Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$

Solution: The $231$ cubes which are not visible must lie below exactly one layer of cubes.  Thus, they form a rectangular solid which is one unit shorter in each dimension.  If the original block has dimensions $l \times m \times n$ , we must have $(l - 1)\times(m-1) \times(n - 1) = 231$ .  The prime factorization of $231 = 3\cdot7\cdot11$ , so we have a variety of possibilities; for instance, $l - 1 = 1$ and $m - 1 = 11$ and $n - 1 = 3 \cdot 7$ , among others.  However, it should be fairly clear that the way to minimize $l\cdot m\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3 \times 7 \times 11$ .  Then the extra layer makes the entire block $4\times8\times12$ , and $N= \boxed{384}$

Answer:  384
Problem: How many positive integers less than 10,000 have at most two different digits

Solution: First, let's count numbers with only a single digit.  We have nine of these for each length, and four lengths, so 36 total numbers.
Now, let's count those with two distinct digits.  We handle the cases "0 included" and "0 not included" separately.
There are ${9 \choose 2}$ ways to choose two digits, $A$ and $B$ .  Given two digits, there are $2^n - 2$ ways to arrange them in an $n$ -digit number, for a total of $(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22$ such numbers (or we can list them: $AB, BA, AAB, ABA, BAA, ABB, BAB, BBA, AAAB, AABA, ABAA,$ $BAAA, AABB, ABAB, BAAB, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA$ ).  Thus, we have ${9 \choose 2} \cdot 22 = 36\cdot22 = 792$ numbers of this form.
Now, suppose 0 is one of our digits.  We have nine choices for the other digit.  For each choice, we have $2^{n - 1} - 1$ $n$ -digit numbers we can form, for a total of $(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11$ such numbers (or we can list them: $A0, A00, A0A, AA0, A000, AA00, A0A0, A00A, AAA0, AA0A, A0AA$ ).  This gives us $9\cdot 11 = 99$ numbers of this form.
Thus, in total, we have $36 + 792 + 99 = \boxed{927}$ such numbers.

Answer:  927
Problem: How many positive integers less than 10,000 have at most two different digits

Solution: We use casework on the number of digits for this problem.
If the number has a single digit, namely the number $n \in [1,9],$ we can clearly all such $n$ work.
If the number has two digits, or the number $n \in [10,99]$ we can clearly see all such $n$ work.
If the number $n$ has three digits, there are a total of $900$ three digit numbers, and there are $9 \cdot 9 \cdot 8$ numbers that have all distinct digits so there are $900 - 9 \cdot 9 \cdot 8$ total three digit numbers that work.
If the number $n$ has four digits, then we have a total of $9 +\binom{9}{2}\left(\binom{4}{1}+\binom{4}{2} +\binom{4}{3}\right) + 9 \cdot \left(\binom{3}{1}+\binom{3}{2} +\binom{3}{3}\right)$ so in total we get $\boxed{927}$ numbers that work.

Answer:  927
Problem: In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional $100$ workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the $800$ workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Solution: A train is traveling at $1000$ miles per hour and has one hour to reach its destination $1000$ miles away. After $15$ minutes and $250$ miles it slows to $900$ mph, and thus takes $\frac{250}{900}(60)=\frac{50}{3}$ minutes to travel the next $250$ miles. Then it slows to $800$ mph, so the next quarter takes $\frac{250}{800}(60)=\frac{150}{8}$ . The train then has $60-15-\frac{50}{3}-\frac{150}{8}=230/24$ minutes left to travel 250 miles, and doing the arithmetic shows that during this last stretch it must travel more than $1565$ mph to make it on time. Therefore the company needs to add $1566-800 = \boxed{766}$ more workers.

Answer:  766
Problem: In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional $100$ workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the $800$ workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Solution: Let each worker's speed be $w$ , the entire time be $t$ , and the total work be $1$
From the initial problem statement, we have $1000w\cdot\frac{1}{4}t=\frac{1}{4}$
We need to find the time the next quarter takes to complete the same amount of work, which is $\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t$
Similarly, we find that the time the third quarter takes is $\frac{1000}{800}\cdot\frac{1}{4}\cdot t=\frac{5}{16}t$
Finally, the time the last quarter takes is $t\left[1-\left(\frac{1}{4}+\frac{5}{18}+\frac{5}{16}\right)\right]=\frac{23}{144}t$
We then let the number of workers needed be $x$ , so we have the equation $\left(800+x\right)w\cdot\frac{23}{144}t=\frac{1}{4}$ . Dividing by the first equation, we have
\[\frac{800+x}{1000}\cdot\frac{\frac{23}{144}}{\frac{1}{4}}=1\] \[\frac{800+x}{1000}\cdot\frac{23}{36}=1\] \[800+x=\frac{36000}{23}\] \[x=\frac{17600}{23}.\]
We can't have a part of a worker, so we take the ceiling of $x$ , which we find to be $\boxed{766}$

Answer:  766
Problem: In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional $100$ workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the $800$ workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Solution: Suppose $1000$ workers can complete one quarter of the job in one day. After the first day, there were $900$ workers remaining so the second quarter was completed in $\frac{10}{9}$ days. Now there are only $800$ workers remaining so the third quarter can be completed in $\frac{10}{8}$ days. It has been $1+\frac{10}{9}+\frac{10}{8}=\frac{121}{36}$ days since the job started, and we still have $\frac{23}{36}$ days to complete the last quarter of the job, so we need $\frac{36}{23}$ of $1000$ workers, or $1565.2173913$ . Since we can’t hire fractional workers, $1566-800=\boxed{766}$ must be the answer.

Answer:  766
Problem: Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$ what is the least possible total for the number of bananas?

Solution: Denote the number of bananas the first monkey took from the pile as $b_1$ , the second $b_2$ , and the third $b_3$ ; the total is $b_1 + b_2 + b_3$ . Thus, the first monkey got $\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3$ , the second monkey got $\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3$ , and the third monkey got $\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3$
Taking into account the ratio aspect, say that the third monkey took $x$ bananas in total. Then,
$x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2 + \frac{11}{48}b_3 = \frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3$
Solve this to find that $\frac{b_1}{11} = \frac{b_2}{13} = \frac{b_3}{27}$ . All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that $b_1$ is divisible by $8$ $b_2$ is divisible by $8$ , and $b_3$ is divisible by $72$ (however, since the denominator contains a $27$ , the factors of $3$ cancel, and it only really needs to be divisible by $8$ ). Thus, the minimal value is when each fraction is equal to $8$ , and the solution is $8(11 + 13 + 27) = \boxed{408}$

Answer:  408
Problem: Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$ what is the least possible total for the number of bananas?

Solution: Let the first monkey take $8x$ bananas, the second monkey take $8y$ , and the third monkey take $24z$ . I chose these numbers to make it so, when each monkey splits his bananas, they will get an integer amount of each variable. So, when the first monkey distributes his bananas, he gets $6x$ , and the other monkeys get $1x$ . So, we can make expressions for how many bananas each monkey gets. Also, the variables have to be integers too.
Monkey 1: $6x+3y+11z$
Monkey 2: $1x+2y+11z$
Monkey 3: $1x+3y+2z$
So, they are in a ratio $3:2:1$ . But, we can turn it into an equation by multiplying the amount of bananas each monkey has by 2, 3, 6. Now, the ratio is $6:6:6$ , so, $12x+6y+22z = 3x+6y+33z = 6x+18y+12z$ . Subtracting $3x+6y+12z$ from all, we get $9x+10z = 21z = 3x+12y$ . Let's split this into 3 equations.
\[9x+10z = 21z\] \[21z = 3x+12y\] \[9x+10z = 3x+12y\]
Let's look at the first equation. Rearranging, it gets us $9x = 11z$
We can rearrange the third equation, then divide by 2, then subtract the second equation. \[21z-6y = 12y-5z\] \[26z = 18y\] \[13z = 9y\]
It is clear $z$ is a multiple of 9, so let $z = 9$ . Then we get the $x = 11$ , and $y = 13$ . Testing, we confirm this will get the first monkey 204 bananas, the second 136, and the third, 68. Adding them up, we get that there were $\boxed{408}$ bananas originally in the pile.

Answer:  408
Problem: Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$ what is the least possible total for the number of bananas?

Solution: In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get
$27x=11z$ $27y=13z$
While solving, make sure to leave a list of numbers that must divide $x$ $y$ , and $z$ . For example, just by looking at the basic fractions you receive from writing the starting equations, 24 divides $z$ , 8 divides $y$ and $x$ . In the expression above, it's also clear that 27 divides $z$ , 13 divides $y$ , and 11 divides $z$ . You might be wondering why I wrote the expression in terms of z. That's because $z$ has the largest divisor. The LCM of all it's divisors shows that $z$ must be divisible by 216. The total amount of bananas can be found to equal to $17z/9$ . This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest possible solution to be $\boxed{408}$

Answer:  408
Problem: Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$ what is the least possible total for the number of bananas?

Solution: Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this your favorite way to get that \[(A,B,C)=\left( \frac{11}{51}, \frac{13}{51}, \frac{9}{17} \right).\] We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \cdot 51 = \boxed{408}.$

Answer:  408
Problem: $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Since $EF$ is the perpendicular bisector of $\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem , we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have \begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC  &= \frac{70}{3} \end{align*} Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$ , and our answer is $m+n=\boxed{293}$

Answer:  293
Problem: $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Let $A = (0,0), B=(0,25)$ , so $E = (0,8)$ and $F = (l,22)$ , and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$ . We know that $EF$ is perpendicular to and bisects $BB'$ . The slope of $BB'$ is thus $\frac{-l}{14}$ , and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$ . Let the point of intersection of $EF, BB'$ be $G$ . Then the y-coordinate of $G$ is $\frac{25}{2}$ , so \begin{align*} \frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ \end{align*} Dividing the two equations yields
The answer is $\boxed{293}$ as above.

Answer:  293
Problem: $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Firstly, observe that if we are given that $AE=8$ and $BE=17$ , the length of the triangle is given and the height depends solely on the length of $CF$ . Let Point $A = (0,0)$ . Since $AE=8$ , point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,-15)$ since $BE=AE$ by symmetry. Draw line segment $BB'$ . Notice that this is perpendicular to $EF$ by symmetry. Next, find the slope of EB, which is $\frac{15}{25}=\frac{3}{5}$ . Then, the slope of $EF$ is - $\frac{5}{3}$
Line EF can be written as y= $-\frac{5}{3}x+b$ . Plug in the point $(8,0)$ , and we get the equation of EF to be y= $_\frac{5}{3}x+\frac{40}{3}$ . Since the length of $AB$ =25, a point on line $EF$ lies on $DC$ when $x=25-3=22$ . Plug in $x=22$ into our equation to get $y=-\frac{70}{3}$ $|y|=BC=\frac{70}{3}$ . Therefore, our answer is $2(AB+BC)=2\left(25+\frac{70}{3}\right)=2\left(\frac{145}{3}\right)=\frac{290}{3}= \boxed{293}$

Answer:  293
Problem: $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Firstly, note that $B'E=BE=17$ , so $AB'=\sqrt{17^2-8^2}=15$ . Then let $\angle BEF=\angle B'EF=\theta$ , so $\angle B'EA = \pi-2\theta$ . Then $\tan(\pi-2\theta)=\frac{15}{8}$ , or
\[\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}\] using supplementary and double angle identities. Multiplying though and factoring yields
\[(3\tan(\theta)-5)(5\tan(\theta)+3)=0\]
It is clear from the problem setup that $0<\theta<\frac\pi2$ , so the correct value is $\tan(\theta)=\frac53$ . Next, extend rays $\overrightarrow{BC}$ and $\overrightarrow{EF}$ to intersect at $C'$ . Then $\tan(\theta)=\frac{BC'}{17}=\frac53$ , so $BC'=\frac{85}{3}$ . By similar triangles, $CC'=\frac{3}{17}BC'=\frac{15}{3}$ , so $BC=\frac{70}{3}$ . The perimeter is $\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}$

Answer:  293
Problem: $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3.
Add all the sides together to get $\boxed{293}$

Answer:  293