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2005AMC10B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日上午10:46

2005AMC10B真题

答案解析请参考文末

Problem 1

A scout troop buys $1000$ candy bars at a price of five for $ $2$ . They sell all the candy bars at a price of two for $ $1$ . What was the profit, in dollars?

$mathrm{(A)} 100 qquad mathrm{(B)} 200 qquad mathrm{(C)} 300 qquad mathrm{(D)} 400 qquad mathrm{(E)} 500$

Problem 2

A positive number $x$ has the property that $x%$ of $x$ is $4$ . What is $x$ ?

$mathrm{(A)} 2 qquad mathrm{(B)} 4 qquad mathrm{(C)} 10 qquad mathrm{(D)} 20 qquad mathrm{(E)} 40$

Problem 3

A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$mathrm{(A)} frac{1}{10} qquad mathrm{(B)} frac{1}{9} qquad mathrm{(C)} frac{1}{3} qquad mathrm{(D)} frac{4}{9} qquad mathrm{(E)} frac{5}{9}$

Problem 4

For real numbers $a$ and $b$ , define $a diamond b = sqrt{a^2 + b^2}$ . What is the value of

$(5 diamond 12) diamond ((-12) diamond (-5))$ ?

$mathrm{(A)} 0 qquad mathrm{(B)} frac{17}{2} qquad mathrm{(C)} 13 qquad mathrm{(D)} 13sqrt{2} qquad mathrm{(E)} 26$

Problem 5

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?

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

Problem 6

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?

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

Problem 7

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

$mathrm{(A)} frac{pi}{16} qquad mathrm{(B)} frac{pi}{8} qquad mathrm{(C)} frac{3pi}{16} qquad mathrm{(D)} frac{pi}{4} qquad mathrm{(E)} frac{pi}{2}$

Problem 8

An $8$ -foot by $10$ -foot ﬂoor is tiled with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the ﬂoor are shaded? $[asy]unitsize(2cm); defaultpen(linewidth(.8pt)); fill(unitsquare,gray); filldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black); filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black); filldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black); filldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black);[/asy]$ $textrm{(A)} 80-20pi qquad textrm{(B)} 60-10pi qquad textrm{(C)} 80-10pi qquad textrm{(D)} 60+10pi qquad textrm{(E)} 80+10pi$

Problem 9

One fair die has faces $1$ , $1$ , $2$ , $2$ , $3$ , $3$ and another has faces $4$ , $4$ , $5$ , $5$ , $6$ , $6$ . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$mathrm{(A)} frac{1}{3} qquad mathrm{(B)} frac{4}{9} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{5}{9} qquad mathrm{(E)} frac{2}{3}$

Problem 10

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 $BD$ ?

$mathrm{(A)} 3 qquad mathrm{(B)} 2sqrt{3} qquad mathrm{(C)} 4 qquad mathrm{(D)} 5 qquad mathrm{(E)} 4sqrt{2}$

Problem 11

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?

$mathrm{(A)} 29 qquad mathrm{(B)} 55 qquad mathrm{(C)} 85 qquad mathrm{(D)} 133 qquad mathrm{(E)} 250$

Problem 12

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$mathrm{(A)} left(frac{1}{12}right)^{12} qquad mathrm{(B)} left(frac{1}{6}right)^{12} qquad mathrm{(C)} 2left(frac{1}{6}right)^{11} qquad mathrm{(D)} frac{5}{2}left(frac{1}{6}right)^{11} qquad mathrm{(E)} left(frac{1}{6}right)^{10}$

Problem 13

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$ ?

$mathrm{(A)} 501 qquad mathrm{(B)} 668 qquad mathrm{(C)} 835 qquad mathrm{(D)} 1002 qquad mathrm{(E)} 1169$

Problem 14

Equilateral $triangle ABC$ has side length $2$ , $M$ is the midpoint of $overline{AC}$ , and $C$ is the midpoint of $overline{BD}$ . What is the area of $triangle CDM$ ? $[asy]defaultpen(linewidth(.8pt)+fontsize(8pt)); pair B = (0,0); pair A = 2*dir(60); pair C = (2,0); pair D = (4,0); pair M = midpoint(A--C); label("$A$",A,NW);label("$B$",B,SW);label("$C$",C, SE);label("$M$",M,NE);label("$D$",D,SE); draw(A--B--C--cycle); draw(C--D--M--cycle);[/asy]$ $textrm{(A)} frac {sqrt {2}}{2}qquad textrm{(B)} frac {3}{4}qquad textrm{(C)} frac {sqrt {3}}{2}qquad textrm{(D)} 1qquad textrm{(E)} sqrt {2}$

Problem 15

An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $ $20$ or more?

$mathrm{(A)} frac{1}{4} qquad mathrm{(B)} frac{2}{5} qquad mathrm{(C)} frac{3}{7} qquad mathrm{(D)} frac{1}{2} qquad mathrm{(E)} frac{2}{3}$

Problem 16

The quadratic equation $x^2 + mx + n = 0$ has roots that are twice those of $x^2 + px + m = 0$ , and none of $m$ , $n$ , and $p$ is zero. What is the value of $n/p$ ?

$mathrm{(A)} 1 qquad mathrm{(B)} 2 qquad mathrm{(C)} 4 qquad mathrm{(D)} 8 qquad mathrm{(E)} 16$

Problem 17

Suppose that $4^a = 5$ , $5^b = 6$ , $6^c = 7$ , and $7^d = 8$ . What is $a cdot b cdot c cdot d$ ?

$mathrm{(A)} 1 qquad mathrm{(B)} frac{3}{2} qquad mathrm{(C)} 2 qquad mathrm{(D)} frac{5}{2} qquad mathrm{(E)} 3$

Problem 18

All of David's telephone numbers have the form $555-abc-defg$ , where $a$ , $b$ , $c$ , $d$ , $e$ , $f$ , and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$ . How many different telephone numbers can David have?

$mathrm{(A)} 1 qquad mathrm{(B)} 2 qquad mathrm{(C)} 7 qquad mathrm{(D)} 8 qquad mathrm{(E)} 9$

Problem 19

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?

$mathrm{(A)} 0 qquad mathrm{(B)} 1 qquad mathrm{(C)} 2 qquad mathrm{(D)} 4 qquad mathrm{(E)} 5$

Problem 20

What is the average (mean) of all $5$ -digit numbers that can be formed by using each of the digits $1$ , $3$ , $5$ , $7$ , and $8$ exactly once?

$mathrm{(A)} 48000 qquad mathrm{(B)} 49999.5 qquad mathrm{(C)} 53332.8 qquad mathrm{(D)} 55555 qquad mathrm{(E)} 56432.8$

Problem 21

Forty slips are placed into a hat, each bearing a number $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b neq a$ . What is the value of $q/p$ ?

$mathrm{(A)} 162 qquad mathrm{(B)} 180 qquad mathrm{(C)} 324 qquad mathrm{(D)} 360 qquad mathrm{(E)} 720$

Problem 22

For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + ldots + n$ ?

$mathrm{(A)} 8 qquad mathrm{(B)} 12 qquad mathrm{(C)} 16 qquad mathrm{(D)} 17 qquad mathrm{(E)} 21$

Problem 23

In trapezoid $ABCD$ we have $overline{AB}$ parallel to $overline{DC}$ , $E$ as the midpoint of $overline{BC}$ , and $F$ as the midpoint of $overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$ ?

$mathrm{(A)} 2 qquad mathrm{(B)} 3 qquad mathrm{(C)} 5 qquad mathrm{(D)} 6 qquad mathrm{(E)} 8$

Problem 24

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ . What is $x + y + m$ ?

$mathrm{(A)} 88 qquad mathrm{(B)} 112 qquad mathrm{(C)} 116 qquad mathrm{(D)} 144 qquad mathrm{(E)} 154$

Problem 25

A subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$ ?

$mathrm{(A)} 50 qquad mathrm{(B)} 51 qquad mathrm{(C)} 62 qquad mathrm{(D)} 65 qquad mathrm{(E)} 68$

2005AMC10B详细解析

$begin{align*} mbox{Expenses} &= 1000 cdot frac25 = 400 mbox{Revenue} &= 1000 cdot frac12 = 500 mbox{Profit} &= mbox{Revenue} - mbox{Expenses} = 500-400 = boxed{(A) ,100}. end{align*}$
Since $x%$ means $0.01x$ , the statement " $x% text{ of } x text{ is 4}$ " can be rewritten as " $0.01x cdot x = 4$ ": $0.01x cdot x=4 Rightarrow x^2 = 400 Rightarrow x = boxed{text{(D)}20}.$
After the first day, there is $1-left(dfrac{1}{3}times1right)$ gallons left, or $dfrac{2}{3}$ gallons. After the second day, there is a total of $dfrac{2}{3}-left(dfrac{1}{3}timesdfrac{2}{3}right)=dfrac{2}{3}-dfrac{2}{9}=dfrac{4}{9}$ . Therefore, the fraction of the original amount of paint that is left is $dfrac{dfrac{4}{9}}{1}=boxed{mathrm{(D)},dfrac{4}{9}}$ Another way to do this is just to simply find the gain everyday and subtract from the remaining you had before.
$(5 diamond 12) diamond ((-12) diamond (-5)) (sqrt{5^2+12^2}) diamond (sqrt{(-12)^2+(-5)^2}) (sqrt{169})diamond(sqrt{169})13diamond13 sqrt{13^2+13^2} sqrt{338} boxed{mathrm{(D),13sqrt{2}}}$
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 $boxed{text{(C)}frac{2}{5}}$ . This was just a simple manipulation of the equation. No solving was needed!
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 $20-18=boxed{text{(B)}2}$ of them. Here, only the A's matter... No complicated stuff!
Let the side of the largest square be $x$ . It follows that the diameter of the inscribed circle is also $x$ . Therefore, the diagonal of the square inscribed inscribed in the circle is $x$ . The side length of the smaller square is $dfrac{x}{sqrt{2}}=dfrac{xsqrt{2}}{2}$ . Similarly, the diameter of the smaller inscribed circle is $dfrac{xsqrt{2}}{2}$ . Hence, its radius is $dfrac{xsqrt{2}}{4}$ . The area of this circle is $left(dfrac{xsqrt{2}}{4}right)^2pi=dfrac{2pi x^2}{16}=dfrac{x^2pi}{8}$ , and the area of the largest square is $x^2$ . The ratio of the areas is $dfrac{dfrac{x^2pi}{8}}{x^2}implies boxed{mathrm{(B)} dfrac{pi}{8}}$ .Let the radius of the smaller circle be $r$ . Then the side length of the smaller square is $2r$ . The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $sqrt{2}r$ . Hence the larger square has sides of length $2sqrt{2}r$ . The ratio of the area of the smaller circle to the area of the larger square is therefore $[frac{pi r^2}{left(2sqrt{2}rright)^2} =frac{pi}{8}implies boxed{mathrm B}.]$ $[asy] draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),14.1),linewidth(0.7)); draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7)); draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7)); draw((0,0)--(-14.1,0),linewidth(0.7)); draw((-7.1,7.1)--(0,0),linewidth(0.7)); label("$sqrt{2}r$",(-6,0),S); label("$r$",(-3.5,3.5),NE); label("$2r$",(-7.1,7.1),W); label("$2sqrt{2}r$",(0,14.1),N); [/asy]$
There are 80 tiles. Each tile has $[mbox{square} - 4 cdot (mbox{quarter circle})]$ shaded. Thus: $begin{align*} mbox{shaded area} &= 80 left( 1 - 4 cdot dfrac{1}{4} cdot pi cdot left(dfrac{1}{2}right)^2right) &= 80left(1-dfrac{1}{4}piright) &= boxed{(A) 80-20pi}. end{align*}$
In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.Case 1: The first die is odd and the second die is even.The probability of this happening is $dfrac{4}{6}timesdfrac{4}{6}=dfrac{16}{36}=dfrac{4}{9}$ Case 2: The first die is even and the second die is odd.The probability of this happening is $dfrac{2}{6}timesdfrac{2}{6}=dfrac{4}{36}=dfrac{1}{9}$ Adding these two probabilities will give us our final answer. $dfrac{4}{9}+dfrac{1}{9}=boxed{mathrm{(D)} dfrac{5}{9}}$
Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . From the Pythagorean Theorem, $CH=sqrt{48}$ . Since $CD=8$ , $HD=sqrt{8^2-48}=sqrt{16}=4$ , and $BD=HD-1$ , so $BD=boxed{text{(A)}3}$ .
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 $4^{text{th}}$ terms will equal $133$ , $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1 (text{mod} 3)$ will produce a result of $250$ . It just so happens that $2005equiv 1 (text{mod} 3)$ , which leads us to the answer of $boxed{mathrm{(E)} 250}$ .
In order for the product of the numbers to be prime, $11$ of the dice have to be a $1$ , and the other die has to be a prime number. There are $3$ prime numbers ( $2$ , $3$ , and $5$ ), and there is only one $1$ , and there are $dbinom{12}{1}$ ways to choose which die will have the prime number, so the probability is $dfrac{3}{6}timesleft(dfrac{1}{6}right)^{11}timesdbinom{12}{1} = dfrac{1}{2}timesleft(dfrac{1}{6}right)^{11}times12=left(dfrac{1}{6}right)^{11}times6=boxed{mathrm{(E)} left(dfrac{1}{6}right)^{10}}$ .
To find the multiples of $3$ or $4$ but not $12$ , you need to find the number of multiples of $3$ and $4$ , and then subtract twice the number of multiples of $12$ , because you overcount and do not want to include them. The multiples of $3$ are $frac{2005}{3} = 668text{ }R1.$ The multiples of $4$ are $frac{2005}{4} = 501 text{ }R1$ . The multiples of $12$ are $frac{2005}{12} = 167text{ }R1.$ So, the answer is $668+501-167-167 = boxed{mathrm{(C)} 835}$ From 1-12, the multiples of 3 or 4 but not 12 are 3, 4, 6, 8, an 9, a total of five numbers. Since $frac{5}{12}$ of positive integers are multiples of 3 or 4 but not 12, the answer is approximately $frac{5}{12} cdot 2005$ = $boxed{mathrm{(C)} 835}$
The area of a triangle can be given by $frac12 ab sin C$ . $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is the same length as $BC$ . Each angle of an equilateral triangle is $60^circ$ so $angle MCD = 120^circ$ . The area is $frac12 (1)(2) sin 120^circ = boxed{textbf{(C)} frac{sqrt{3}}{2}}$ .In order to calculate the area of $triangle CDM$ , we can use the formula $A=dfrac{1}{2}bh$ , where $overline{CD}$ is the base. We already know that $overline{CD}=2$ , so the formula now becomes $A=h$ . We can drop verticals down from $A$ and $M$ to points $E$ and $F$ , respectively. We can see that $triangle AEC sim triangle MFC$ . Now, we establish the relationship that $dfrac{AE}{MF}=dfrac{AC}{MC}$ . We are given that $overline{AC}=2$ , and $M$ is the midpoint of $overline{AC}$ , so $overline{MC}=1$ . Because $triangle AEB$ is a $30-60-90$ triangle and the ratio of the sides opposite the angles are $1-sqrt{3}-2$ $overline{AE}$ is $sqrt{3}$ . Plugging those numbers in, we have $dfrac{sqrt{3}}{MF}=dfrac{2}{1}$ . Cross-multiplying, we see that $2timesoverline{MF}=sqrt{3}times1implies overline{MF}=dfrac{sqrt{3}}{2}$ Since $overline{MF}$ is the height $triangle CDM$ , the area is $boxed{mathrm{(C)} dfrac{sqrt{3}}{2}}$ .Draw a line from $M$ to the midpoint of $overline{BC}$ . Call the midpoint of $overline{BC}$ $P$ . This is an equilateral triangle, since the two segments $overline{PC}$ and $overline{MC}$ are identical, and $angle C$ is 60°. Using the Pythagorean Theorem, point $M$ to $overline{BC}$ is $dfrac{sqrt{3}}{2}$ . Also, the length of $overline{CD}$ is 2, since $C$ is the midpoint of $overline{BD}$ . So, our final equation is $dfrac{sqrt{3}}{2}times2over2$ , which just leaves us with $boxed{mathrm{(C)} dfrac{sqrt{3}}{2}}$ .Drop a vertical down from $M$ to $BC$ . WLOG, let us call the point of intersection $X$ and the midpoint of $BC$ , $Y$ . We can observe that $triangle AYC$ and $triangle MXC$ are similar. By the Pythagorean theorem, $AY$ is $sqrt3$ . Since $AC:MC=2:1,$ we find $MX=frac{sqrt3}{2}.$ Because $C$ is the midpoint of $BD,$ and $BC=2,$ $CD=2.$ Using the area formula, $frac{CD*MX}{2}=frac{sqrt3}{2},$ $boxed{mathrm{(C)} dfrac{sqrt{3}}{2}}.$
The only way to get a total of $ $20$ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of $dbinom{8}{2}=dfrac{8times7}{2times1}=28$ ways to choose $2$ bills out of $8$ . There are $12$ ways to choose a twenty and some other non-twenty bill. There is $1$ way to choose both twenties, and also $1$ way to choose both tens. Adding these up, we find that there are a total of $14$ ways to attain a sum of $20$ or greater, so there is a total probability of $dfrac{14}{28}=boxed{mathrm{(D)} dfrac{1}{2}}$ .Another way to do this problem is to use complementary counting, i.e. how many ways that the sum is less than 20. Now, you do not have to consider the 2 twenties, so you have 6 bills left. $dbinom{6}{2} = dfrac{6times5}{2times1} = 15$ ways. However, you counted the case when you have 2 tens, so you need to subtract 1, and you get 14. Finding the ways to get 20 or higher, you subtract 14 from 28 and get 14. So the answer is $dfrac{14}{28} = boxed{mathrm{(D)} dfrac{1}{2}}$ .
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 $frac{n}{p} = frac{4ab}{-(a+b)} = frac{4m}{frac{m}{2}} = boxed{mathrm{(D)} 8}$ .Indeed, consider the quadratics $x^2 + 8x + 16 = 0, x^2 + 16x + 64 = 0$ .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 equations, $[2a + 2b = -m]$ $[a + b = -p]$ $[2a(2b) = n]$ $[a(b) = m]$ Therefore, substituting the second equation into the first equation gives $[m = 2(p)]$ and substituting the fourth equation into the third equation gives $[n = 4(m)]$ Therefore, $n = 8p$ , so $frac{n}{p} = 8 = boxed{textbf{D}}$
$[8=7^d]$ $[8=left(6^cright)^d]$ $[8=left(left(5^bright)^cright)^d]$ $[8=left(left(left(4^aright)^bright)^cright)^d]$ $[8=4^{acdot bcdot ccdot d}]$ $[2^3=2^{2cdot acdot bcdot ccdot d}]$ $[3=2cdot acdot bcdot ccdot d]$ $[acdot bcdot ccdot d=boxed{mathrm{(B)} dfrac{3}{2}}]$ We can write $a$ as $log_4 5$ , $b$ as $log_5 6$ , $c$ as $log_6 7$ , and $d$ as $log_7 8$ . We know that $log_b a$ can be rewritten as $frac{log a}{log b}$ , so $a*b*c*d=$ $[frac{log5}{log4}cdotfrac{log6}{log5}cdotfrac{log7}{log6}cdotfrac{log8}{log7}]$ $[frac{log8}{log4}]$ $[frac{3log2}{2log2}]$ $[boxed{frac{3}{2}}]$
The only digits available to use in the phone number are $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , and $9$ . There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$ . The answer is just $dbinom{8}{7}=dfrac{8!}{7!,(8-7)!}=boxed{mathrm{(D)} 8}$ Alternatively, we could just choose $1$ out of the $8$ numbers not to be used. There are obviously $boxed{8}$ ways to do so.
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 $10^{text{th}}$ and $11^{text{th}}$ terms are both $85$ . The mean is $dfrac{70,(2)+80,(5)+85,(4)+90,(3)+95,(6)}{20}=dfrac{1720}{20}=86$ . The difference between the mean and median, therefore, is $boxed{mathrm{(B)} 1}$ .
We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 times (1+3+5+7+8) times (11111) = 24 times 24 times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 div 120$ to get $boxed{mathrm{(C)}53332.8}$ We can first solve for the mean for the digits 1, 3, 5, 7, and 9 since each is 2 away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the 8 with a 9. We can start how be calculating the gain when the 8 was in the ones digit. Since there are $4! = 24$ numbers with the 8 in the ones digit and 1 was gain from each of them, 24 is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the 9 instead of the 8 to get $6666600-266664=6399936$ . Finally, we divide this by 120 to get the average. $6399936/120= boxed{mathrm{(C)}53332.8}$
There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 cdot 39 cdot 38 cdot 37$ total ways to draw four slips. Thus $p = frac{10cdot 4!}{40 cdot 39 cdot 38 cdot 37}$ .There are ${10 choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 times 3 times 4 times 3$ ways to pick the slips, so $q = frac{45 cdot 6 cdot 4^2 cdot 3^2}{40 cdot 39 cdot 38 cdot 37}$ .Hence, the answer is $frac{q}{p} = frac{45 cdot 6 cdot 4^2 cdot 3^2}{10cdot 4!} = boxed{ mathbf{(A)}162}$ .For probability $p$ , there are $binom{10}{1}=10$ ways to choose the card you want to show up $4$ times.Hence, the probability is $frac{10}{binom{40}{4}}$ .For probability $q$ , there are $binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $binom{4}{2}cdotbinom{4}{2}$ to pick which cards you want out of the $4$ of each.Hence, the probability is $frac{45cdot6cdot6}{binom{40}{4}}$ Hence, $frac{q}{p}=frac{45cdot6cdot6}{10}=9cdot6cdot3=162implies boxed{A}$ .
Since $1 + 2 + cdots + n = frac{n(n+1)}{2}$ , the condition is equivalent to having an integer value for $frac{n!}{frac{n(n+1)}{2}}$ . This reduces, when $nge 1$ , to having an integer value for $frac{2(n-1)!}{n+1}$ . This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 25, so there are $24 - 8 = boxed{text{(C)}16}$ numbers less than or equal to 24 that satisfy the condition.
Since the height of both trapezoids are equal, and the area of is twice the area of ,., so. is exactly halfway between and , so ., so, and.
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Mark $DC=z$ , $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$ .

Then, we have that $frac{x+y}{2}cdot h=2(frac{y+z}{2} cdot h)$ .

From this, we get that $x=2z+y$ .

We also get that $frac{x+z}{2} cdot h= 3(frac{y+z}{2} cdot h)$ .

Simplifying, we get that $2x=z+3y$

Notice that we want $frac{AB}{DC}=frac{x}{z}$ .

Dividing the first equation by $z$ , we get that $frac{x}{z}=2+frac{y}{z}implies 3(frac{x}{z})=6+3(frac{y}{z})$ .

Dividing the second equation by $z$ , we get that $2(frac{x}{z})=1+3(frac{y}{z})$ .

Now, when we subtract the top equation from the bottom, we get that $frac{x}{z}=5$

Hence, the answer is $boxed{5}$
Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero. Then $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 cdot 11$ , so $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum of $a+b$ is $9+8=17$ , so $a+b=11$ . Then we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 Longrightarrow a=frac{15}{2}, b=frac{7}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=65+56+33=154 mathbf{(E)}$ .The first steps are the same as above. Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . This is where the solution diverges.We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get $3^2cdot 11$ . In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals 11 while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ . The sum of the numbers is $boxed{textbf{(E) }154}$ Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$ . Therefore, $(a+b)(a-b) = frac{m^2}{99}$ and $frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 cdot 11.$ Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 cdot 11^2$ gives us 1089 as $n$ and $m = sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 cdot 3^2$ . A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $boxed{textbf{(E) }154}$ . Note that all of the solutions used $a+b$ or $a-b$ as part of their solution.
The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ : $25+100$ , $26+99$ , $27+98$ , $ldots$ , $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = boxed{mathrm{(C)} 62}$ . Also, it is possible to see that since the numbers $1$ to $24$ are in the set there are only the numbers $25$ to $100$ to consider. As $62+63$ gives $125$ , the numbers $25$ to $62$ can be put in subset $B$ without having two numbers add up to $125$ . In this way, subset $B$ will have the numbers $1$ to $62$ , and so $boxed{mathrm{(C)} 62}$ ."Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $boxed{mathrm{(C)} 62}$ .