As stated above, all numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \cdot p$ can be expressed as ( $p$ $q$ )+( $p$ $q$ ) for some positive, even $q$ less then $p$
If $p$ $1 \bmod{30}$ , p±4 would both be composite
If $p$ $7 \bmod{30}$ , p±2 would both be composite
If $p$ $11 \bmod{30}$ , p±14 would both be composite
If $p$ $13 \bmod{30}$ , p±8 would both be composite
If $p$ $17 \bmod{30}$ , p±8 would both be composite
If $p$ $19 \bmod{30}$ , p±14 would both be composite
If $p$ $23 \bmod{30}$ , p±2 would both be composite
If $p$ $29 \bmod{30}$ , p±4 would both be composite
So $p$ < 30
From here, just try all possible p and find the answer is $\boxed{038}$

Answer:  038
Problem: Determine $x^2+y^2+z^2+w^2$ if

Solution: Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\] This equation is satisfied when $t \in \{4, 16, 36, 64\}$ . After clearing fractions, for each of the values $t=4,16,36,64$ , we have the equation \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$
Since the polynomials on each side are equal at $t=4,16,36,64$ , we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$ , so \begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$
Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$ . In each case, most terms drop, and we end up with \begin{align*}  x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*} Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$

Answer:  036