Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$

Solution: In order to find the area of rectangle $ABCD$ , we need to find $AB$ first. Using the distance formula , we can derive:
$AB = \sqrt{(2006-6)^2 + (178-(-22))^2} = \sqrt{(2000)^2 + (200)^2 } = 200\sqrt{101}$
Now we can look at the answer choices. Because all of them are integers, then we know that $AD$ has to also contain $\sqrt{101}$ to ensure that $AB*AD$ is an integer. Once you know that, you can guarantee that the answer must be a multiple of 101. (If you're wondering why there can't be a fraction like $1/101$ , there can't because $y$ is an integer the number under the square root in $AD$ , is an integer.) Having that information narrows down the answer choices to just $(B)$ $4040$ , and $(E)$ $40400$
Looking back at what we found for $AB$ , which is $200\sqrt{101}$ , and what we know about $AD$ , that it has to contain $\sqrt{101}$ . we know that the answer is at least $200\sqrt{101}*\sqrt{101}=20200$ , which is already greater than answer choice $(B)$ , which is $4040$ . From this, we can conclude that the answer is $\boxed{40400}$

Answer: E 40400
Problem: A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8);  pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep );  draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );  label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$

Solution: Label the points in the figure as shown below, and draw the segment $CF$ . This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$
[asy] unitsize(2cm); defaultpen(.8);  pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep );  draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );  label("$7$",(1.45,0.15)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35));  draw( C -- F, dashed );  label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",Ep,NW); label("$F$",F,S);  label("$x$",(1,1)); label("$y$",(1.6,1)); [/asy]
Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$
Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$
Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$ , their bases must be in the same ratio, hence $EF:FB = 3:7$
Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$ , their areas must be in the same ratio, hence $x:(y+7) = 3:7$ , which gives us $7x = 3(y+7)$
Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$ , which solves to $x=\frac{15}{2}$ . Then $y=x+3 = \frac{15}{2}+3 = \frac{21}{2}$ , and the total area of the quadrilateral is $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{18}$

Answer: D 18
Problem: A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8);  pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep );  draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );  label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$

Solution: Connect points $E$ and $D$ . Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ratio $3:7$ . Their bases, $EF$ and $FB$ , must be in the same $3:7$ ratio.
Triangles $EFD$ and $FBD$ share an altitude and their bases are in a $3:7$ ratio. Therefore, their areas are in a $3:7$ ratio and the area of triangle $EFD$ is $3$
Triangle $CED$ and $DEA$ share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases $CE$ and $EA$ . The ratio is $A:(3+3) \Rightarrow A:6$ where $A$ is the area of triangle $CED$
Triangles $CEB$ and $EAB$ also share an altitude. The ratio of their areas is also equal to the ratio of bases $CE$ and $EA$ . The ratio is $(A+3+7):(3+7) \Rightarrow (A+10):10$
Because the two ratios are equal, we get the equation $\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15$ . We add the area of triangle $EDF$ to get that the total area of the quadrilateral is $\boxed{18}$

Answer: D 18
Problem: A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8);  pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep );  draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );  label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$

Solution: We use mass points (similar to above). Let the triangle be $ABC$ with cevians (lines to opposite side) from $B$ and $C$ . Let the points opposite $B$ and $C$ be $D$ and $F$ respectively and the intersection as $P$
Assign masses of 1 at $B$ and $D$ since $[BPC] = [DPC]$ . Then the mass at $P$ is 2. To find masses at $F$ and $C$ , we let the mass at $F$ be x and the mass at $C$ be y. Then $3x = 7y$ and $y = \frac{3}{7}x$ . Then $\frac{10}{7}x = 2$ since we add the masses for the fulcrum mass, and $x = \frac{7}{5}$ and $y = \frac{3}{5}$
To calculate the mass at a, it is merely $\frac{7}{5} - 1 = \frac{2}{5}$ which means $\frac{[BCF]}{[ACF]} = \frac{2}{5}$ or $[ACF] = 25$ . It is easy to see the answer is $\boxed{18}$

Answer: D 18
Problem: Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

Solution: Let $S$ be the set of the ages of Mr. Jones' children (in other words $i \in S$ if Mr. Jones has a child who is $i$ years old). Then $|S| = 8$ and $9 \in S$ . Let $m$ be the positive integer seen on the license plate. Since at least one of $4$ or $8$ is contained in $S$ , we have $4 | m$
We would like to prove that $5 \not\in S$ , so for the sake of contradiction, let us suppose that $5 \in S$ . Then $5\cdot 4 = 20 | m$ so the units digit of $m$ is $0$ . Since the number has two distinct digits, each appearing twice, another digit of $m$ must be $0$ . Since Mr. Jones can't be $00$ years old, the last two digits can't be $00$ . Therefore $m$ must be of the form $d0d0$ , where $d$ is a digit. Since $m$ is divisible by $9$ , the sum of the digits of $m$ must be divisible by $9$ (see Divisibility rules for 9 ). Hence $9 | 2d$ which implies $d = 9$ . But $m = 9090$ is not divisible by $4$ , contradiction. So $5 \not\in S$ and $5$ is not the age of one of Mr. Jones' kids. $\boxed{5}$

Answer: B 5
Problem: Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

Solution: Alternatively, we can see that if one of Mr. Jones' children is of the age 5, then the license plate will have to end in the digit $0$ . The license plate cannot end in the digit $5$ as $2$ is a factor of the number, so it must be even. This means that the license plate would have to have two $0$ digits, and would either be of the form $XX00$ or $X0X0$ (X being the other digit in the license plate) . The condition $XX00$ is impossible as Mr. Jones can't be $00$ years old. If we separate the second condition, $X0X0$ into its prime factors, we get $X\cdot 10\cdot 101$ $101$ is prime, and therefore can't account for allowing $X0X0$ being evenly divisible by the childrens' ages.  The $10$ accounts for the 5 and one factor of 2. This leaves $X$ , but no single digit number contains all the prime factors besides 5 and 2 of the childrens' ages, thus $5$ $\boxed{5}$ can't be one of the childrens' ages.

Answer: B 5
Problem: Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

Solution: Another way to do the problem is by the process of elimination. The only possible correct choices are the highest powers of each prime, $2^3=8$ $3^2=9$ $5^1=5$ , and $7^1=7$ , since indivisibility by any powers lower than these means indivisibility by a higher power of the prime (for example, indivisibility by $2^2=4$ means indivisibility by $2^3=8)$ . Since the number is divisible by $9$ , it is not the answer, and the digits have to add up to $9$ . Let $(a,b)$ be the digits: since $2\cdot(a+b) | 9$ $(a+b)|9$ . The only possible choices for $(a,b)$ are $(0,9)$ $(1,8)$ $(2,7)$ $(3,6)$ , and $(4,5)$ . The only number that works is $5544$ , but it is not divisible by $5$ $\boxed{5}$

Answer: B 5
Problem: Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

Solution: Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is $\boxed{5}$ .
-liu4505
~edited by mobius247

Answer: B 5
Problem: While eating out, Mike and Joe each tipped their server $2$ dollars. Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars between their bills?
$\textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20$

Solution: Let $m$ be Mike's bill and $j$ be Joe's bill.
$\frac{10}{100}m=2$
$m=20$
$\frac{20}{100}j=2$
$j=10$
So the desired difference is $m-j=20-10=10 \Rightarrow \boxed{10}$

Answer: D 10
Problem: For each pair of real numbers $a \neq b$ , define the operation $\star$ as
$(a \star b) = \frac{a+b}{a-b}$
What is the value of $((1 \star 2) \star 3)$
$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\, defined.}$

Solution: $((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{0}$

Answer: C 0
Problem: The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$

Solution: $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$

Answer: B 4
Problem: A store normally sells windows at $$100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?
$\textbf{(A) } 100\qquad \textbf{(B) } 200\qquad \textbf{(C) } 300\qquad \textbf{(D) } 400\qquad \textbf{(E) } 500$

Solution: The store's offer means that every $5$ th window is free.
Dave would get $\left\lfloor\frac{7}{5}\right\rfloor=1$ free window.
Doug would get $\left\lfloor\frac{8}{5}\right\rfloor=1$ free window.
This is a total of $2$ free windows.
Together, they would get $\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3$ free windows.
So they get $3-2=1$ additional window if they purchase the windows together.
Therefore they save $1\cdot100= \boxed{100}$

Answer: A 100
Problem: The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?
$\textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27$

Solution: Since the average of the first $20$ numbers is $30$ , their sum is $20\cdot30=600$
Since the average of $30$ other numbers is $20$ , their sum is $30\cdot20=600$
So the sum of all $50$ numbers is $600+600=1200$
Therefore, the average of all $50$ numbers is $\frac{1200}{50}=\boxed{24}$

Answer: B 24
Problem: Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?
$\textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8$

Solution: Let $m$ be the distance in miles that Mike rode.
Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.
Since their combined distance was $13$ miles,
$\frac{8}{5}m + m = 13$
$\frac{13}{5}m = 13$
$m = \boxed{5}$

Answer: B 5
Problem: In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$ . What is the area of the inner square $EFGH$
AMC102005Aq.png
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

Solution: We see that side $BE$ , which we know is $1$ , is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$ . Then $HB = HE + BE = HE + 1$ , and $HE$ is one of the sides of the square whose area we want to find. So:
\[1^2 + (HE+1)^2=\sqrt{50}^2\]
\[1 + (HE+1)^2=50\]
\[(HE+1)^2=49\]
\[HE+1=7\]
\[HE=6\] So, the area of the square is $6^2=\boxed{36}$

Answer: C 36
Problem: There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$

Solution: quadratic equation has exactly one root if and only if it is a perfect square .  So set
$4x^2 + ax + 8x + 9 = (mx + n)^2$
$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$
Two polynomials are equal only if their coefficients are equal, so we must have
$m^2 = 4, n^2 = 9$
$m = \pm 2, n = \pm 3$
$a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12$
$a = 4$ or $a = -20$
So the desired sum is $(4)+(-20)=\boxed{16}$

Answer: A 16
Problem: There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$

Solution: Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have \[(a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80.\] We can use the quadratic formula to solve for its roots (we can ignore the expression in the radical sign as it will cancel out due to the $\pm$ sign when added). So we must have \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] $\frac{-32}{2} = \boxed{16}$

Answer: A 16
Problem: There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$

Solution: There is only one positive value for $k$ such that the quadratic equation would have only one solution. $k-8$ and $-k-8$ are the values of $a$ $-8-8=-16$ , so the answer is $\boxed{16}$

Answer: A 16
Problem: A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$
$\textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7$

Solution: Since there are $n^2$ little faces on each face of the big wooden cube , there are $6n^2$ little faces painted red.
Since each unit cube has $6$ faces, there are $6n^3$ little faces total.
Since one-fourth of the little faces are painted red,
$\frac{6n^2}{6n^3}=\frac{1}{4}$
$\frac{1}{n}=\frac{1}{4}$
$n=\boxed{4}$

Answer: B 4
Problem: How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution: We're given $(130n)^{50} > n^{100} > 2^{200}$ , so
$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus
$130n > n^2 > 2^4$
$130n > n^2 > 16$
Solving each part separately:
$n^2 > 16 \Longrightarrow n > 4$
$130n > n^2 \Longrightarrow 130 > n$
So $4 < n < 130$
Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{125}$

Answer: E 125
Problem: How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution: We're given $\left(130n\right)^{50}>n^{100}>2^{200}$
Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$ . Because the exponents are equal, we can ignore them and solve for $n$ $130n>n^{2}$ , or $n<130$
The second half: $n^{100}>2^{200}$ , or $n^{100}>4^{100}$ , which means $n>4$
Therefore $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{125}$

Answer: E 125
Problem: How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$

Solution: If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework
If the middle digit is $1$ , possible numbers range from $111$ to $210$ . So there are $2$ numbers in this case.
If the middle digit is $2$ , possible numbers range from $123$ to $420$ . So there are $4$ numbers in this case.
If the middle digit is $3$ , possible numbers range from $135$ to $630$ . So there are $6$ numbers in this case.
If the middle digit is $4$ , possible numbers range from $147$ to $840$ . So there are $8$ numbers in this case.
If the middle digit is $5$ , possible numbers range from $159$ to $951$ . So there are $9$ numbers in this case.
If the middle digit is $6$ , possible numbers range from $369$ to $963$ . So there are $7$ numbers in this case.
If the middle digit is $7$ , possible numbers range from $579$ to $975$ . So there are $5$ numbers in this case.
If the middle digit is $8$ , possible numbers range from $789$ to $987$ . So there are $3$ numbers in this case.
If the middle digit is $9$ , the only possible number is $999$ . So there is $1$ number in this case.
So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=\boxed{45}$

Answer: E 45
Problem: How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$

Solution: Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into $2$ cases:
If both the first digit and the last digit are odd, then we have $1, 3, 5, 7,$ or $9$ as choices for each of these digits, and there are $5\cdot5=25$ numbers in this case.
If both the first and last digits are even, then we have $2, 4, 6, 8$ as our choices for the first digit and $0, 2, 4, 6, 8$ for the third digit. There are $4\cdot5=20$ numbers here.
The total number, then, is $20+25=\boxed{45}$

Answer: E 45
Problem: How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$

Solution: As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have $9$ possibilities $(1-9)$ , and the third digit can have $10$ possibilities $(0-9)$ . This means there can be $9\cdot10=90$ possible two-digit numbers in which the first digit and the third digit are digits. Exactly half of these would have their sum be divisible by $2$ (since $90$ is even), so our answer is $\frac{90}{2}=\boxed{45}$

Answer: E 45