Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$

Solution: Factorizing the numerator, $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ then becomes $\frac{\frac{5}{2}}{a^{2}}$ which is equal to $\frac{5}{2}\cdot 2^2$ which is $\boxed{10}$

Answer: D 10
Problem: What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$

Solution: Substituting $\frac{1}{2}$ for $a$ in $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ gives us $\boxed{10}$ . ~peelybonehead

Answer: D 10
Problem: Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution: Substituting carefully, $\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)$
becomes $|4032-2016|+2016=2016+2016=4032$ which is $\boxed{4032}$

Answer: D 4032
Problem: Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution: Solution by e_power_pi_times_i
Substitute $-y = x = -2016$ into the equation. Now, it is $\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y$ . Since $y = 2016$ , it is a positive number, so $|y| = y$ . Now the equation is $\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y$ . This further simplifies to $2y-y+y = 2y$ , so the answer is $\boxed{4032}$

Answer: D 4032
Problem: The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$

Solution: The sum of the ages of the cousins is $4$ times the mean, or $32$ .
There are an even number of cousins, so there is no single median, so $5$ must be the mean of the two in the middle.
Therefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\textbf{(D)}\ \boxed{22}$

Answer:  22
Problem: Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution: Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$ . The hundreds digits of these numbers must be at least $1$ and $2$ , so $a\ge 100$ and $b\ge 200$
Say $a=100+p$ and $b=200+q$ ; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$ .There are many solutions, but $p=3$ and $q=97$ give $103+297=400$ which proves that $\boxed{4}$ is attainable.

Answer: B 4
Problem: Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution: For this problem, to find the $3$ -digit integer with the smallest sum of digits, one should make the units and tens digit add to $0$ . To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $7$ works best for the top number which makes the bottom digit $3$ . The tens digits need to add to $9$ because of the $1$ that needs to be carried from the addition of the units digits. We see that $5$ and $4$ work the best as we can't use $6$ and $3$ . Finally, we use $2$ and $1$ for our hundreds place digits.
Adding the numbers $257$ and $143$ , we get $400$ which means our answer is $\boxed{4}$

Answer: B 4
Problem: The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

Solution: We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into the second equation yields \[\begin{split} 90 - x & = 2 (90 - 1.25x) \\ 1.5x & = 90 \\ x & = 60 \end{split}\]
Substituting this $x$ value back into the first equation yields $y$ $=$ $75$ , leaving $x$ $+$ $y$ equal to $\boxed{135}$

Answer: C 135
Problem: The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

Solution: We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$ . We know $90-4x>90-5x$ since this inequality gives $x>0$ , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so
$90-4x=2(90-5x)$
$90-4x=180-10x$
$6x=90$
$x=15$
Therefore, the angles are $75$ and $60$ , which sum to $\boxed{135}$

Answer: C 135
Problem: What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$

Solution: Notice that, for $n\ge 2$ $2015^n\equiv 15^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself).  Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$
So the answer is $\textbf{(A)}\ \boxed{0}$

Answer:  0
Problem: What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$

Solution: Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in $mod 100$ .) We will use the " $\equiv$ " sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else.
\[\begin{split} 15^{1}\equiv15 \\ 15^{2}\equiv25 \end{split}\] From here we only need to multiply $15\cdot25$ and we can ignore the hundreds digits. \[\begin{split} 15^{3}\equiv75 \\ 15^{4}\equiv25 \\ 15^{5}\equiv75 \end{split}\] Notice that for every $x\neq1$ $15^{x}\equiv25$ if $x$ is even, and $15^{x}\equiv75$ if $x$ is odd. Since $2015^{2016}$ has an even exponent, we conclude that the last two digits will be $25$ , and subtracting $25-17=0 \Longrightarrow \boxed{0}$ .
~JH. L

Answer: A 0
Problem: All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

Solution: Let the point where the height of the triangle intersects with the base be $D$ . Now we can guess what $x$ is and find $y$ . If $x$ is $3$ , then $y$ is $9$ . The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$ , respectively. The distance between $B$ and $C$ is $6$ , meaning the area would be $\frac{6 \cdot 9}{2}=27$ , not $64$ . Now we let $x=4$ $y$ would be $16$ . The cords of $B$ and $C$ would be $(-4,16)$ and $(4,16)$ , respectively. $BC$ would be $8$ , and the height would be $16$ . The area would then be $\frac{8 \cdot 16}{2}$ which is $64$ , so $BC$ is $\boxed{8}$

Answer: C 8
Problem: A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution: We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$

Answer: D 33.3
Problem: A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution: Also recall that the area of an equilateral triangle is $\frac{a^2\sqrt3}{4}$ so we can give a ratio as follows:
$\frac{\frac{9\sqrt3}{4}}{12}$ $=$ $\frac{\frac{25\sqrt3}{4}}{x}$
Cross multiplying and simplifying, we get $12 \cdot \frac{25}{9}$
Which is $33.\overline{3}$ $\approx$ $\boxed{33.3}$

Answer: D 33.3
Problem: A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution: Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is the same. The ratio of their areas is equal to the square of the ratio of their sides. So if $x$ denotes the  weight of the second triangle, we have \[\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}\] Solving gives us $x \approx 33.33$ so the answer is $\boxed{33.3}$

Answer: D 33.3
Problem: Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$

Solution: If the dimensions are $4a\times 4b$ , then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).
The total number of posts is $2(a+b)=20$
This diagram represents the number of posts around the garden. [asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E); label("$a$",(1,1));label("$b$",(4,5)); [/asy]
We solve the system $\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}$ to get $a=3,\ b=7$ . Then the area is $4a\cdot 4b=336$ which is $\boxed{336}$

Answer: B 336
Problem: Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$

Solution: To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only $16$ posts. Now there are twice as many posts on the longer side then the shorter side. From this, we can see that we can put $8$ posts on the longer side and $4$ posts on the shorter side. On the shorter side, we have $3$ spaces between the $4$ posts. On the longer side, we have $7$ spaces between the $8$ fence posts. There are $4$ yards between each post. Therefore, the answer is $12\times28=\boxed{336}$

Answer: B 336
Problem: Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

Solution: The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $\frac{\tbinom32}{\tbinom52}=\frac3{10}$ , so the answer is $1-0.3$ which is $\boxed{0.7}$

Answer: D 0.7
Problem: Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

Solution: There are $2$ cases to get an even number. Case 1: $\text{Even} \times \text{Even}$ and Case 2: $\text{Odd} \times \text{Even}$ . Thus, to get an $\text{Even} \times \text{Even}$ , you get $\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}$ . And to get $\text{Odd} \times \text{Even}$ , you get $\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}$ $\frac {1}{10}+\frac {6}{10}=\frac {7}{10}$ which is $0.7$ and the answer is $\boxed{0.7}$

Answer: D 0.7
Problem: Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

Solution: Note that we have three cases to get an even number: even $\times$ even, odd $\times$ even and even $\times$ odd.
The probability of case 1 is $\dfrac{2}{5}\cdot\dfrac{1}{4}$ , the probability of case 2 is $\dfrac{2}{5}\cdot\dfrac{3}{4}$ and the probability of case 3 is $\dfrac{3}{5}\cdot\dfrac12$
Adding these up we get $\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{10} = \boxed{0.7}.$

Answer: D 0.7
Problem: At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$

Solution: We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \[\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}\]
Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields \[\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}\]
Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$ , but rather $4c$ . Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$ , leaving us with the number of babies born as quadruplets equal to $\boxed{100}$

Answer: D 100
Problem: At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$

Solution: Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\cdot2=24x$ twins, $4x\cdot3=12x$ triplets, and $x\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\frac{1}{10}(1000)=\boxed{100}$

Answer: D 100
Problem: How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution: The region is a right triangle which contains the following lattice points: $(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$
[asy]size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow); draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray); [/asy]
Squares $1\times 1$ :
Suppose that the top-right corner is $(x,y)$ , with $2\le x\le 5$ . Then to include all other corners, we need $1\le y\le 3(x-1)$ .
This produces $3+6+9+12=30$ squares.
Squares $2\times 2$ :
Here $3\le x\le 5$ . To include all other corners, we need $2\le y\le 3(x-2)$ .
This produces $2+5+8=15$ squares.
Squares $3\times 3$ :
Similarly, this produces $5$ squares.
No other squares will fit in the region. Therefore the answer is $\boxed{50}$

Answer: D 50
Problem: How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution: The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ .
This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of $1\cdot1$ $2\cdot2$ , and $3\cdot3$ squares and getting $30$ $15$ , and $5$ respectively, and we end up with $\boxed{50}$

Answer: D 50
Problem: How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution: The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases.
Case 1: Square has length $\bf1$
The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12 = 30$ squares for this case.
Case 2: Square has length $\bf2$
The $x$ coordinates must be $(1,3)$ or $(2,4)$ or $(3,15)$ and so now it starts at $y=2.$ It ends up being $2+5+8 = 15.$
Case 3: Square has length $\bf3$
The $x$ coordinates must be $(1,4)$ or $(2,5)$ so there is $1+4 = 5$ squares for this case.
Therefore, $30+15+5 = \boxed{50}$

Answer: D 50
Problem: All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity: [asy]size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7));[/asy] But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $\boxed{7}$

Answer: C 7
Problem: All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Quick testing shows that \[3~2~1\] \[4~7~8\] \[5~6~9\] is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\boxed{7}$ .. — @adihaya talk ) 12:27, 21 February 2016 (EST) ~edited

Answer: C 7
Problem: All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: First let the numbers be \[1   ~8~   7\] \[2  ~ 9   ~6\] \[3  ~ 4~   5\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$ . ~edited

Answer: C 7
Problem: All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is \[1 ~8~ 9\] \[2 ~ 7 ~6\] \[3 ~ 4~ 5\]
$\boxed{7}$

Answer: C 7