id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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2,401 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24 | 2 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem ... | 75 |
2,402 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24 | 3 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | The ratio of the $\overline{BG}$ to $\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\bigtriangleup ABG$ is $\frac{5}{5+1}\times120=100$ . Since $\overline{DE}$ is the midsegment... | 75 |
2,403 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24 | 4 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$ . Notice, $\overline{DE} || \overline{BC}$ , so $\bigtriangleup ABG \sim \bigtriangleup ADF$ , and since $\overline{AD}:\overline{AB}=1:2$ , the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$ . Given... | 75 |
2,404 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24 | 5 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\triangle ADE$ is equal to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ and the area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A$ . The rati... | 75 |
2,405 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_25 | 1 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | By geometric series, we have \begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\r... | 18 |
2,406 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_25 | 2 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | Immediately start trying $n = 1$ and $n = 2$ . These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$ ). These imply that $a^2 = 9c$ , so the possible $(a, c)$ pairs are $(9, 9)$ $(6, 4)$ , and $(3, 1)$ . The first puts $b$ out of range but the second make... | 18 |
2,407 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_25 | 3 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | The given equation can be written as \[c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.\] Divide by $\overbrace{11 \ldots 11}... | 18 |
2,408 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_25 | 4 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | By PaperMath’s sum , the answer is $6+8+4=\boxed{18}$ | 18 |
2,409 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_1 | 1 | Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$ | The area of the pan is $20\cdot18=360$ . Since the area of each piece is $2\cdot2=4$ , there are $\frac{360}{4} = \boxed{90}$ pieces. | 90 |
2,410 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_1 | 2 | Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$ | By dividing each of the dimensions by $2$ , we get a $10\times9$ grid that makes $\boxed{90}$ pieces. | 90 |
2,411 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_3 | 1 | In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?
$\textbf{(A) }2 \qquad \textbf{(B) }3\qqua... | We have $\binom{4}{2}$ ways to choose the pairs, and we have $2!$ ways for the values to be rearranged, hence $\frac{6}{2}=\boxed{3}$ | 3 |
2,412 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_3 | 3 | In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?
$\textbf{(A) }2 \qquad \textbf{(B) }3\qqua... | There are exactly $4!$ ways to arrange the numbers and $2!2!2!$ overcounts per way due to commutativity. Therefore, the answer is $\frac{4!}{2!2!2!}=\boxed{3}$ | 3 |
2,413 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_5 | 1 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$ | Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations.
$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$ . Using the answer choices, the only multiple of 15 is $\boxed{240}$ | 240 |
2,414 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_5 | 2 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$ | Subsets of $\{2,3,4,5,6,7,8,9\}$ include a single digit up to all eight numbers. Therefore, we must add the combinations of all possible subsets and subtract from each of the subsets formed by the composite numbers.
Hence:
$\binom{8}{1} - \binom{4}{1} + \binom{8}{2} - \binom{4}{2} + \binom{8}{3} - \binom{4}{3} + \binom... | 240 |
2,415 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_5 | 3 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$ | Total subsets is $(2^8) = 256$ Using complementary counting and finding the sets with composite numbers:
only 4,6,8 and 9 are composite. Each one can be either in the set or out: $2^4$ = 16 $256-16=240$ $\boxed{240}$ | 240 |
2,416 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_5 | 4 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$ | We multiply the number of possibilities of the set having prime numbers and the set having composites.
The possibilities of primes are $2^4-1=15$ (As there is one solution not containing any primes)
The possibilities of the set containing composites are $2^4=16$ (There can be a set with no composites)
Multiplying this ... | 240 |
2,417 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_7 | 1 | In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. T... | Use the answer choices and calculate them. The one that works is $\bold{\boxed{19}$ | 19 |
2,418 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_7 | 2 | In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. T... | Let the number of semicircles be $n$ and let the radius of each semicircle be $r$ . To find the total area of all of the small semicircles, we have $n \cdot \frac{\pi \cdot r^2}{2}$
Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be $n \cdot r$ . So, the are... | 19 |
2,419 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_7 | 3 | In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. T... | Each small semicircle is $\frac{1}{N^2}$ of the large semicircle. Since $N$ small semicircles make $\frac{1}{19}$ of the large one, $\frac{N}{N^2} = \frac1{19}$ . Solving this, we get $\boxed{19}$ . ... | 19 |
2,420 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_8 | 1 | Sara makes a staircase out of toothpicks as shown:
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } } [/asy]
This is a 3-step stairca... | A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n + 2$ toothpicks. This can be rewritten as $(n+1)(n+2) -2$
So, $(n+1)(n+2) - 2 = 180$
So, $(n+1)(n+2) = 182.$
Inspection could tell us that $13 \cdot 14 = 182$ , so the answer is $13 - 1 = \boxed{12}$ | 12 |
2,421 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_8 | 2 | Sara makes a staircase out of toothpicks as shown:
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } } [/asy]
This is a 3-step stairca... | Layer $1$ $4$ steps
Layer $1,2$ $10$ steps
Layer $1,2,3$ $18$ steps
Layer $1,2,3,4$ $28$ steps
From inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by $2$ . Using this pattern:
$4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180$
F... | 12 |
2,422 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_8 | 3 | Sara makes a staircase out of toothpicks as shown:
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } } [/asy]
This is a 3-step stairca... | We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be $2$ and the leading coefficient is $1$ . The function is $f(n)=n^2+3n$ where $n$ is the layer and $f(n)$ is the number of toothpicks.
We have to solve for $n$ when $n^2+3n=180\Rightarr... | 12 |
2,423 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | 1 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the n... | 39 |
2,424 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | 2 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | Let's call the unknown value $x$ . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and $x$ . So,
$10 - 7 = 42- x$
$x = 39$ and our answer is $\boxed{39}$ By: Soccer_JAMS | 39 |
2,425 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | 3 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | For the sums to have equal probability, the average sum of both sets of $7$ dies has to be $(6+1)\cdot 7 = 49$ . Since having $10$ is similar to not having $10$ , you just subtract 10 from the expected total sum. $49 - 10 = 39$ so the answer is $\boxed{39}$ | 39 |
2,426 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | 4 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | The expected value of the sums of the die rolls is $3.5\cdot7=24.5$ , and since the probabilities should be distributed symmetrically on both sides of $24.5$ , the answer is $24.5+(24.5-10)=39$ , which is $\boxed{39}$ | 39 |
2,427 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | 5 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | Another faster and easier way of doing this, without using almost any math at all, is realizing that the possible sums are ${7,8,9,10,...,39,40,41,42}$ . By symmetry, (and doing a few similar problems in the past), you can realize that the probability of obtaining $7$ is the same as the probability of obtaining $42$ $P... | 39 |
2,428 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | 2 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is $\frac{1}{3}Bh$ , with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is $\frac{1}{3} (3 \times 2/2 \... | 2 |
2,429 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | 3 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | You can calculate the volume of the rectangular pyramid by using the formula, $\frac{Ah}{3}$ $A$ is the area of the base, $BCHE$ , and is equal to $BC * BE$ . The height, $h$ , is equal to the height of triangle $FBE$ drawn from $F$ to $BE$
$BE=\sqrt{BF^2 + EF^2}=\sqrt{13}$ Area of $BCHE = BC * BE = \sqrt{13}$
$h = 2 *... | 2 |
2,430 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | 4 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | We can start by identifying the information we need. We need to find the area of rectangle $EHCB$ and the height of rectangular prism $EHCBM$
In order to find the area of $EHCB,$ we can use the Pythagorean Theorem. We find that $EB = \sqrt{13}$ , so the area of rectangle $EHCB = \sqrt{13}$ . We shall refer to this as $... | 2 |
2,431 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | 5 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | We start by setting the formula for the volume of a rectangular pyramid: $\frac{1}{3}Bh$ . By the Pythagorean Theorem, we know that $BE = \sqrt{BF^2 + EF^2} = \sqrt{13}$ . Therefore, the area of the base is $1 \times \sqrt{13} = \sqrt{13}$ . Next, we would like to know the height of the pyramid. We can observe that the... | 2 |
2,432 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | 6 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | AMC 10B 10 2018.jpg
Using the Pythagorean Theorem, we can easily find that $EB = \sqrt{2^2 + 3^2} = \sqrt{13}$ . Quickly computing, we find the area of the base, $BCHE = \sqrt{13} \cdot 1 = \sqrt{13}$ . Now we can make the following adjustments to our 3d shape as shown in the diagram. All we need now is to solve for th... | 2 |
2,433 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_12 | 1 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the ... | For each $\triangle ABC,$ note that the length of one median is $OC=12.$ Let $G$ be the centroid of $\triangle ABC.$ It follows that $OG=\frac13 OC=4.$
As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively: [asy] /* Made by MRENTHUSIASM */... | 50 |
2,434 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_12 | 2 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the ... | We assign coordinates. Let $A = (-12,0)$ $B = (12,0)$ , and $C = (x,y)$ lie on the circle $x^2 +y^2 = 12^2$ . Then, the centroid of $\triangle ABC$ is $G = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)$ . Thus, $G$ traces out a circle with a radius $\frac13$ of the radius of ... | 50 |
2,435 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_12 | 3 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the ... | First we can draw a few conclusions from the given information. Firstly we can see clearly that the distance from the centroid to the center of the circle will remain the same no matter $C$ is on the circle. Also we can see that because the two legs of every triangles will always originate on the diameter, using inscri... | 50 |
2,436 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_13 | 1 | How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$ | The number $10^n+1$ is divisible by 101 if and only if $10^n\equiv -1\pmod{101}$ . We note that $(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}$ , so the powers of 10 are 4-periodic mod 101.
It follows that $10^n\equiv -1\pmod{101}$ if and only if $n\equiv 2\pmod 4$
In the given list, $10^2+1,10^3+1,10^4+1,\dots,10^... | 505 |
2,437 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_13 | 2 | How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$ | Note that $10^{2k}+1$ for some odd $k$ will suffice $\mod {101}$ . Each $2k \in \{2,6,10,\dots,2018\}$ , so the answer is $\boxed{505}$ | 505 |
2,438 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_13 | 3 | How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$ | If we divide each number by $101$ , we see a pattern occuring in every 4 numbers. $101, 1000001, 10000000001, \dots$ . We divide $2018$ by $4$ to get $504$ with $2$ left over. Looking at our pattern of four numbers from above, the first number is divisible by $101$ . This means that the first of the $2$ left over will ... | 505 |
2,439 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_13 | 4 | How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$ | Note that $909$ is divisible by $101$ , and thus $9999$ is too. We know that $101$ is divisible and $1001$ isn't so let us start from $10001$ . We subtract $9999$ to get 2. Likewise from $100001$ we subtract, but we instead subtract $9999$ times $10$ or $99990$ to get $11$ . We do it again and multiply the 9's by $10$ ... | 505 |
2,440 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_13 | 5 | How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$ | Note that $101=x^2+1$ and $100...0001=x^n+1$ , where $x=10$ . We have that $\frac{x^n+1}{x^2+1}$ must have a remainder of $0$ . By the remainder theorem, the roots of $x^2+1$ must also be roots of $x^n+1$ . Plugging in $i,-i$ to $x^n+1$ yields that $n\equiv2\mod{4}$ . Because the sequence starts with $10^2+1$ , the ans... | 505 |
2,441 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_14 | 1 | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \boxed{225}.$ | 225 |
2,442 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_14 | 2 | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | As in Solution 1, we want to maximize the number of time each number appears to do so. We can set up an equation $10 + 9( x - 1 )\geq2018,$ where $x$ is the number of values. Notice how we can then rearrange the equation into $1 + 9 ( 1 )+9 ( x - 1 )\geq2018,$ which becomes $9 x\geq2017,$ or $x\geq224\frac19.$ We canno... | 225 |
2,443 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | 1 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Verify that $a^3 \equiv a \pmod{6}$ manually for all $a\in \mathbb{Z}/6\mathbb{Z}$ . We check: $0^3 \equiv 0 \pmod{6}$ $1^3 \equiv 1 \pmod{6}$ $2^3 \equiv 8 \equiv 2 \pmod{6}$ $3^3 \equiv 27 \equiv 3 \pmod{6}$ $4^3 \equiv 64 \equiv 4 \pmod{6}$ , and $5^3 \equiv 125 \equiv 5 \pmod{6}$ . We conclude that $a^3 \equiv a \p... | 4 |
2,444 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | 2 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2\\ +\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k$
Note that $a_1^3+a_2^3+\cdots+a_{2018}^... | 4 |
2,445 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | 3 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, we will assume that $a_1 = 1$
We first note that $1^3+2^3+...+n^3 = (1+2+...+n)^2$ . So what we are trying to find is what $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6... | 4 |
2,446 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | 4 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} \pmod{6}$ (which happens to be $4$ ). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 \pmod{6}$ or jus... | 4 |
2,447 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | 5 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just $2018^{2018} \pmod{6}$ , which is $\boxed{4}$ | 4 |
2,448 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_17 | 3 | In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octag... | Let the octagon's side length be $x$ . Then $PH = \frac{6 - x}{2}$ and $PA = \frac{8 - x}{2}$ . By the Pythagorean theorem, $PH^2 + PA^2 = HA^2$ , so $\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2$ . By expanding the left side and combining the like terms, we get $\frac{x^2}{2} - 7x + 25 = x^... | 7 |
2,449 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_17 | 4 | In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octag... | Let $AB$ , or the side of the octagon, be $x$ . Then, $BQ = \left(\frac{8-x}{2}\right)$ and $CQ = \left(\frac{6-x}{2}\right)$ . By the Pythagorean Theorem $BQ^2+CQ^2=x^2$ , or $\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2$ . Multiplying this out, we have $x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}$ . Simp... | 7 |
2,450 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 1 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.
Case $1$ :
Second Row: a b c
Third Row: b c a
Case $2$ :
Second Row: a c b
Third Row: c b a
Case $3$ :
Second Row: a... | 96 |
2,451 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 2 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | Lets call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$ . We can split our problem into two cases:
There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.
Starting off with the first case, we see that there are $3!=6$ ways to arrange... | 96 |
2,452 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 3 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | Call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it ( $\circ$ denotes an empty seat):
\[A_1 \circ \circ\] \[\circ \ \circ \ \circ\]
Then there are 4 choices fo... | 96 |
2,453 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 4 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | WLOG, define the three pairs of siblings to be: $A$ $B$ , and $C$ . Now, notice that you can only form a correct grouping either like this:
\[A B C\]
\[B C A\]
or this:
\[C B A\]
\[A C B\]
However, we need to consider the different orders. There are $3!$ ways to order each pair (eg. the same letters) and $2^3$ ways to ... | 96 |
2,454 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 5 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | Let the families be $A$ $B$ $C$ . In any given possible arrangement, there are $3! = 6$ ways to arrange the families and $2 \cdot 2 \cdot 2 = 8$ ways to arrange the siblings. This means the answer has to be divisble by $6 \cdot 8 = 48$ . The only answer choice that satisfies this is $\boxed{96}$ | 96 |
2,455 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | 6 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | If a pair of siblings are in the same row, the other $2$ pairs of siblings cannot fit in the remaining $4$ seats to meet the requirements. Therefore, the siblings must be in different rows.
Let the first pair of siblings be $a_1$ $a_2$ , the second pair of siblings be $b_1$ $b_2$ , and the third pair of siblings be $c_... | 96 |
2,456 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19 | 1 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime ... | 11 |
2,457 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19 | 2 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solut... | 11 |
2,458 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19 | 3 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zo... | 11 |
2,459 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19 | 4 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$... | 11 |
2,460 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 1 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | For all integers $n \geq 7,$ note that \begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ &=-f(n-3)+2n-1 \\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ &=-f(n-4)+f(n-5)+n+2 \\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ &=f(n-6)+6. \end{align*} It follows that \begin{align*} f(2018)&=f(2012)+6 \\ &=f(2006)+12 \\ &=f... | 17 |
2,461 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 2 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | For all integers $n\geq3,$ we rearrange the given equation: \[f(n)-f(n-1)+f(n-2)=n. \hspace{28.25mm}(1)\] For all integers $n\geq4,$ it follows that \[f(n-1)-f(n-2)+f(n-3)=n-1. \hspace{15mm}(2)\] For all integers $n\geq4,$ we add $(1)$ and $(2):$ \[f(n)+f(n-3)=2n-1. \hspace{38.625mm}(3)\] For all integers $n\geq7,$ it ... | 17 |
2,462 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 3 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | Preamble: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$ th term of the sequence $A.$ This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more m... | 17 |
2,463 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 4 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | Start out by listing some terms of the sequence. \begin{align*} f(1)&=1 \\ f(2)&=1 \\ f(3)&=3 \\ f(4)&=6 \\ f(5)&=8 \\ f(6)&=8 \\ f(7)&=7 \\ f(8)&=7 \\ f(9)&=9 \\ f(10)&=12 \\ f(11)&=14 \\ f(12)&=14 \\ f(13)&=13 \\ f(14)&=13 \\ f(15)&=15 \\ & \ \vdots \end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple o... | 17 |
2,464 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 5 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | Writing out the first few values, we get \[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\ldots.\] We see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\pmod{6}$ and less than $2018$ is $2017,$ so we have $f(2017)=f(2018)=\boxed{20... | 17 |
2,465 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 6 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | \begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ f(n-1)&=f(n-2)-f(n-3)+n-1 \end{align*} Subtracting those two and rearranging gives \begin{align*} f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\ f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 \end{align*} Subtracting those two gives $f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.$
The characteristic polynomial is ... | 17 |
2,466 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | 7 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | We utilize patterns to solve this equation: \begin{align*} f(3)&=3, \\ f(4)&=6, \\ f(5)&=8, \\ f(6)&=8, \\ f(7)&=7, \\ f(8)&=8. \end{align*} We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now... | 17 |
2,467 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21 | 3 | Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \t... | The prime factorization of $323$ is $17 \cdot 19$ . Our answer must be a multiple of either $17$ or $19$ or both. Since $17 < 19$ , the next smallest divisor that is divisble by $17$ would be $323 + 17 = \boxed{340}$ | 340 |
2,468 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21 | 5 | Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \t... | Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$ -digit $n$ . So, we need a choice that shares a factor(s) with $323$ , such that the factors we'll need to add to the prime factorization of $n$ (in result to adding the chosen divisor) won't cause our number to multiply to more than $4$ di... | 340 |
2,469 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 1 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visual... | 199 |
2,470 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 2 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Same as the first solution, $x^2=10,000\{x\}$
We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in $\{x\}$ in terms of $\lfloor x \rfloor$ \[\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0\]
We use the quadratic formula to solve for $\{x\}$ \[\{x\} = \frac {-... | 199 |
2,471 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 3 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$ .
We can then rewrite the problem below:
$(a+k)^2 + 10000a = 10000(a+k)$
From here, we get
$(a+k)^2 + 10000a = 10000a + 10000k$
Solving for $a+k = x$
$(a+k)^2 = 10000k$
$x = a+k = \pm100\sqrt{k}$
Because $0 \leq k < 1$ , we know th... | 199 |
2,472 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 4 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Notice the given equation is equivalent to $(\lfloor x \rfloor+\{x\})^2=10,000\{x\}$
Now we know that $\{x\} < 1$ so plugging in $1$ for $\{x\}$ we can find the upper and lower bounds for the values.
$(\lfloor x \rfloor +1)^2 = 10,000(1)$
$(\lfloor x \rfloor +1) = \pm 100$
$\lfloor x \rfloor = 99, -101$
And just like $... | 199 |
2,473 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 5 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Firstly, if $x$ is an integer, then $10,000\lfloor x \rfloor=10,000x$ , so $x$ must be $0$
If $0<x<1$ , then we know the following:
$0<x^2<1$
$10,000\lfloor x \rfloor =0$
$0<10,000x<10,000$
Therefore, $0<x^2+10,000\lfloor x \rfloor <1$ , which overlaps with $0<10,000x<10,000$ . This means that there is at least one rea... | 199 |
2,474 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 6 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | General solution to this type of equation $f(x, \lfloor x \rfloor) = 0$
$x^2 - 10000x + 10000 \lfloor x \rfloor =0$
$x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}$ $\lfloor x \rfloor \le 2500$
$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$
If $x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}$ $x \ge 5000$ , it contradi... | 199 |
2,475 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | 7 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Subtracting $10000\lfloor x\rfloor$ from both sides gives $x^2=10000(x-\lfloor x\rfloor)=10000\{x\}$ . Dividing both sides by $10000$ gives $\left(\frac{x}{100}\right)^2=\{x\}<1$ $\left(\frac{x}{100}\right)^2<1$ when $-100<x<100$ so the answer is $\boxed{199}$ | 199 |
2,476 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 1 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: \[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{127}\] | 127 |
2,477 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 2 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$ . This is always $1$ less than a power of $2$ . The only admissible answer choice by this rule is thus $\boxed{127}$ | 127 |
2,478 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 3 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{127}$ | 127 |
2,479 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 4 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | If you distribute this you get a sum of the powers of $2$ . The largest power of $2$ in the series is $64$ , so the sum is $\boxed{127}$ | 127 |
2,480 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 5 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ $=(2(2(2(2(2(3)+1)+1)+1)+1)+1)$ $=(2(2(2(2(6+1)+1)+1)+1)+1)$ $=(2(2(2(2(7)+1)+1)+1)+1)$ $=(2(2(2(14+1)+1)+1)+1)$ $=(2(2(2(15)+1)+1)+1)$ $=(2(2(30+1)+1)+1)$ $=(2(2(31)+1)+1)$ $=(2(62+1)+1)$ $=(2(63)+1)$ $=(126+1)$ $=127 \Longrightarrow \boxed{127}$ | 127 |
2,481 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | 6 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | Notice that $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1$ . Substituting $2$ for $x$ , we get \[2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \Longrightarrow \boxed{127}\] | 127 |
2,482 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_2 | 1 | Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$ -popsicle boxes for $$2$ each, and $5$ -popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textb... | $$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$ , we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ popsicles... | 13 |
2,483 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_3 | 1 | Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
[asy] draw((0,0)--(0,10)--(15,10)--(15,0)--cycle); fill((0,0)--(0,10)--(15,10)--(15,0)--c... | Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.
Since the width of Tamara's garden contains three margins, the total width is $2\cdot 6+3\cdot 1 = 15$ feet.
Similarly, the height of Tamara's garden is $3... | 78 |
2,484 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_4 | 1 | Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys... | Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $... | 14 |
2,485 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5 | 1 | The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\frac{x}{xy}+\frac{y}{xy}=4.$
\[\frac{1}{y}+\frac{1}{x}=\boxed{4}.\] | 4 |
2,486 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5 | 2 | The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.
See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{4}.$ | 4 |
2,487 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5 | 3 | The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | Notice that from the information given above, $x+y=4xy$
Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$
We can solve this by substituting $x+y\implies 4xy$
Our answer is simply $\frac{4xy}{xy}\implies4$
Therefore, the answer is... | 4 |
2,488 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_7 | 1 | Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to... | Let $j$ represent how far Jerry walked, and $s$ represent how far Silvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since S... | 30 |
2,489 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8 | 1 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\... | Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\cdot20 = 200$ . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$ , or $\binom{10}{2}... | 245 |
2,490 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8 | 2 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\... | We can also use complementary counting. First of all, $\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\dbinom{20}{2}=190$ hugs. $435-190= \boxed{... | 245 |
2,491 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8 | 3 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\... | We can focus on how many handshakes the $10$ people who don't know anybody get.
The first person gets $29$ handshakes with other people not him/herself, the second person gets $28$ handshakes with other people not him/herself and not the first person, ..., and the tenth receives $20$ handshakes with other people not hi... | 245 |
2,492 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8 | 4 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\... | First, we can find out the number of handshakes that the $10$ people who don't know anybody share with the $20$ other people. This is simply $10 \cdot 20 = 200$ . Next, we need to find out the number of handshakes that are shared within the $10$ people who don't know anybody. Here, we can use the formula $\frac{n(n-1)}... | 245 |
2,493 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_10 | 1 | Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose... | The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{17}$ | 17 |
2,494 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_11 | 1 | The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$ | In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within $3$ units of a segment. I... | 20 |
2,495 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_11 | 2 | The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$ | Because this is just a cylinder and $2$ hemispheres ("half spheres"), and the radius is $3$ , the volume of the $2$ hemispheres is $\frac{4(3^3)\pi}{3} = 36 \pi$ . Since we also know that the volume of this whole thing is $216 \pi$ , we do $216-36$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180... | 20 |
2,496 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_13 | 1 | Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf... | A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\boxed{9}$ | 9 |
2,497 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14 | 1 | Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To ... | Let $m$ = cost of movie ticket Let $s$ = cost of soda
We can create two equations:
\[m = \frac{1}{5}(A - s)\] \[s = \frac{1}{20}(A - m)\]
Substituting we get:
\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\] which yields: \[m = \frac{19}{99}A\]
Now we can find s and we get:
\[s = \frac{4}{99}A\]
Since we want to find what... | 23 |
2,498 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14 | 2 | Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To ... | We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$ .
Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$ .
Adding, we get $\frac{2300}{99}$ , which is closest to $23$ which is ... | 23 |
2,499 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14 | 3 | Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To ... | Let $m$ be the price of a movie ticket and $s$ be the price of a soda.
Then,
\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]
Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$ . Assume WLOG that $s=4$ $m=19$ , thus making a solution for this equati... | 23 |
2,500 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16 | 1 | There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start runn... | If we have horses, $a_1, a_2, \ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LC... | 3 |
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