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2002AMC10B真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月5日下午4:08

2002AMC10B真题

答案解析请参考文末

Problem 1

The ratio $frac{2^{2001}cdot3^{2003}}{6^{2002}}$ is:

$mathrm{(A) } frac{1}{6}qquad mathrm{(B) } frac{1}{3}qquad mathrm{(C) } frac{1}{2}qquad mathrm{(D) } frac{2}{3}qquad mathrm{(E) } frac{3}{2}$

Problem 2

For the nonzero numbers $a, b,$ and $c,$ define $[(a,b,c)=frac{abc}{a+b+c}]$ Find $(2,4,6)$ .

$mathrm{(A) } 1qquad mathrm{(B) } 2qquad mathrm{(C) } 4qquad mathrm{(D) } 6qquad mathrm{(E) } 24$

Problem 3

The arithmetic mean of the nine numbers in the set ${9,99,999,9999,ldots,999999999}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ does not contain the digit

$mathrm{(A) } 0qquad mathrm{(B) } 2qquad mathrm{(C) } 4qquad mathrm{(D) } 6qquad mathrm{(E) } 8$

Problem 4

What is the value of

$[(3x-2)(4x+1)-(3x-2)4x+1]$

when $x=4$ ?

$mathrm{(A) } 0qquad mathrm{(B) } 1qquad mathrm{(C) } 10qquad mathrm{(D) } 11qquad mathrm{(E) } 12$

Problem 5

Circles of radius $2$ and $3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.

$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r1=3; real r2=2; real r3=5; pair A=(-2,0), B=(3,0), C=(0,0); pair X=(1,0), Y=(5,0); path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r3); fill(circleC,gray); fill(circleA,white); fill(circleB,white); draw(circleA); draw(circleB); draw(circleC); draw(A--X); draw(B--Y); pair[] ps={A,B}; dot(ps); label("$3$",midpoint(A--X),N); label("$2$",midpoint(B--Y),N); [/asy]$ $mathrm{(A) } 3piqquad mathrm{(B) } 4piqquad mathrm{(C) } 6piqquad mathrm{(D) } 9piqquad mathrm{(E) } 12pi$

Problem 6

For how many positive integers $n$ is $n^2-3n+2$ a prime number?

$mathrm{(A) } text{none}qquad mathrm{(B) } text{one}qquad mathrm{(C) } text{two}qquad mathrm{(D) } text{more than two, but finitely many}qquad mathrm{(E) } text{infinitely many}$

Problem 7

Let $n$ be a positive integer such that $frac{1}{2}+frac{1}{3}+frac{1}{7}+frac{1}{n}$ is an integer. Which of the following statements is not true?

$mathrm{(A) } 2text{ divides }nqquad mathrm{(B) } 3text{ divides }nqquad mathrm{(C) } 6text{ divides }nqquad mathrm{(D) } 7text{ divides }nqquad mathrm{(E) } n>84$

Problem 8

Suppose July of year $N$ has five Mondays. Which of the following must occur five times in the August of year $N$ ? (Note: Both months have $31$ days.)

$textbf{(A)} text{Monday} qquad textbf{(B)} text{Tuesday} qquad textbf{(C)} text{Wednesday} qquad textbf{(D)} text{Thursday} qquad textbf{(E)} text{Friday}$

Problem 9

Using the letters $A$ , $M$ , $O$ , $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position

$mathrm{(A) } 112qquad mathrm{(B) } 113qquad mathrm{(C) } 114qquad mathrm{(D) } 115qquad mathrm{(E) } 116$

Problem 10

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2+ax+b=0$ has solutions $a$ and $b$ . what is the pair $(a,b)$ ?

$mathrm{(A) } (-2,1)qquad mathrm{(B) } (-1,2)qquad mathrm{(C) } (1,-2)qquad mathrm{(D) } (2,-1)qquad mathrm{(E) } (4,4)$

Problem 11

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$mathrm{(A) } 50qquad mathrm{(B) } 77qquad mathrm{(C) } 110qquad mathrm{(D) } 149qquad mathrm{(E) } 194$

Problem 12

For which of the following values of $k$ does the equation $frac{x-1}{x-2} = frac{x-k}{x-6}$ have no solution for $x$ ?

$textbf{(A) } 1qquad textbf{(B) } 2qquad textbf{(C) } 3qquad textbf{(D) } 4qquad textbf{(E) } 5$

Problem 13

Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$ .

$textbf{(A) } frac23 qquad textbf{(B) } frac32 text{ or } -frac14 qquad textbf{(C) } -frac23 text{ or } -frac14 qquad textbf{(D) } frac32 qquad textbf{(E) } -frac32 text{ or } -frac14$

Problem 14

The number $25^{64}cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is

$mathrm{(A) } 7qquad mathrm{(B) } 14qquad mathrm{(C) } 21qquad mathrm{(D) } 28qquad mathrm{(E) } 35$

Problem 15

The positive integers $A$ , $B$ , $A-B$ , and $A+B$ are all prime numbers. The sum of these four primes is

$mathrm{(A) } text{even}qquad mathrm{(B) } text{divisible by }3qquad mathrm{(C) } text{divisible by }5qquad mathrm{(D) } text{divisible by }7qquad mathrm{(E) } text{prime}$

Problem 16

For how many integers $n$ is $frac{n}{20-n}$ the square of an integer?

$textbf{(A) } 1qquad textbf{(B) } 2qquad textbf{(C) } 3qquad textbf{(D) } 4qquad textbf{(E) } 10$

Problem 17

A regular octagon $ABCDEFGH$ has sides of length two. Find the area of $triangle ADG$ .

$textbf{(A) } 4 + 2sqrt2 qquad textbf{(B) } 6 + sqrt2qquad textbf{(C) } 4 + 3sqrt2 qquad textbf{(D) } 3 + 4sqrt2 qquad textbf{(E) } 8 + sqrt2$

Problem 18

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$textbf{(A) } 8qquad textbf{(B) } 9qquad textbf{(C) } 10qquad textbf{(D) } 12qquad textbf{(E) } 16$

Problem 19

Suppose that ${a_n}$ is an arithmetic sequence with $[a_1+a_2+cdots+a_{100}=100 text{ and } a_{101}+a_{102}+cdots+a_{200}=200.]$ What is the value of $a_2 - a_1 ?$

$mathrm{(A) } 0.0001qquad mathrm{(B) } 0.001qquad mathrm{(C) } 0.01qquad mathrm{(D) } 0.1qquad mathrm{(E) } 1$

Problem 20

Let $a, b,$ and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7.$ Then $a^2-b^2+c^2$ is

$mathrm{(A) } 0qquad mathrm{(B) } 1qquad mathrm{(C) } 4qquad mathrm{(D) } 7qquad mathrm{(E) } 8$

Problem 21

Andy's lawn has twice as much area as Beth's lawn and three times as much as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?

$mathrm{(A) } text{Andy}qquad mathrm{(B) } text{Beth}qquad mathrm{(C) } text{Carlos}qquad mathrm{(D) } text{Andy and Carlos tie for first.}qquad mathrm{(E) } text{All three tie.}$

Problem 22

Let $triangle{XOY}$ be a right-angled triangle with $mangle{XOY}=90^circ$ . Let $M$ and $N$ be the midpoints of the legs $OX$ and $OY$ , respectively. Given $XN=19$ and $YM=22$ , find $XY$ .

$mathrm{(A) } 24qquad mathrm{(B) } 26qquad mathrm{(C) } 28qquad mathrm{(D) } 30qquad mathrm{(E) } 32$

Problem 23

Let ${a_k}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is

$mathrm{(A) } 45qquad mathrm{(B) } 56qquad mathrm{(C) } 67qquad mathrm{(D) } 78qquad mathrm{(E) } 89$

Problem 24

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?

$mathrm{(A) } 5qquad mathrm{(B) } 6qquad mathrm{(C) } 7.5qquad mathrm{(D) } 10qquad mathrm{(E) } 15$

Problem 25

When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?

$mathrm{(A) } 4qquad mathrm{(B) } 5qquad mathrm{(C) } 6qquad mathrm{(D) } 7qquad mathrm{(E) } 8$

2002AMC10B详细解析

$frac{2^{2001}cdot3^{2003}}{6^{2002}}=frac{6^{2001}cdot 3^2}{6^{2002}}=frac{9}{6}=frac{3}{2}$ or $mathrm{ (E) }$
$frac{2cdot 4cdot 6}{2+4+6}=frac{48}{12}=4Longrightarrowmathrm{ (C) }$
We wish to find $frac{9+99+cdots +999999999}{9}$ , or $frac{9(1+11+111+cdots +111111111)}{9}=123456789$ . This does not have the digit 0, so the answer is $boxed{mathrm{(A)} 0}$ Notice that the final number is guaranteed to have the digits ${1, 3, 5, 7, 9}$ and that each of these digits can be paired with an even number adding up to 9. $boxed{mathrm{(A)} 0}$ can be taken out, with the other digits fulfilling divisibility by 9.
$(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = boxed{mathrm{(D)} 11}$ .
A line going through the centers of the two smaller circles also go through the diameter. The length of this line within the circle is $3+3+2+2=10.$ Because this is the length of the larger circle's diameter, the length of its radius is $5.$ The area of the large circle is $25pi$ , and the area of the two smaller circles is $9pi + 4pi = 13pi.$ To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. $longrightarrow 25pi - 13pi = boxed{mathrm{(E) } 12pi}$
Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$ . Either $n-1$ or $n-2$ is odd, and the other is even. Their product must yield an even number. The only prime that is even is $2$ , which is when $n$ is $3$ . The answer is $boxed{mathrm{(B)} text{one}}$ .Considering parity, we see that $n^2 - 3n + 2$ is always even. The only even prime is $2$ , and so $n^2-3n=0$ whence $n=3Rightarrowboxed{mathrm{(B)} text{one}}$ .
Since $frac 12 + frac 13 + frac 17 = frac {41}{42}$ , $[0 < lim_{n rightarrow infty} left(frac{41}{42} + frac{1}{n}right) < frac {41}{42} + frac 1n < frac{41}{42} + frac 11 < 2]$ From which it follows that $frac{41}{42} + frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $boxed{mathrm{(E)} n>84}$ .
If there are five Mondays, there are only three possibilities for their dates: $(1,8,15,22,29)$ , $(2,9,16,23,30)$ , and $(3,10,17,24,31)$ .In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August.The only day of the week that is guaranteed to appear five times is therefore $boxed{textrm{(D)} text{Thursday}}$ .
There are $4!cdot 4$ "words" beginning with each of the first four letters alphabetically. From there, there are $3!cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next "word" is $USAMO$ , hence our answer is $4cdot 4!+3cdot 3!+1=boxed{115Rightarrowtext{(D)}}$ .Let $A = 1, ldots, U = 5$ . Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$ , so the answer is $boxed{115Rightarrowtext{(D)}}$
Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$ , it follows by comparing coefficients that $-a - b = a$ and that $ab = b$ . Since $b$ is nonzero, $a = 1$ , and $-1 - b = 1 Longrightarrow b = -2$ . Thus $(a,b) = boxed{mathrm{(C)} (1,-2)}$ .Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the $x$ coefficient, since the leading coefficient is 1; in other words, $a + b = -a$ and the product of the solutions is equal to the constant term (i.e, $a*b = b$ ). Since $b$ is nonzero, it follows that $a = 1$ and therefore (from the first equation), $b = -2a = -2$ . Hence, $(a,b) = boxed{mathrm{(C)} (1,-2)}$ Note that for roots $a$ and $b$ , $ab = b$ . This implies that $a$ is $1$ , and there is only one answer choice with $1$ in the position for $a$ , hence, $(a,b) = boxed{mathrm{(C)} (1,-2)}$
Let the three consecutive positive integers be $a-1$ , $a$ , and $a+1$ . So, $a(a-1)(a+1)=24aRightarrow(a-1)(a+1)=24$ . $24=4times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=boxed{mathrm{ (B)} 77}$ .
The domain over which we solve the equation is $mathbb{R} - {2,6}$ .We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$ .Simplifying that, we get $7x-6 = (k+2)x - 2k$ . Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $boxed{mathrm{ (E)} 5}$ For other $k$ , one can solve for $x$ : $x(5-k) = 6-2k$ , hence $x=frac {6-2k}{5-k}$ . We can easily verify that for none of the other four possible values of $k$ is this equal to $2$ or $6$ , hence there is a solution for $x$ in each of the other cases.
We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$ .As $(4y + 1)(2x - 3) = 0$ must be true for all $y$ , we must have $2x - 3 = 0$ , hence $boxed{x = frac 32 mathrm{ (D)}}$ .Since we want only the $y$ -variable to be present, we move the terms only with the $y$ -variable to one side, thus constructing $8xy - 12y + 2x - 3 = 0$ to $8xy + 2x = 12y + 3$ . For there to be infinite solutions for $y$ and there is no $x$ , we simply find a value of $x$ such that the equation is symmetrical. Therefore, $2x = 3$ $8xy = 12y$ There is only one solution, namely $x = boxed{dfrac{3}{2}}$ or $text{(D)}$
Taking the root, we get $N=sqrt{25^{64}cdot 64^{25}}=5^{64}cdot 8^{25}$ .Now, we have $N=5^{64}cdot 8^{25}=5^{64}cdot (2^{3})^{25}=5^{64}cdot 2^{75}$ .Combining the $2$ 's and $5$ 's gives us $(2cdot 5)^{64}cdot 2^{(75-64)}=(2cdot 5)^{64}cdot 2^{11}=10^{64}cdot 2^{11}$ .This is the number $2048$ with a string of sixty-four $0$ 's at the end. Thus, the sum of the digits of $N$ is $2+4+8=14Longrightarrowboxed{mathrm{ (B)} 14}$
Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $A-B$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$ , it follows that $A$ (as a prime greater than $2$ ) is odd. Thus $B = 2$ , and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$ , from which it follows that $A-2 = 3$ and $A = 5$ . The sum of these numbers is thus $17$ , a prime, so the answer is $boxed{mathrm{(E)} text{prime}}$ .In order for both $A - B$ and $A + B$ to be prime, one of $A, B$ must be 2, or else both $A - B$ , $A + B$ would be even numbers.If $A = 2$ , then $A < B$ and $A - B < 0$ , which is not possible. Thus $B = 2$ .Since $A$ is prime and $A > A - B > 2$ , we can infer that $A > 3$ and thus $A$ can be expressed as $6n pm 1$ for some natural number $n$ .However in either case, one of $A - B$ and $A + B$ can be expressed as $6n pm 3 = 3(2n pm 1)$ which is a multiple of 3. Therefore the only possibility that works is when $A - B = 3$ and $[A + B + (A - B) + (A + B) = 5 + 2 + 3 + 7 = 17]$ Which is a prime number. $boxed{(E)}$
Let $x^2 = frac{n}{20-n}$ , with $x ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = frac{20x^2}{x^2 + 1}$ . Since $text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $boxed{mathrm{(D)} 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $nin{1,dots,9}$ it is between 0 and 1.Thus we only need to examine $n=0$ and $nin{10,dots,19}$ .For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.This leaves $nin{12,14,15,16,18}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $boxed{mathrm{(D)} 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).If $frac{n}{20-n} = k^2 ge 0$ , then $n ge 0$ and $20-n > 0$ , otherwise $frac{n}{20-n}$ will be negative. Thus $0 le n le 19$ and $[0 = frac{0}{20-(0)} le frac{n}{20-n} le frac{19}{20-(19)} = 19]$ Checking all $k$ for which $0 le k^2 le 19$ , we have $0$ , $1$ , $2$ , $3$ as the possibilities. $boxed{(D)}$
The area of the triangle $ADG$ can be computed as $frac{DG cdot AP}2$ . We will now find $DG$ and $AP$ .Clearly, $PFG$ is a right isosceles triangle with hypotenuse of length $2$ , hence $PG=sqrt 2$ . The same holds for triangle $QED$ and its leg $QD$ . The length of $PQ$ is equal to $FE=2$ . Hence $GD = 2 + 2sqrt 2$ , and $AP = PD = 2 + sqrt 2$ .Then the area of $ADG$ equals $frac{DG cdot AP}2 = frac{(2+2sqrt 2)(2+sqrt 2)}2 = frac{8+6sqrt 2}2 = boxed{textbf{(C)}4+3sqrt 2}$ .
For any given pair of circles, they can intersect at most $2$ times. Since there are ${4choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 cdot 2 = 12$ . We can construct such a situation as below, so the answer is $boxed{mathrm{(D)} 12}$ .Because a pair or circles can intersect at most $2$ times, the first circle can intersect the second at $2$ points, the third can intersect the first two at $4$ points, and the fourth can intersect the first three at $6$ points. This means that our answer is $2+4+6=boxed{mathrm{(D)} 12}.$ Pick a circle any circle- $4$ ways. Then, pick any other circle- $3$ ways. For each of these circles, there will be $2$ intersections for a total of $4*3*2$ = $24$ intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of $frac{24}{2}=boxed{12}$ , which corresponds to $text{(D)}$ .
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like......we get the value of the common difference of every hundred terms hundred times. So we have to divide the answer by hundred to get ......the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...
Adding the two given equations together gives

$a_1+a_2+...+a_{200}=300$ .

Now, let the common difference be $d$ . Notice that $a_2-a_1=d$ , so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is

$frac{n}{2}(2a_1+d(n-1))$ ,

where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$ . Therefore, we have

$50(2a_1+99d)=100$ ,

or

$2a_1+99d=2$ . *(1)

For the sum of the equations (shown at the beginning of the solution) we have $n=200$ , so

$100(2a_1+199d)=300$

or

$2a_1+199d=3$ *(2)

Now we have a system of equations in terms of $a_1$ and $d$ . Subtracting (1) from (2) eliminates $a_1$ , yielding $100d=1$ , and $d=a_2-a_1=frac{1}{100}=boxed{(text{C}) .01}$ .

Subtracting the 2 given equations yields

$(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$

Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms

$((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$

Simplifying and canceling $a_1$ and $x$ terms gives

$100x+100x+100x+...+100x=100$

$100xtimes100=100$

$100x=1$
Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$ Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=boxed{(text{B})1}$ .The easiest way is to assume a value for $a$ and then solving the system of equations. For $a = 1$ , we get the equations $-7b + 8c = 3$ and $4b - c = -1$
Multiplying the second equation by $8$ , we have

$32b - 8c = -8$

Adding up the two equations yields

$25b = -5$ , so $b = -frac{1}{5}$

We obtain $c = frac{1}{5}$ after plugging in the value for $b$ .

Therefore, $a^2-b^2+c^2 = 1-frac{1}{25}+frac{1}{25}=boxed{1}$ which corresponds to $text{(B)}$ .
We say Andy's lawn has an area of $x$ . Beth's lawn thus has an area of $frac{x}{2}$ , and Carlos's lawn has an area of $frac{x}{3}$ .We say Andy's lawn mower cuts at a speed of $y$ . Carlos's cuts at a speed of $frac{y}{3}$ , and Beth's cuts at a speed $frac{2y}{3}$ .Each person's lawn is cut at a speed of $frac{text{area}}{text{rate}}$ , so Andy's is cut in $frac{x}{y}$ time, as is Carlos's. Beth's is cut in $frac{3}{4}timesfrac{x}{y}$ , so the first one to finish is $boxed{mathrm{(B)} text{Beth}}$ .
Let $OM = x$ , $ON = y$ . By the Pythagorean Theorem on $triangle XON, MOY$ respectively, $begin{align*} (2x)^2 + y^2 &= 19^2\ x^2 + (2y)^2 &= 22^2end{align*}$ Summing these gives $5x^2 + 5y^2 = 845 Longrightarrow x^2 + y^2 = 169$ .By the Pythagorean Theorem again, we have $[(2x)^2 + (2y)^2 = XY^2 Longrightarrow XY = sqrt{4(x^2 + y^2)} = sqrt{4(169)} = sqrt{676} = boxed{mathrm{(B)} 26}]$ Alternatively, we could note that since we found $x^2 + y^2 = 169$ , segment $MN=13$ . Right triangles $triangle MON$ and $triangle XOY$ are similar by Leg-Leg with a ratio of $frac{1}{2}$ , so $XY=2(MN)=boxed{mathrm{(B)} 26}$
When $m=1$ , $a_{n+1}=1+a_n+n$ . Hence, $[a_{2}=1+a_1+2]$ $[a_{3}=1+a_2+3]$ $[a_{4}=1+a_3+4]$ $[dots]$ $[a_{12}=1+a_{11}+11]$ Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=boxed{78}$ ~AopsUser101Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ : $a_{m+1}=a_m+a_{1}+m$ . Since $a_1 = 1$ , $a_{m+1}=a_m+m+1$ . Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$ , and so on until $a_2 = a_1 + 2$ . Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + cdots + a_2$ ; adding the Right Hand Sides of these equations gives $(a_{m-1} + a_{m-2} + a_{m-3} + cdots + a_1) + (m + (m-1) + (m-2) + cdots + 2)$ . These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + cdots + a_1) + (m + (m-1) + (m-2) + cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + cdots + 2)$ . Substituting $a_1 = 1$ : $a_m = 1 + (m + (m-1) + (m-2) + cdots + 2) = 1+2+3+4+ cdots +m = frac{(m+1)(m)}{2}$ . Thus we have a general formula for $a_m$ and substituting $m=12$ : $a_{12} = frac{(13)(12)}{2} = (13)(6) = boxed{mathrm{(D) } 78}$ .We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since $a_{m+n} = a_m+a_n +mn$ , we know $a_2=a_1+a_1+1cdot1=1+1+1=3$ . After this, we can use $a_2$ to find $a_4$ . $a_4=a_2+a_2+2cdot 2 = 3+3+4 = 10$ . Now, we can use $a_2$ and $a_4$ to find $a_6$ , or $a_6=a_4+a_2+4cdot 2 = 10+3+8=21$ . Lastly, we can use $a_6$ to find $a_{12}$ . $a_{12} = a_6+a_6+6cdot 6 = 21+21+36= boxed{(D) 78}$
We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $triangle OBC$ is a $30-60-90$ triangle. That means $angle BOC = 60^circ$ and the ferris wheel has made $frac{60}{360} = frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $frac{1}{6}text{th}$ of a minute, or $10$ seconds. The answer is $boxed{mathrm{(D) } 10}$ . Alternatively, we could also say that $triangle ABC$ is congruent to $triangle OBC$ by SAS, so $AC$ is 20, and $triangle AOC$ is equilateral, and $angle BOC = 60^circ$
Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $dfrac{x+15}{y+1}=dfrac{x}{y}+2$ and $dfrac{x+15+1}{y+1+1}=dfrac{x+16}{y+2}=frac{x}{y}+2-1=frac{x}{y}+1$ . With a lot of algebra, the solution is found to be $y= boxed{textbf{(A)} 4}$ .

We let $n$ be the original number of elements in the set and we let $m$ be the original average of the terms of the original list. Then we have $mn$ is the sum of all the elements of the list. So we have two equations: $[mn+15=(m+2)(n+1)=mn+m+2n+2]$ and $[mn+16=(m+1)(n+2)=mn+2m+n+2.]$ Simplifying both equations and we get, $[13=m+2n]$ $[14=2m+n]$ Solving for $m$ and $n$ , we get $m=5$ and $n=boxed{textbf{(A)}4}$ .

Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in probem 25 it is funny to see this.

Plug and chug random numbers with the answer choices, starting with the choice of $4$ numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math $[frac{5+5+5+5+15}{5}=7]$ adding in 1 to the set you result in the mean to be 6. $[frac{5+5+5+5+15+1}{6}=6]$ Thus we conclude that 4 is the correct choice or $boxed{textbf{(A)}}$