Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$

Solution: It's obvious that we cannot have any number $\ge 7777$ because $7777 > 7000$ so the max number that an occur is $777$
Let's say we have $a$ 777's , $b$ 77's and $c$ 7's
From here we get our required equation as $777a + 77b + 7c = 7000$
Now comes the main problem , one might think that if we find number of $(a,b,c)$ then we're done , but in reality we are over-counting our number of $n$ 's. This is because $n$ is the total number of 7's and from our equation we'll get $n$ as $3a + 2b + c$ (because there are three 7's , two 7's and one 7)
The reason why we're over-counting is because , say $a_1 , b_1 , c_1$ be a solution of our original equation and $a_2 , b_2 , c_2$ be another solution of our original equation , then there can be a possibility that $3a_1 + 2b_1 + c_1 = 3a_2 + 2b_2 + c_2$ where $a_1 \neq a_2 , b_1 \neq b_2 , c_1 \neq c_2$ (example : $2 + 3 + 4 = 1 + 5 + 3$ but $2 \neq 1 , 3 \neq 5 , 4 \neq 3$
We know that $0 \le a \le 9$ $0 \le b \le 90$ $0 \le c \le 1000$ The bound on $a$ is easier to handle with , so lets start putting values on $a$ and calculate $b , c , n$ by making cases
Reduced equation : $111a + 11b + c = 1000$
Case 1 : $a = 9$
We get $11b + c = 1 \implies b = 0 , c = 1$ is our only solution thus only $\boxed{1}$ value of $n$

Answer:  1
Problem: A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a $256$ by $1$ strip of quadruple thickness. This process is repeated $8$ more times. After the last fold, the strip has become a stack of $1024$ unit squares. How many of these squares lie below the square that was originally the $942$ nd square counting from the left?

Solution: Number the squares $0, 1, 2, 3, ... 2^{k} - 1$ .  In this case $k = 10$ , but we will consider more generally to find an inductive solution.  Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$
Now, consider the strip of length $1024$ .  The problem asks for $s_{941, 10}$ .  We can derive some useful recurrences for $s_{n, k}$ as follows:  Consider the first fold.  Each square $s$ is now paired with the square $2^{k} - s - 1$ .  Now, imagine that we relabel these pairs with the indices $0, 1, 2, 3... 2^{k - 1} - 1$ - then the $s_{n, k}$ value of the pairs correspond with the $s_{n, k - 1}$ values - specifically, double, and maybe $+ 1$ (if the member of the pair that you're looking for is the top one at the final step).
So, after the first fold on the strip of length $1024$ , the $941$ square is on top of the $82$ square.  We can then write
\[s_{941, 10} = 2s_{82, 9} + 1\]
(We add one because $941$ is the odd member of the pair, and it will be on top.  This is more easily visually demonstrated than proven.)  We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two):
\[s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127 - 82, 6} = 8s_{45, 6}\]
\[s_{45, 6} = 2s_{63 - 45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31 - 18, 4} + 1 = 4s_{13, 4} + 1\]
\[s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1\]
We can easily calculate $s_{2, 3} = 4$ from a diagram.  Plugging back in,
\begin{align*} s_{13, 4} &= 9 \\ s_{45, 6} &= 37 \\ s_{82, 9} &= 296 \\ s_{941, 10} &= \boxed{593}

Answer:  593
Problem: A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a $256$ by $1$ strip of quadruple thickness. This process is repeated $8$ more times. After the last fold, the strip has become a stack of $1024$ unit squares. How many of these squares lie below the square that was originally the $942$ nd square counting from the left?

Solution: We can keep track of the position of the square labeled 942 in each step. We use an $(x,y)$ coordinate system, so originally the 942 square is in the position $(942,1)$ . In general, suppose that we've folded the strip into an array $r=2^k$ squares wide and $c=1024/r=2^{10-k}$ squares tall (so we've made $10-k$ folds). Then if a square occupies the location $(x,y)$ , we find that after the next fold, it will be in the location described by the procedure \[(x,y)\to\begin{cases}(x,y)&\text{if }x\le 2^{k-1}\\ (r+1-x,2c+1-y)&\text{otherwise}.\end{cases}\] Therefore, we can keep track of the square's location in the following table. \[\begin{array}{c|c|c} (x,y)&\text{rows}&\text{columns}\\\hline (942,1)&1024&1\\ (83,2)&512&2\\ (83,2)&256&4\\ (83,2)&128&8\\ (46,15)&64&16\\ (19,18)&32&32\\ (14,47)&16&64\\ (3,82)&8&128\\ (3,82)&4&256\\ (2,431)&2&512\\ (1,594)&1&1024.\\ \end{array}\] Therefore, at the end of the process, the square labeled 942 will be in the position $(1,594)$ , i.e., it will be above $\boxed{593}$ squares.

Answer:  593
Problem: Given that
where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$

Solution: Note that \[{{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.\] Because $120\cdot719!<720!$ , we can conclude that $n < 720$ . Thus, the maximum value of $n$ is $719$ . The requested value of $k+n$ is therefore $120+719=\boxed{839}$

Answer:  839
Problem: One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius $100$ can be expressed as $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m + n.$

Solution: To get the green area, we can color all the circles of radius $100$ or below green, then color all those with radius $99$ or below red, then color all those with radius $98$ or below green, and so forth.  This amounts to adding the area of the circle of radius $100$ , but subtracting the circle of radius $99$ , then adding the circle of radius $98$ , and so forth.
The total green area is thus given by $100^{2} \pi - 99^{2} \pi + 98^{2} \pi - \ldots - 1^{2} \pi$ , while the total area is given by $100^{2} \pi$ , so the ratio is \[\frac{100^{2}\pi - 99^{2}\pi + 98^{2}\pi - \ldots - 1^{2}\pi}{100^{2}\pi}\]
For any $a$ $a^{2}-(a-1)^{2}=a^{2}-(a^{2}-2a+1)=2a-1$ .  We can cancel the factor of pi from the numerator and denominator and simplify the ratio to
\[\frac{(2\cdot100 - 1)+(2\cdot98 - 1) + \ldots + (2\cdot 2 - 1)}{100^{2}} = \frac{2\cdot(100 + 98 + \ldots + 2) - 50}{100^2}.\]
Using the formula for the sum of an arithmetic series , we see that this is equal to
\[\frac{2(50)(51)-50}{100^{2}}=\frac{50(101)}{100^{2}}=\frac{101}{200},\]
so the answer is $101 + 200 =\boxed{301}$

Answer:  301
Problem: One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius $100$ can be expressed as $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m + n.$

Solution: We want to find $\frac{\sum\limits_{n=1}^{50} (4n-1)\pi}{10000\pi}=\frac{\sum\limits_{n=1}^{50} (4n-1)}{10000}=\frac{(\sum\limits_{n=1}^{50} (4n) )-50}{10000}=\frac{101}{200} \rightarrow 101+200=\boxed{301}$

Answer:  301
Problem: One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius $100$ can be expressed as $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m + n.$

Solution: The sum of the areas of the green regions is
\[\left[(2^2-1^2)+(4^2-3^2)+(6^2-5^2)+\cdots+(100^2-99^2)\right]\pi\] \[=\left[(2+1)+(4+3)+(6+5)+\cdots+(100+99)\right]\pi\] \[={1\over2}\cdot100\cdot101\pi.\]
Thus the desired ratio is \[{1\over2}\cdot{{100\cdot101\pi}\over{100^2\pi}}={101\over200},\] and $m+n=\boxed{301}$

Answer:  301
Problem: Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.

Solution: Order the numbers in the set from greatest to least to reduce error: $\{34, 21, 13, 8, 5, 3, 2, 1\}.$ Each element of the set will appear in $7$ two-element subsets , once with each other number.
Therefore the desired sum is $34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}$

Answer:  484
Problem: Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.

Solution: Thinking of this problem algorithmically, one can "sort" the array to give: \[{1, 2, 3, 5, 8, 13, 21, 34}\]
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we set the sum of the whole array to $x,$ we get out answer to be
\[(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125\]
Finding $7x$ isn't hard, and we see that it is equal to $609$
\[609 - 125 = \boxed{484}\]

Answer:  484
Problem: Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

Solution: Using the properties of logarithms , we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$ . Therefore, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]
Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} 10) \\ \log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\ \sin x + \cos x &= \sqrt{\frac{n}{10}} \\ (\sin x + \cos x)^{2} &= \left(\sqrt{\frac{n}{10}}\right)^2 \\ \sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\ \end{align*}
By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$ , and we can substitute the value for $\sin x \cos x$ from $(*)$ $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$

Answer:  012
Problem: Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

Solution: Examining the first equation, we simplify as the following: \[\log_{10} \sin x \cos x = -1\] \[\implies \sin x \cos x = \frac{1}{10}\]
With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties): \[\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10)\] \[\implies \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} \frac{n}{10})\] \[\implies \log_{10} (\sin x + \cos x) = \log_{10} \sqrt{\frac{n}{10}}\]
From here, we may divide both sides by $\log_{10} (\sin x + \cos x)$ and then proceed with the change-of-base logarithm property: \[1 = \frac{\log_{10} \sqrt{\frac{n}{10}}}{\log_{10} (\sin x + \cos x)}\] \[\implies 1 = \log_{\sin x + \cos x} \sqrt{\frac{n}{10}}\]
Thus, exponentiating both sides results in $\sin x + \cos x = \sqrt{\frac{n}{10}}$ . Squaring both sides gives us \[\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{n}{10}\]
Via the Pythagorean Identity, $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x$ is simply $\frac{1}{5}$ , via substitution. Thus, substituting these results into the current equation: \[1 + \frac{1}{5} = \frac{n}{10}\] \[\implies \frac{6}{5} = \frac{n}{10}\]
Using simple cross-multiplication techniques, we have $5n = 60$ , and thus $\boxed{012}$ .
~ nikenissan

Answer:  012
Problem: Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

Solution: By the first equation, we get that $\sin(x)*\cos(x)=10^{-1}$ . We can let $\sin(x)=a$ $\cos(x)=b$ . Thus $ab=\frac{1}{10}$ . By the identity $\sin^2x+\cos^2x=1$ , we get that $a^2+b^2=1$ . Solving this, we get $a+b=\sqrt{\frac{12}{10}}$ . So we have
\[\log\left(\sqrt{\frac{12}{10}}\right)=\frac12(\log(n)-1)\] \[2\log\left(\sqrt{\frac{12}{10}}\right)=\log(n)-1\] \[\log\left(\frac{12}{10}\right)+1=\log(n)\] \[\log\left(\frac{12}{10}\right)+\log(10)=\log(n)\] \[\log\left(\frac{12}{10}\times 10\right)=\log(12)=\log(n)\]
From here it is obvious that $\boxed{012}$

Answer:  012
Problem: Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers , and $n$ and $p$ are relatively prime , find $m + n + p.$

Solution: The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$ , the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent pair of spheres.
The combined volume of these parts is $60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3}$ . Thus, the answer is $m+n+p = 462+40+3 = \boxed{505}$

Answer:  505
Problem: The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers . Find $m + n + p.$

Solution: Since there are $8$ vertices of a cube , there are ${8 \choose 3} = 56$ total triangles to consider.  They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal ), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.
Each face of the cube contains ${4\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$ , so each triangle of the first type has area $\frac 12$
Each edge of the cube is a side of exactly $2$ of the triangles of the second type, and there are $12$ edges, so there are $24$ triangles of the second type. Each of these is a right triangle with legs of length $1$ and $\sqrt 2$ , so each triangle of the second type has area $\frac{\sqrt{2}}{2}$
Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are $8$ vertices of the cube, so there are $8$ triangles of the third type. Each of the these is an equilateral triangle with sides of length $\sqrt 2$ , so each triangle of the third type has area $\frac{\sqrt 3}2$
Thus the total area of all these triangles is $24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}$ and the answer is $12 + 288 + 48 = \boxed{348}$

Answer:  348
Problem: Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers . Let $s$ be the sum of all possible perimeters of $\triangle ACD$ . Find $s.$

Solution: Denote the height of $\triangle ACD$ as $h$ $x = AD = CD$ , and $y = BD$ . Using the Pythagorean theorem , we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$ . Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$ . The LHS is difference of squares , so $(x + y)(x - y) = 189$ . As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$
The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$ . This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$ . The last is not a possibility since it simply degenerates into a line . The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$

Answer:  380
Problem: In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms.

Solution: Denote the first term as $a$ , and the common difference between the first three terms as $d$ . The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$
Since the first and fourth terms differ by $30$ , we have that $\frac{(a + 2d)^2}{a + d} - a = 30$ . Multiplying out by the denominator, \[(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).\] This simplifies to $3ad + 4d^2 = 30a + 30d$ , which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$
Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative , then $d > 10$ and $d < \frac{15}{2}$ , which is a contradiction). Then, $d = 8, 9$ . Directly substituting and testing shows that $d \neq 8$ , but that if $d = 9$ then $a = 18$ . Alternatively, note that $3|2d$ or $3|2d-15$ implies that $3|d$ , so only $9$ may work. Hence, the four terms are $18,\ 27,\ 36,\ 48$ , which indeed fits the given conditions. Their sum is $\boxed{129}$

Answer:  129
Problem: In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms.

Solution: The sequence is of the form $a-d,$ $a,$ $a+d,$ $\frac{(a+d)^2}{a}$ . Since the first and last terms differ by 30, we have \[\frac{(a+d)^2}{a}-a+d=30\] \[d^2+3ad=30a\] \[d^2+3ad-30a=0\] \[d=\frac{-3a + \sqrt{9a^2+120a}}{2}.\] Let $9a^2+120a=x^2$ , where $x$ is an integer. This yields the following: \[9a^2+120a-x^2=0\] \[a=\frac{-120 + \sqrt{14400+36x^2}}{18}\] \[a=\frac{-20 + \sqrt{400+x^2}}{3}.\] We then set $400+x^2=y^2$ , where $y$ is an integer. Factoring using difference of squares, we have \[400=2^4 \cdot 5^2=(y+x)(y-x).\] Then, noticing that $y+x > y-x$ , we set up several systems of equations involving the factors of $400$ . The second system we set up in this manner, \[y+x=2^3 \cdot 5^2\] \[y-x=2,\] yields the solution $y=101, x=99$ . Plugging back in, we get that $a=27 \implies d=9$ , so the sequence is $18,$ $27,$ $36,$ $48,$ and the answer is $\boxed{129}.$

Answer:  129
Problem: In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms.

Solution: We represent the values as $a-d$ $a$ $a+d$ , and $\frac{(a+d)^2}{a}$ Take the difference between the first and last values \[\frac{(a+d)^2}{a}-a+d=30\] Manipulating the values by expanding and then long division we see \[\frac{a^2+2ad+d^2}{a}-a+d=30\] \[\frac{(a+2d)a+d^2}{a}-a+d=30\] \[a+2d+\frac{d^2}{a}-a+d=30\] Combining like terms we get \[3d+\frac{d^2}{a}=30\] And looking at the values we know that since the second term must be positive (since both a and d are positive), $d$ must be a maximum of 9, which offers 9 possible values (since $d$ must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of $d=9$ at which $a=27$ . The solution is thus $\boxed{129}.$

Answer:  129
Problem: An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

Solution: If the common sum of the first two and last two digits is $n$ , such that $1 \leq n \leq 9$ , there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit).  This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers.  If the common sum of the first two and last two digits is $n$ , such that $10 \leq n \leq 18$ , there are $19 - n$ choices for both pairs.  This gives $\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285$ balanced numbers.  Thus, there are in total $330 + 285 = \boxed{615}$ balanced numbers.

Answer:  615
Problem: An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

Solution: Call the number $\overline{abcd}$ . Then $a+b=c+d$ . Set $a+b=x$
Clearly, $0\le x \le18$
If $x=0$ $0000$ is not acceptable.
If $x=1$ : The only case is $1001$ or $1010$ . 2 choices.
If $x=2$ : then since $a\neq0$ $a=1=b$ or $a=2, b=0$ . There are 3 choices for $(c,d)$ $(2,0), (0, 2), (1, 1)$ $2*3=6$ here.
If $x=3$ : Clearly, $a\neq b$ because if so, the sum will be even, not odd. Counting $(a,b)=(3,0)$ , we have $4$ choices. Subtracting that, we have $3$ choices. Since it doesn't matter whether $c=0$ or $d=0$ , we have 4 choices for $(c,d)$ . So $3*4=12$ here.
If $x=4$ : Continue as above. $4$ choices for $(a,b)$ $5$ choices for $(c,d)$ $4*5=20$ here.
If $x=5$ : You get the point. $5*6=30$
If $x=6$ $6*7=42$
If $x=7$ $7*8=56$
If $x=8$ $8*9=72$
If $x=9$ $9*10=90$
Now we need to be careful because if $x=10$ $(c,d)=(0,10)$ is not valid. However, we don't have to worry about $a\neq0$
If $x=10$ $(a,b)=(1,9), (2, 8), ..., (9, 1)$ . Same thing for $(c,d)$ $9*9=81$
If $x=11$ : We start at $(a,b)= (2,9)$ . So $8*8$
Continue this pattern until $x=18: 1*1=1$ . Add everything up: we have $\boxed{615}$

Answer:  615
Problem: An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

Solution: We ignore the requirement that the first digit is non-zero, and do casework on the sum of the sum of the pairs of digits.
If two digits $a$ and $b$ sum to $0$ , we have $1$ possibility: $(a,b) = (0,0)$
If $a+b = 1$ , we have $2$ possibilities: $(a,b) = (0,1)$ and $(a,b) = (1,0)$
$a+b = 2$ $(0,2)$ $(2,0)$ , and $(1,1)$ are the only $3$ possibilities.
We notice a pattern: for all $k \leq 9$ , there are $k+1$ ordered pairs of digits $(a,b)$ such that $a+b = k$ . Then, testing for $10 \leq k \leq 18$ , we notice that there are $(18 - k) + 1$ ordered pairs with a sum of $k$
So the number of ways to pick $4$ digits satisfying the constraint is \[(1)(1) + (2)(2) \dots (10)(10) + (9)(9) + \dots + (1)(1) = \frac{(10)(11)(21)}{6} + \frac{(9)(10)(19)}{6} = 670\]
We have to subtract out the cases where the first digit is $0$ , which is $1+2+\dots+10  = 55$
So our answer is $670-55 = \boxed{615}$

Answer:  615