AMC10 2023 B

The first three glasses each have a full glass. Let's assume that each glass has "1 unit" of juice. It won't matter exactly how much juice everyone has because we're dealing with ratios, and that wouldn't affect our answer. The fourth glass has a glass that is one third. So the total amount of juice will be $1+1+1+\dfrac{1}{3} = \dfrac{10}{3}$. If we divide the total amount of juice by 4, we get $\dfrac{5}{6}$, which should be the amount of juice in each glass. This means that each of the first three glasses will have to contribute $1 - \dfrac{5}{6} = \boxed{\textbf{(C) }\dfrac16}$ to the fourth glass.

The discounted price of the shoes is 20\% off the original price. This means the customer pays \[ 1 - 0.20 = 0.80 \] (or 80\%) of the original price. There is also a 7.5\% sales tax, so the final amount paid is \[ 0.80 \times 1.075 = 0.86 \] (or 86\%) of the original price. Let $x$ be the original price. Then \[ 0.86x = 43. \] Solving for $x$, \[ x = \frac{43}{0.86} = 50. \]

Because the triangles are right triangles, we know the hypotenuses are diameters of circles $A$ and $B$. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply $\pi$ to get $6.25\pi$ and $42.25\pi$ as the areas of the circles. Multiply 4 on both numbers to get $25\pi$ and $169\pi$. Cancel out the $\pi$, and lastly, divide, to get your answer $=\boxed{\textbf{(D) }\frac{25}{169}}.$

$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$, so the answer is $\boxed{\textbf{(C) 1,625}}$.

Let there be $n$ numbers in the list of numbers, and let their sum be $S$. Then we have the following \[S+3n=45\] \[3S=45\] From the second equation, $S=15$. So, $15+3n=45$ $\Rightarrow$ $n=\boxed{\textbf{(A) }10}.$

We calculate more terms: \[1,3,4,7,11,18,\ldots.\] We find a pattern: if $n+2$ is a multiple of $3$, then the term is even, or else it is odd. There are $\left\lfloor \frac{2023}{3} \right\rfloor =\boxed{\textbf{(E) }674}$ multiples of $3$ from $1$ to $2023$.

First, let's call the center of both squares $I$. Then, $\angle{AIE} = 20$, and since $\overline{EI} = \overline{AI}$, $\angle{AEI} = \angle{EAI} = 80$. Then, we know that $AI$ bisects angle $\angle{DAB}$, so $\angle{BAI} = \angle{DAI} = 45$. Subtracting $45$ from $80$, we get $\boxed{\text{(B)} 35}$

$2022^{2023} + 2023^{2022} \equiv 2^3 + 3^2 \equiv 17 \equiv 7$ (mod 10). $\boxed{\text{A}}$

Let $x$ be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$. Thus, $x \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation. A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1.

Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles. Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$. Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$

We let the number of $\$20$, $\$50$, and $\$100$ bills be $a,b,$ and $c,$ respectively. We are given that $20a+50b+100c=800.$ Dividing both sides by $10$, we see that $2a+5b+10c=80.$ We divide both sides of this equation by $5$: $\dfrac25a+b+2c=16.$ Since $b+2c$ and $16$ are integers, $\dfrac25a$ must also be an integer, so $a$ must be divisible by $5$. Let $a=5d,$ where $d$ is some positive integer. We can then write $2\cdot5d+5b+10c=80.$ Dividing both sides by $5$, we have $2d+b+2c=16.$ We divide by $2$ here to get $d+\dfrac b2+c=8.$ $d+c$ and $8$ are both integers, so $\dfrac b2$ is also an integer. $b$ must be divisible by $2$, so we let $b=2e$. We now have $2d+2e+2c=16\implies d+e+c=8$. Every substitution we made is part of a bijection (i.e. our choices were one-to-one); thus, the problem is now reduced to counting how many ways we can have $d,e,$ and $c$ such that they add to $8$. We still have another constraint left, that each of $d,e,$ and $c$ must be at least $1$. For $n\in\{d,e,c\}$, let $n'=n-1.$ We are now looking for how many ways we can have $d'+e'+c'=8-1-1-1=5.$ We use a classic technique for solving these sorts of problems: stars and bars. We have $5$ stars and $3$ groups, which implies $2$ bars. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.

First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a side length of $\sqrt{2}$ (diagonal $2$). The area of the square is $\sqrt{2}^2 = 2.$ Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis. So now we have $2$ squares. Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis. Concluding, we have $4$ congruent squares. Thus, the total area is $4\cdot2 =$ $\boxed{\text{(B) 8}}$

Let's use 10th grade math to solve this. After all, it is called the AMC 10 for a reason! We have \[m^2 + mn + n^2 = m^2n^2.\] We subtract mn on both sides to get m2+n2=m2n2−mn. Fun Fact! You can write m2+n2 as (m+n)2−2mn! Let's use this! We convert the Left Hand Side into (m+n)2−2mn to get (m+n)2−2mn=m2n2−mn. Adding by 2mn gives us (m+n)2=m2n2+mn. We aren't done yet though! m2n2+mn can be simplified into mn(mn+1), giving us (m+n)2=mn(mn+1). Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property! How? I'll show you! We subtract mn(mn+1) on both sides of the equation to get (m+n)2−mn(mn+1)=0. Now we do a bit of casework. Notice how (m+n)2−mn(mn+1)=0 is just (m+n)(m+n)−mn(mn+1)=0. So, either m+n=0 and mn=0, or m+n=0 and mn+1=0. Let's look at it through both cases. Case 1: m+n=0 and mn=0. If m+n=0 and mn=0, then that must mean that either m=0 or n=0, and if we substitute either m=0 or n=0 in, we still get either m=0 or n=0, so therefore we have 1 ordered pair, (0,0). Case 2: m+n=0 and mn+1=0. mn+1=0 means that mn=−1. So, for this to be possible, either m=−1 and n=1, or m=1 and n=−1. Let's check for contradictions quickly. We see that m+n=0, and −1+1=0 and 1−1=0, so we know the ordered pairs (−1,1) and (1,−1) both work. We have a total of $\boxed{\textbf{(C) 3}}.$ ordered pairs. (−1,1), (0,0), and (1,−1).

We want $m\cdot2!\cdot3!\cdot4!\cdot\dots\cdot16!$ to be a perfect square. Notice that we can rewrite and pair up certain elements: \[m\cdot2\cdot3!\cdot(4\cdot3!)\cdot5!\cdot(6\cdot5!)\cdot\dots\cdot15!\cdot(16\cdot15!)=m\cdot2\cdot4\cdot6\cdot\dots\cdot16\cdot(3!)^2(5!)^2\cdot\dots\cdot(15!)^2.\] Note here that this is equivalent to simply $m\cdot2\cdot4\cdot\dots\cdot16$ being a perfect square (this is intuitively obvious: i.e. if $a=bc$ is a perfect square and so is $b$, then of course $c$ must be a perfect square too). We can rewrite this as the following: \[m\cdot2^8\cdot(1\cdot2\cdot3\cdot\dots\cdot8)\equiv m\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\equiv m\cdot70.\] The smallest $m$ s.t. $70m$ is a perfect square is, of course, $70$ itself. QED.

First, we know that $D$ is greater than $U$, since there are less $upnos$ than $downnos$. To see why, we examine what determines an upno or $downno$. We notice that, given any selection of unique digits (notice that "unique" constrains this to be a finite number), we can construct a unique downno. Similarly, we can also construct an $upno$, but the selection can not include the digit $0$ since that isn't valid. We then have $[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]$ as the pool of numbers that we can pick from for $downnos$, and {1, 2, 3, 4, 5, 6, 7, 8, 9} as the pool of numbers that we can pick for $upnos$. There are only two states each number can be in: appearing and not appearing in the arrangement. (That is why we use the number 2 as the base in the exponent!) Thus, there are $2^{10}$ total $downnos$ and $2^9$ total $upnos$. However, we are told that each $upno$ or $downno$ must be at least $2$ digits, so we subtract out the $0$-digit and $1$-digit cases (Referring back to Paragraph 2, this is when every number's state is nonappearing or every number except one has the state of nonappearing). For the $downnos$, there are $10$ $1$-digit cases, and for the $upnos$, there are $9$ $1$-digit cases. There is $1$ $0$-digit case for both $upnos$ and $downnos$. Thus, the difference is $\left(\left(2^{10}-10-1\right)-\left(2^9-9-1\right)\right)=2^9-1=\boxed{\textbf{(E) }511}.$

We can create three equations using the given information. \[4a+4b+4c = 13\] \[2ab+2ac+2bc=\frac{11}{2}\] \[abc=\frac{1}{2}\] We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.

We examine each of the conditions. The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $115$. Since $\gcd(3,14)=1$, condition $I$ would imply that $\gcd(c,210)=1.$ However, $\gcd(115,210)$ is clearly not $1$ (they share a common factor of $5$). Condition $I$ is false so that we can rule out choices $A,B,$ and $C$. We now decide between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to check $II$ simply. We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd(a,14)\neq1$ and $\gcd(b,15)\neq1$, then $\gcd(c,210)\neq1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor that is not $1$ with $210$. Let's say that $a=a'\cdot n$, where $a'$ is a factor of $14$ not equal to $1$. (So $a'$ is the common factor.) We can rewrite the given equation as $15a+14b=c\implies15(a'n)+14b=c.$ We can express $14$ as $a'\cdot n'$, for some positive integer $n'$ (this $n'$ can be $1$). We can factor $a'$ out to get $a'(15n+bn')=c.$ Since all values in this equation are integers, $c$ must be divisible by $a'$. Since $a'$ is a factor of $14$, $a'$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a'$), so $\gcd(c,210)\neq1$. This is what $II$ states, so therefore $II$ is true. Thus, our answer is $\boxed{\textbf{(E) }\text{II and III only}}.$

WLOG, we assume Sonya jumps $0.5$ units every time, since that is her expected value. If Sonya is within $0.5$ blocks of an edge, she can jump off the board. Let us examine the region that is at most $0.5$ blocks from exactly one edge. If Sonya starts in this region, she has a $\dfrac14$ chance of landing outside (there's exactly one direction she can hop to get out). The total area of this region is $4\cdot0.5\cdot5=10.$ For this region, Sonya has a $\dfrac14$ chance, so we multiply $10$ by $\dfrac14$ to get $2.5.$ If Sonya is in one of the corner squares, she can go two directions to get out, so she has a $\dfrac24=\dfrac12$ chance to get out. The total area is $0.5\cdot0.5\cdot4=1$, so this region yields $\dfrac12\cdot1=\dfrac12.$ Adding the two, we get $3$. This is out of $36$ square units of area, so our answer is thus $\boxed{\textbf{(B) }\tfrac{1}{12}}.$ Note: When Sonya is within 0 to 1 units away from the border in a given direction, the probability that she will jump outside the border can be thought of as a function of distance from the border. It is easy to see that if Sonya is x units away from the border, a jump distance in the interval [x,1] will move her outside. Thus the probability is a function P(x)=1−x. The probability in an entire region 0 to 1 units away from the border is an integral of the function, ∫01(1−x)dx=[x−12x2]01=1/2. This is why the probability is 1/2.

Focus on 2 of the points. Let the center of the Sphere be A. Label two points that form the diameter of one of the four semicircles M and C respectively. Triangle AMC is a right triangle through the inscribed right triangle theorem, with AM=AC=2. This is a 45-45-90 triangle, so the length of the diameter MC is just the hypotenuse of the triangle AMC, which is 22. This means the radius is 2. The circumference of the semicircle is 12⋅22⋅π=π2, and because there are four congruent semicircles, the length of the semicircular region is 42π=π⋅42. The solution is in the form πn, so we convert π⋅42 to π162 to get π32. n=$\boxed{\textbf{(A) 32}}$.

Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.

To further grasp at this equation, we rearrange the equation into \[\lfloor{x}\rfloor^2=3x-2.\] Thus, $3x-2$ is a perfect square and nonnegative. It is now much more apparent that $x \ge 2/3,$ and that $x = 2/3$ is a solution. Additionally, by observing the RHS, $x<4,$ as \[\lfloor{4}\rfloor^2 > 3\cdot4,\] since squares grow quicker than linear functions. Now that we have narrowed down our search, we can simply test for intervals $[2/3,1], [1,2],[2,3],[3,4).$ This intuition to use intervals stems from the fact that $x=1,2$ are observable integral solutions. Notice how there is only one solution per interval, as $3x-2$ increases while $\lfloor{x}\rfloor^2$ stays the same. Finally, we see that $x=3$ does not work, however, through setting $\lfloor{x}\rfloor^2 = 9,$ $x = 11/3$ is a solution and within our domain of $[3,4).$ This provides us with solutions $\left(\frac23, 1, 2, \frac{11}{3}\right),$ thus the final answer is $\boxed{(\text{B}) \ 4}.$

Since one of the terms was either $1$ more or $1$ less than it should have been, the sum should have been $222-1=221$ or $222+1=223.$ The formula for an arithmetic series is $an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(2a+d(n-1)\right).$ This can quickly be rederived by noticing that the sequence goes $a,a+d,a+2d,a+3d,\dots,a+(n-1)d$, and grouping terms. We know that $\dfrac n2(2a+d(n-1))=221$ or $223$. Let us now show that $223$ is not possible. If $\dfrac n2(2a+d(n-1))=223$, we can simplify this to be $n(2a+d(n-1))=223\cdot2.$ Since every expression here should be an integer, we know that either $n=2$ and $2a+d(n-1)=223$ or $n=223$ and $2a+d(n-1)=2.$ The latter is not possible, since $n\ge3,d>1,$ and $a>0.$ The former is also impossible, as $n\ge3.$ Thus, $\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221$. (Alternatively, we have $S=mn$ with $n$ terms and arithmetic mean $m$, and $223=mn$ does not satisfy both $m > 1$ and $n > 1$ because it is prime.) We can factor $221$ as $13\cdot17$. Using similar reasoning, we see that $221\cdot2$ cannot be paired as $2$ and $221$, but rather must be paired as $13$ and $17$ with a factor of $2$ somewhere. Let us first try $n=13.$ Our equation simplifies to $2a+12d=34\implies a+6d=17.$ We know that $d>1,$ so we try the smallest possible value: $d=2.$ This would give us $a=17-2\cdot6=17-12=5.$ (Indeed, this is the only possible $d$.) There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer $a+d+n=5+2+13=\boxed{\textbf{(B) }20.}$ For the sake of completeness, we can explore $n=17.$ It turns out that we reach a contradiction in this case, so we are done.

Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$. We first fix $u$ and $v$ to $0$, and graph $(-3w,4w)$ from $0\le w\le1$. When $w$ is $0$, we have the point $(0,0)$, and when $w$ is $1$, we have the point $(-3,4)$. We see that since this is a directly proportional function, we can just connect the dots like this: Now, when we vary $2u$ from $0$ to $2$, this line is translated to the right $2$ units: We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$, so we shade in the region: We can also shift this quadrilateral one unit up, because of $v$. Thus, this is our figure: The length of the boundary is simply $1+2+5+1+2+5$ ($5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$.). This equals $\boxed{\textbf{(E) }16.}$

Since $A$ is folded onto $O$, $AM = MO$ where $M$ is the intersection of $AO$ and the creaseline between $A$ and $O$. Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry. Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by $\left(\frac{OM}{ON}\right)^{2} = \left(\frac{\frac{OA}{2}}{OA\sin (\angle OAE)}\right)^{2} = \frac{1}{4\sin^{2}54}$ Remember that $\sin54 = \frac{1+\sqrt5}{4}$. $\cos54 = \sin36$ $4\cos^{3}18-3\cos18 = 2\sin18\cos18$ $4(1-\sin^{2}18)-3-2\sin18=0$ $4\sin^{2}18+2\sin18-1=0$ $\sin18 = \frac{-1+\sqrt5}{4}$ $\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$ $\sin^{2}54 =\frac{3+\sqrt5}{8}$ Let the inner pentagon be $Z$. $[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$ $= \frac{2(1+\sqrt5)}{3+\sqrt5}$ $= \sqrt5-1$ So the answer is $\boxed{B}$