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 |
|---|---|---|---|---|---|
4,901 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_3 | 2 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct. | 18 |
4,902 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | 1 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$ | 23 |
4,903 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | 2 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$ | 23 |
4,904 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | 3 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this v... | 23 |
4,905 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_5 | 1 | Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. ... | Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{4}$ more dollars than Doug. | 4 |
4,906 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | 1 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,... | 6 |
4,907 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | 2 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$... | 6 |
4,908 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | 3 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$ | 6 |
4,909 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_8 | 1 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the n... | 3 |
4,910 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_8 | 2 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] ... | 3 |
4,911 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_10 | 1 | For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
$\textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{... | 11 |
4,912 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_12 | 1 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring ... | The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\frac{18 \cdot 19}{2} + 2 = \boxed{173}$ | 173 |
4,913 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_12 | 2 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring ... | Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{173}$ | 173 |
4,914 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_14 | 1 | problem_id
ff8b260c6e48fc087b54f3971592eb5d Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us tha... | 30 |
4,915 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_14 | 2 | problem_id
ff8b260c6e48fc087b54f3971592eb5d Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. S... | 30 |
4,916 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_21 | 1 | Let $S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$ and $S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$
What is the ratio of the area of $S_2$ to the area of $S_1$
$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$ | Looking at the constraints of $S_1$
$x+y > 0$
$\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$
$1+x^2+y^2 \le 10x+10y$
$x^2-10x+y^2-10y \le -1$
$x^2-10x+25+y^2-10y+25 \le 49$
$(x-5)^2 + (y-5)^2 \le (7)^2$
$S_1$ is a circle wit... | 102 |
4,917 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | 1 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\q... | This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:
How many ways are there to choose $k$ elements from an ordered $n$ element set without choosing two consecutive members?
You want to choose $k$ numbers ou... | 405 |
4,918 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | 2 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\q... | Let $s$ be the numbers of elements in a subset. First we examine the second condition. No elements less than $s$ can be put in a subset of size $s$ , therefore the "lowest" element that can go into the subset is $s$ , whereas the "highest" element that can go into the subset is $15$ . This is a total of $15-s+1$ or $16... | 405 |
4,919 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | 3 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\q... | We will split the problem into cases, and maybe one could then generalize this to arbitrary $n$
$\bold{Case 1}:$ $n=15, k=1$ ). Then this is easy. We have $\binom{15}{1}=15$
$\bold{Case 2}:$ $n=14, k=2$ ). Now we have something tricky. To get a good grasp on this case, let us consider the smallest element; $2$ , in the... | 405 |
4,920 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_3 | 1 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$ | If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\Rightarrow \boxed{10}$ | 10 |
4,921 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_3 | 2 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$ | Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 \\ \rightarrow 2x+14=34 \\ \rightarrow 2x=20 \\ \rightarrow x = 10$
and the answer is $\boxed{10}$ | 10 |
4,922 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_6 | 1 | Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?
$\text {(A) } 129 \qquad \text {(B) } 137 \qq... | Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ by $3$... | 137 |
4,923 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_7 | 1 | Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?
$\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \te... | There are only two possible occupants for the driver's seat. After the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus, there are $2\cdot 3\cdot 2 = \boxed{12}$ possible seating arrangements. ~ aopsav (Credit to AoPS Alc... | 12 |
4,924 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | 1 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Let the integer have digits $a$ $b$ , and $c$ , read left to right. Because $1 \leq a<b<c$ , none of the digits can be zero and $c$ cannot be 2. If $c=4$ , then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$ , and for each choice there is one ac... | 34 |
4,925 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | 2 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.
If it was 3, there are 2 possibilities for th... | 34 |
4,926 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | 3 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | The last digit is 4, 6, or 8.
If the last digit is $x$ , the possibilities for the first two digits correspond to 2-element subsets of $\{1,2,\dots,x-1\}$
Thus the answer is ${3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}$ | 34 |
4,927 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | 5 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Casework:
For the sake of simplicity, we are going to call the number $\overline{abc}$
1. If $a=1$
a. $c=2$ . No such number exists.
b. $c=4$ $b=2, 3$
c. $c=6$ $b=2, 3, 4, 5$
d. $c=8$ $b=2, 3, 4, 5, 6, 7$
2. If $a=2$ : continue as above.
We can count up that there are 34 such integers, so select $\boxed{34}$ | 34 |
4,928 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_10 | 1 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$ | If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the fir... | 43 |
4,929 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_12 | 1 | The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $p\ne 0$ . What is $b$
$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$ | A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\ne 0$ , it follows that $a=-2/p$ . Thus \[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\boxed{4}$ | 4 |
4,930 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | 1 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathr... | Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies... | 121 |
4,931 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | 2 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathr... | $10$ moves results in a lot of possible endpoints, so we try small cases first.
If the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to.
If the object makes $2$ moves, it can move to $(0, 2)$ $(1, 1)$ $(2, 0)$ $(1, -1)$ $(0, -2)$ $(-1, -1)$ $(-2, 0)$ as well as... | 121 |
4,932 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | 1 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 ... | Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$
The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$ . The second one gives us $10^k \leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if a... | 16 |
4,933 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | 2 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 ... | The largest value for $x$ is $10^{0}$ . If $x > 10^{-1}$ , then $\lfloor\log_{10}4x\rfloor$ doesn't fulfill the condition unless $10^{-2} \leq x < 0.25 * 10^{-1}$ . The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n \leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.
Recognize ... | 16 |
4,934 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | 1 | Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 50... | The power of $10$ for any factorial is given by the well-known algorithm \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher power... | 492 |
4,935 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | 2 | Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 50... | Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides
it. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$
Consider $n! = 1\cdot 2 \cdot \dots \cdot n$ . Each fifth term is divisible by $5$ , each $25$ -th one
by $25$ , and so on. Hence the total po... | 492 |
4,936 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | 1 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); ... | [asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f)... | 127 |
4,937 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | 2 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); ... | Let point $P$ have coordinates $(x,y)$ and $C$ have coordinates $(0,0).$ Then, $A$ has $(s,0)$ and $B$ has $(0,s)$
By distance formula, we have \[x^2+y^2=36 \tag{1}.\] \[x^2+(y-s)^2=49 \tag{2}.\] \[(x-s)^2+y^2=121 \tag{3}\]
Expanding $(2)$ and $(3)$ gives \[x^2+y^2-2ys+s^2=49,\] and \[x^2+y^2-2sx+s^2=121,\] respectivel... | 127 |
4,938 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_25 | 1 | A sequence $a_1,a_2,\dots$ of non-negative integers is defined by the rule $a_{n+2}=|a_{n+1}-a_n|$ for $n\geq 1$ . If $a_1=999$ $a_2<999$ and $a_{2006}=1$ , how many different values of $a_2$ are possible?
$\mathrm{(A)}\ 165 \qquad \mathrm{(B)}\ 324 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 499 \qquad \mathrm{(E)}\... | We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.
Note that if $d = \gcd(999, a_2) \neq 1$ , then $d|a_n$ for all $n \geq 1$ , and $d$ does not divide $1$ , so if $\gcd(999, a_2) \neq 1$ , then $(a_n)$ does ... | 324 |
4,939 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_2 | 1 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$ | 4 |
4,940 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_13 | 1 | In the five-sided star shown, the letters $A$ $B$ $C$ $D$ and $E$ are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ form an
arithmetic sequen... | Let the terms in the arithmetic sequence be $a$ $a + d$ $a + 2d$ $a + 3d$ , and $a + 4d$ . We seek the middle term $a + 2d$
These five terms are $A + B$ $B + C$ $C + D$ $D + E$ , and $E + A$ , in some order. The numbers $A$ $B$ $C$ $D$ , and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 ... | 12 |
4,941 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_19 | 1 | A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 180... | Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$ ; only, in base $9$ , the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=... | 462 |
4,942 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_19 | 2 | A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 180... | This is very analogous to base $9$ . But, in base $9$ , we don't have a $9$ . So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$ $1463 - 1 = 1462$
Therefore, our answer is $\boxed{1462}$ | 462 |
4,943 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_22 | 1 | A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$ | Box P has dimensions $l$ $w$ , and $h$ .
Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]
The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh)... | 10 |
4,944 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_22 | 2 | A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$ | As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$ , and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$
Now, since this is competition math, we only need to find the space diagonal of any one box that fits the ... | 10 |
4,945 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_24 | 1 | Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$ | We can write the problem as
Since $\deg P(x) = 3$ and $\deg R(x) = 3$ $\deg P(x)\cdot R(x) = 6$ . Thus, $\deg P(Q(x)) = 6$ , so $\deg Q(x) = 2$
Hence, we conclude $Q(1)$ $Q(2)$ , and $Q(3)$ must each be $1$ $2$ , or $3$ . Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quad... | 22 |
4,946 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_24 | 2 | Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$ | We see that \[P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).\] Therefore, $P(x) | P(Q(x))$ . Since $\deg Q = 2,$ we must have $x-1, x-2, x-3$ divide $P(Q(x))$ . So, we pair them off with one of $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ to see that there are $3!+3 \cdot 2 \cdot \binom{3}{2} = 24$ witho... | 22 |
4,947 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_25 | 1 | Let $S$ be the set of all points with coordinates $(x,y,z)$ , where $x$ $y$ , and $z$ are each chosen from the set $\{0,1,2\}$ . How many equilateral triangles all have their vertices in $S$
$(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88$ | For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left.
First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex , and connecting the three adjacent vertices into a triangle.... | 80 |
4,948 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_1 | 1 | A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?
$\textbf{(A) }\ 100 \qquad \textbf{(B) }\ 200 \qquad \textbf{(C) }\ 300 \qquad \textbf{(D) }\ 400 \qquad \textbf{(E) }\ 500$ | \begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{100} Note: Revenue is a gain. | 100 |
4,949 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_2 | 1 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Since $x\%$ means $0.01x$ , the statement " $x\% \text{ of } x \text{ is 4}$ " can be rewritten as " $0.01x \cdot x = 4$ ":
$0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.$ | 20 |
4,950 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_2 | 2 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\boxed{20}.$ Very fast. | 20 |
4,951 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_3 | 1 | Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qq... | Let $m =$ Brianna's money. We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ . Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{25}$ | 25 |
4,952 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_4 | 1 | At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$
$\textbf{(A) }\ 1 \qquad... | Lisa's goal was to get an $A$ on $80\% \cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\boxed{2}$ of them. | 2 |
4,953 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | 1 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\sqrt{48}$ . Since $CD=8$ $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{3}$ | 3 |
4,954 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | 2 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one ... | 3 |
4,955 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | 3 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Let $BD=k$ . Then, by Stewart's Theorem
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$ | 3 |
4,956 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_7 | 1 | What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$ , then $a$ is either $b$ or $-b$ ):
\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}
We can then put these equations in slope-intercept form in order to graph them.
\begin{align*} ... | 24 |
4,957 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_7 | 2 | What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is
\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{24}\] | 24 |
4,958 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_8 | 1 | For how many values of $a$ is it true that the line $y = x + a$ passes through the
vertex of the parabola $y = x^2 + a^2$
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 10 \qquad \mathrm{(E)}\ \text{infinitely many}$ | We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$ . The y-intercept of the line $y = x + a$ is $(0,\,a)$ . We want to find the values (if any) such that $a=a^2$ . Solving for $a$ , the only values that satisfy this are $0$ and $1$ , so the answer is $\boxed{2}$ | 2 |
4,959 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_9 | 1 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^{\text... | 1 |
4,960 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_9 | 2 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | The remaining $30\%$ of the students got $95$ points.
The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$ .
The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$ , the median is $85$ . The di... | 1 |
4,961 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_10 | 1 | The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$ | Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ $3^{\text{rd}}$ , and $4^{\text{th}}$ t... | 250 |
4,962 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_12 | 1 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$ . Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \... | 8 |
4,963 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_12 | 2 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \[2a + 2b = -m\] \[a + b = -p\] \[2a(2b) = n\] \[a(b) = m\] Therefore, substituting the second equation into the first equation gives \[m = 2(p)\] and substituting the fourth equation ... | 8 |
4,964 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | 1 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | $221$ can be written as the sum of four two-digit numbers, let's say $\overline{ae}$ $\overline{bf}$ $\overline{cg}$ , and $\overline{dh}$ . Then $221= 10(a+b+c+d)+(e+f+g+h)$ . The last digit of $221$ is $1$ , and $10(a+b+c+d)$ won't affect the units digits, so $(e+f+g+h)$ must end with $1$ . The smallest value $(e+f+g... | 4 |
4,965 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | 2 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d$ , where $d$ is some digit. Clearly, $\boxed{4}$ | 4 |
4,966 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_17 | 1 | How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with
\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 17 \qquad \mathrm{(D)}\ 2004 \qquad \mathrm{(E)}\ \text{infinitely many}$ | Using the laws of logarithms , the given equation becomes
\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\] \[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\] \[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]
As $a,b,c,d$ must all be rational, and there are no pow... | 1 |
4,967 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_18 | 1 | Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 39 \qquad \mathrm{(C)}\ 51 \qquad \... | [asy] Label f; f.p=fontsize(6); xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0)); pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE); D(A--B); filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray); filldraw(CP(0.5(A+B),A),white); D(A); D(B); [/asy]
For angle $A$ and $B$ to be acute, $C$ must be between the two lines... | 51 |
4,968 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_20 | 1 | Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$ | The sum of the set is $-7-5-3-2+2+4+6+13=8$ , so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$ . Considering the set of four terms containing $13$ , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5... | 34 |
4,969 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_20 | 2 | Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$ | Trying out the values for a bit leads to us getting $34$ with $(-3, -2, 2, 6)$ in 1 set for a total of 9 and the other numbers giving a total of 25.
We start off with trying to get a $2(4^2) = 32$ solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).
We can add 7 to e... | 34 |
4,970 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_21 | 1 | A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$
$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$ | We may let $n = 7^k \cdot m$ , where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$ . Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$ . These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k ... | 2 |
4,971 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_24 | 1 | All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15... | Let the three points be at $A = (x_1, x_1^2)$ $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.
Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + ... | 14 |
4,972 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_24 | 2 | All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15... | Using the slope formula and differences of squares, we find:
$a+b$ = the slope of $AB$
$b+c$ = the slope of $BC$
$a+c$ = the slope of $AC$
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$ . Without loss of generality, let $AB$ be the side that has the smallest... | 14 |
4,973 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_1 | 1 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | $20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{29}$ | 29 |
4,974 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_1 | 2 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{29}$ | 29 |
4,975 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_3 | 1 | For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$
$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$ | If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (... | 49 |
4,976 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | 1 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$ | 26 |
4,977 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | 2 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$ | 26 |
4,978 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | 3 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters. Adding them together gives the answer of $\boxed{26}$ | 26 |
4,979 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | 1 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens,... | We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.
After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respecti... | 37 |
4,980 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | 2 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens,... | Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.
First round: $15,14,13$ (given)
Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$
We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So... | 37 |
4,981 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | 3 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens,... | Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either dec... | 37 |
4,982 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | 1 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ $[\triangle CDB]$ . So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$ . Therefore, $[\triangle ... | 4 |
4,983 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | 2 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | Since $AE \perp AB$ and $BC \perp AB$ $AE \parallel BC$ . By alternate interior angles and $AA\sim$ , we find that $\triangle ADE \sim \triangle CDB$ , with side length ratio $\frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \... | 4 |
4,984 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | 3 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | Let $[X]$ represent the area of figure $X$ . Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{4}$ | 4 |
4,985 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | 4 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$ .
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen ea... | 4 |
4,986 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_9 | 1 | A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } ... | When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$ | 36 |
4,987 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_11 | 1 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$... | 0 |
4,988 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_11 | 2 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using ... | 0 |
4,989 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 1 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and th... | 1 |
4,990 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 2 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ . Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first ... | 1 |
4,991 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 3 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplif... | 1 |
4,992 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 4 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$ | 1 |
4,993 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 5 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and ... | 1 |
4,994 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | 6 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$ | 1 |
4,995 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_15 | 1 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\... | Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $... | 350 |
4,996 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_15 | 2 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\... | The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meet... | 350 |
4,997 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_25 | 1 | For each integer $n\geq 4$ , let $a_n$ denote the base- $n$ number $0.\overline{133}_n$ . The product $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$ , where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$
$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text... | Note that \[0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},\] by geometric series.
Thus, we're aiming to find the value of \[\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.\] Expanding the product out, this is equivalent to \[\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ...... | 962 |
4,998 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_1 | 1 | At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm ... | Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \Rightarrow \mathrm{(A)}$
Because there are five days, or four transformations between days (day 1 $\right... | 3 |
4,999 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_2 | 1 | In the expression $c\cdot a^b-d$ , the values of $a$ $b$ $c$ , and $d$ are $0$ $1$ $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$ | If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ If $b=0$ , the expression evaluates to $c-d\leq 2$ Case $d=0$ remains.
In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$ . Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$ , where $c\cdot a^b=1\cdot 3^2=\boxed{... | 9 |
5,000 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_3 | 1 | If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$ | $1296 = 2^4 3^4$ and $4+4=\boxed{8}$ | 8 |
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