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2000AMC10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4:33

2000AMC10真题

答案解析请参考文末

Problem 1

In the year 2001, the United States will host the International Mathematical Olympiad. Let $I$ , $M$ , and $O$ be distinct positive integers such that the product $I cdot M cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$ ?

$mathrm{(A)} 23 qquadmathrm{(B)} 55 qquadmathrm{(C)} 99 qquadmathrm{(D)} 111 qquadmathrm{(E)} 671$

Problem 2

$2000({2000}^{2000}) =$

$mathrm{(A)} {2000}^{2001} qquad mathrm{(B)} {4000}^{2000} qquad mathrm{(C)} {2000}^{4000} qquad mathrm{(D)} {4,000,000}^{2000} qquadmathrm{(E)} {2000}^{4,000,000}$

Problem 3

Each day, Jenny ate $20%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?

$mathrm{(A)} 40 qquadmathrm{(B)} 50 qquadmathrm{(C)} 55 qquadmathrm{(D)} 60 qquadmathrm{(E)} 75$

Problem 4

Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $$12.48$ , but in January her bill was $$17.54$ because she used twice as much connect time as in December. What is the fixed monthly fee?

$mathrm{(A)} $2.53 qquadmathrm{(B)} $5.06 qquadmathrm{(C)} $6.24 qquadmathrm{(D)} $7.42 qquadmathrm{(E)} $8.77$

Problem 5

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?

(a) the length of the segment $MN$

(b) the perimeter of $triangle PAB$

(d) the area of trapezoid $ABNM$

$[asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N); [/asy]$

$mathrm{(A)} 0 qquadmathrm{(B)} 1 qquadmathrm{(C)} 2 qquadmathrm{(D)} 3 qquadmathrm{(E)} 4$

Problem 6

The Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, ldots$ starts with two $1$ s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

$mathrm{(A)} 0 qquadmathrm{(B)} 4 qquadmathrm{(C)} 6 qquadmathrm{(D)} 7qquadmathrm{(E)} 9$

Problem 7

In rectangle $ABCD$ , $AD=1$ , $P$ is on $overline{AB}$ , and $overline{DB}$ and $overline{DP}$ trisect $angle ADC$ . What is the perimeter of $triangle BDP$ ?

$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); [/asy]$

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

Problem 8

At Olympic High School, $frac{2}{5}$ of the freshmen and $frac{4}{5}$ of the sophomores took the AMC 10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?

$mathrm{(A)}$ There are five times as many sophomores as freshmen.

$mathrm{(B)}$ There are twice as many sophomores as freshmen.

$mathrm{(C)}$ There are as many freshmen as sophomores.

$mathrm{(D)}$ There are twice as many freshmen as sophomores.

$mathrm{(E)}$ There are five times as many freshmen as sophomores.

Problem 9

If $|x-2|=p$ , where $x<2$ , then $x-p=$

$mathrm{(A)} -2 qquadmathrm{(B)} 2 qquadmathrm{(C)} 2-2p qquadmathrm{(D)} 2p-2 qquadmathrm{(E)} |2p-2|$

Problem 10

The sides of a triangle with positive area have lengths $4$ , $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ , $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$ ?

$mathrm{(A)} 2 qquadmathrm{(B)} 4 qquadmathrm{(C)} 6 qquadmathrm{(D)} 8 qquadmathrm{(E)} 10$

Problem 11

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$mathrm{(A)} 22 qquadmathrm{(B)} 60 qquadmathrm{(C)} 119 qquadmathrm{(D)} 180 qquadmathrm{(E)} 231$

Problem 12

Figures $0$ , $1$ , $2$ , and $3$ consist of $1$ , $5$ , $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

$[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]$

$mathrm{(A)} 10401 qquadmathrm{(B)} 19801 qquadmathrm{(C)} 20201 qquadmathrm{(D)} 39801 qquadmathrm{(E)} 40801$

Problem 13

There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

$[asy] unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((0,3)); dot((1,3)); dot((0,4)); [/asy]$

$mathrm{(A)} 0 qquadmathrm{(B)} 1 qquadmathrm{(C)} 5!cdot 4!cdot 3!cdot 2!cdot 1! qquadmathrm{(D)} frac{15!}{5!cdot 4!cdot 3!cdot 2!cdot 1!} qquadmathrm{(E)} 15!$

Problem 14

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were $71$ , $76$ , $80$ , $82$ , and $91$ . What was the last score Mrs. Walter entered?

$mathrm{(A)} 71 qquadmathrm{(B)} 76 qquadmathrm{(C)} 80 qquadmathrm{(D)} 82 qquadmathrm{(E)} 91$

Problem 15

Two non-zero real numbers, $a$ and $b$ , satisfy $ab=a-b$ . Find a possible value of $frac{a}{b}+frac{b}{a}-ab$ .

$mathrm{(A)} -2 qquadmathrm{(B)} -frac{1}{2} qquadmathrm{(C)} frac{1}{3} qquadmathrm{(D)} frac{1}{2} qquadmathrm{(E)} 2$

Problem 16

The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$ . Find the length of segment $AE$ .

$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); [/asy]$

$mathrm{(A)} frac{4sqrt{5}}{3} qquadmathrm{(B)} frac{5sqrt{5}}{3} qquadmathrm{(C)} frac{12sqrt{5}}{7} qquadmathrm{(D)} 2sqrt{5} qquadmathrm{(E)} frac{5sqrt{65}}{9}$

Problem 17

Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?

$mathrm{(A)}$ $$3.63$

$mathrm{(B)}$ $$5.13$

$mathrm{(C)}$ $$6.30$

$mathrm{(D)}$ $$7.45$

$mathrm{(E)}$ $$9.07$

Problem 18

Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?

$mathrm{(A)} 24 qquadmathrm{(B)} 27 qquadmathrm{(C)} 39 qquadmathrm{(D)} 40 qquadmathrm{(E)} 42$

Problem 19

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

$mathrm{(A)} frac{1}{2m+1} qquadmathrm{(B)} m qquadmathrm{(C)} 1-m qquadmathrm{(D)} frac{1}{4m} qquadmathrm{(E)} frac{1}{8m^2}$

Problem 20

Let $A$ , $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $Acdot Mcdot C+Acdot M+Mcdot C+Ccdot A$ ?

$mathrm{(A)} 49 qquadmathrm{(B)} 59 qquadmathrm{(C)} 69 qquadmathrm{(D)} 79 qquadmathrm{(E)} 89$

Problem 21

If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?

$[textrm{I. All alligators are creepy crawlers.}]$ $[textrm{II. Some ferocious creatures are creepy crawlers.}]$ $[textrm{III. Some alligators are not creepy crawlers.}]$

$mathrm{(A)} text{I only} qquadmathrm{(B)} text{II only} qquadmathrm{(C)} text{III only} qquadmathrm{(D)} text{II and III only} qquadmathrm{(E)} text{None must be true}$

Problem 22

One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

$mathrm{(A)} 3 qquadmathrm{(B)} 4 qquadmathrm{(C)} 5 qquadmathrm{(D)} 6 qquadmathrm{(E)} 7$

Problem 23

When the mean, median, and mode of the list $[10,2,5,2,4,2,x]$ are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x$ ?

$mathrm{(A)} 3 qquadmathrm{(B)} 6 qquadmathrm{(C)} 9 qquadmathrm{(D)} 17 qquadmathrm{(E)} 20$

Problem 24

Let $f$ be a function for which $fleft(frac{x}{3}right)=x^2+x+1$ . Find the sum of all values of $z$ for which $f(3z)=7$ .

$mathrm{(A)} -frac{1}{3} qquadmathrm{(B)} -frac{1}{9} qquadmathrm{(C)} 0 qquadmathrm{(D)} frac{5}{9} qquadmathrm{(E)} frac{5}{3}$

Problem 25

In year $N$ , the $300^text{th}$ day of the year is a Tuesday. In year $N+1$ , the $200^text{th}$ day is also a Tuesday. On what day of the week did the $100^text{th}$ day of year $N-1$ occur?

$mathrm{(A)} text{Thursday} qquadmathrm{(B)} text{Friday} qquadmathrm{(C)} text{Saturday} qquadmathrm{(D)} text{Sunday} qquadmathrm{(E)} text{Monday}$

2000AMC10详细解析

$1 cdot 3 cdot 667 = 2001$ and $1+3+667=671 Longrightarrow boxed{text{(E)}}$ .
$2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2001} Rightarrow boxed{A}$
Since Jenny eats $20%$ of her jelly beans per day, $80%=frac{4}{5}$ of her jelly beans remain after one day.Let $x$ be the number of jelly beans in the jar originally. $frac{4}{5}cdotfrac{4}{5}cdot x=32$ $frac{16}{25}cdot x=32$ $x=frac{25}{16}cdot32= 50 Rightarrow boxed{B}$
Let $x$ be the fixed fee, and $y$ be the amount she pays for the minutes she used in the first month. $x+y=12.48$ $x+2y=17.54$ $y=5.06$ $x=7.42$ We want the fixed fee, which is $boxed{text{D}}$
(a) Clearly $AB$ does not change, and $MN=frac{1}{2}AB$ , so $MN$ doesn't change either.(b) Obviously, the perimeter changes. For example, imagine if P was extremely far to the left.(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the area of the trapezoid remains the same.Only $1$ quantity changes, so the correct answer is $boxed{text{B}}$ .
Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $bmod{10}$ : $1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$ The last digit to appear in the units position of a number in the Fibonacci sequence is $6 Longrightarrow boxed{mathrm{C}}$ .
$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); label("$1$",(0,1),W); label("$2$",(1.7,1),SE); label("$frac{sqrt{3}}{3}$",(0.65,2),N); label("$frac{2sqrt{3}}{3}$",(0.85,1),NW); label("$frac{2sqrt{3}}{3}$",(2.35,2),N); [/asy]$ $AD=1$ .Since $angle ADC$ is trisected, $angle ADP= angle PDB= angle BDC=30^circ$ .Thus, $PD=frac{2sqrt{3}}{3}$ $DB=2$ $BP=sqrt{3}-frac{sqrt{3}}{3}=frac{2sqrt{3}}{3}$ .
Adding, we get $boxed{textbf{(B) } 2+frac{4sqrt{3}}{3}}$ .
Let $f$ be the number of freshman and $s$ be the number of sophomores. $frac{2}{5}f=frac{4}{5}s$ $2f = 4s$ $f=2s$ There are twice as many freshmen as sophomores. $boxed{text{D}}$
When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$ .Thus $x-p = (2-p)-p = 2-2p$ . $boxed{text{C}}$
Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ . $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, the answer is $boxed{bold{D}}$ .
All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B, D, and A. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $boxed{textbf{(C) }119}$ .Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick, we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $boxed{textbf{(C) }119}$ .The answer must be in the form $pq - p - q$ = $(p - 1)(q - 1) - 1$ . Since $p - 1$ and $q - 1$ are both even, $(p - 1)(q - 1) - 1$ is $3 pmod 4$ , and the only answer that is $3 pmod 4$ is $boxed{textbf{(C) }119}$ .
We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$ : $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$ , so for figure $n$ , there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $boxed{textbf{(C) }20201}$ .Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ . Thus, the original sequence can be generated from a quadratic function.If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ , $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables: $f(0) = 1$ gives $c = 1$
$f(1) = 5$ gives $a + b + c = 5$

$f(2) = 13$ gives $4a + 2b + c = 13$

Plugging in $c=1$ into the last two equations gives

$a + b = 4$

$4a + 2b = 12$

Dividing the second equation by 2 gives the system:

$a + b = 4$

$2a + b = 6$

Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ . Thus, our quadratic function is:

$f(n) = 2n^2 + 2n + 1$

Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $boxed{textbf{(C) }20201}$ .

We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $sum_{n=1}^{100} n = 5050$ squares. $4 cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = boxed{textbf{(C) }20201}$ .

Let $a_n$ be the number of squares in figure $n$ . We can easily see that $[a_0=4cdot 0+1]$ $[a_1=4cdot 1+1]$ $[a_2=4cdot 3+1]$ $[a_3=4cdot 6+1.]$ Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number. Hence, $a_{100}=4cdot frac{100cdot 101}{2}+1=boxed{textbf{(C) }20201}$ . ~qkddud~

Let $f_n$ denote the number of unit cubes in a figure. We have $[f_0=1]$ $[f_1=5]$ $[f_2=13]$ $[f_3=25]$ $[f_4=41]$ $[...]$

Computing the difference between the number of cubes in each figure yields $[4,8,12,16,...]$ It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$

From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields

$[S_{100}=frac{100[2cdot 4+(100-1)4]}{2}]$ $[=50(2cdot 4+99cdot 4)]$ $[=50(101cdot 4)]$ $[=200cdot 101]$ $[=20200]$

So $f_{100}=1+20200=boxed{textbf{(C) }20201}$ unit cubes.
In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown: $[asy] unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); label("Y",(4,-.35),N); dot((0,1)); dot((1,1)); dot((2,1)); label("Y",(3,.6),N); dot((0,2)); dot((1,2)); label("Y",(2,1.6),N); dot((0,3)); label("Y",(1,2.6),N); label("Y",(0,3.6),N); [/asy]$ After this we can proceed to fill in the whole pegboard, so there is only $1$ arrangement of the pegs. The answer is $boxed{text{B}}$ $[asy] unitsize(20); label("O",(0,-.35),N); label("B",(1,-.35),N); label("G",(2,-.35),N); label("R",(3,-.35),N); label("Y",(4,-.35),N); label("B",(0,.6),N); label("G",(1,.6),N); label("R",(2,.6),N); label("Y",(3,.6),N); label("G",(0,1.6),N); label("R",(1,1.6),N); label("Y",(2,1.6),N); label("R",(0,2.6),N); label("Y",(1,2.6),N); label("Y",(0,3.6),N); [/asy]$ If (but this is not the case) we were told to distinguish pegs of the same color, the answer would be C.
The first number is divisible by 1.The sum of the first two numbers is even.The sum of the first three numbers is divisible by 3.The sum of the first four numbers is divisible by 4.The sum of the first five numbers is 400.Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.
Case 1: 76 is the last number entered.

Since $400equiv 76equiv 1pmod{3}$ , the fourth number must be divisible by 3, but none of the scores are divisible by 3.

Case 2: 80 is the last number entered.

Since $80equiv 2pmod{3}$ , the fourth number must be $2pmod{3}$ . That number is 71 and 71 only. The next number must be 91, since the sum of the first two numbers is even. So the only arrangement of the scores $76, 82, 91, 71, 80$ $Rightarrow text{(C)}$

We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers $pmod{3}$ , which gives 2,1,2,1,1, respectively. Clearly, the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is $mathrm{C}$ .
$frac {a}{b} + frac {b}{a} - ab = frac{a^2 + b^2}{ab} - (a - b) = frac{a^2 + b^2}{a-b} - frac{(a-b)^2}{(a-b)} = frac{2ab}{a-b} = frac{2(a-b)}{a-b} =2 Rightarrow text{(E)}$ .Another way is to solve the equation for $b,$ giving $b = frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$ This simplifies to $ab+b-a=0 Rightarrow (a+1)(b-1) = -1$ . The two integer solutions to this are $(-2,2)$ and $(0,0)$ . The problem states than $a$ and $b$ are non-zero, so we consider the case of $(-2,2)$ . So, we end up with $frac{-2}{2} + frac{2}{-2} - 2 cdot -2 = 4 - 1 - 1 = boxed{2}~ textbf{E}$
Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$ . If we set the bottom left point at $(0,0)$ , then $A=(0,3)$ , $B=(6,0)$ , $C=(4,2)$ , and $D=(2,0)$ .The line $l_1$ is given by the equation $y=m_1x+b_1$ . The $y$ -intercept is $A=(0,3)$ , so $b_1=3$ . We are given two points on $l_1$ , hence we can compute the slope, $m_1$ to be $frac{0-3}{6-0}=frac{-1}{2}$ , so $l_1$ is the line $y=frac{-1}{2}x+3$ Similarly, $l_2$ is given by $y=m_2x+b_2$ . The slope in this case is $frac{2-0}{4-2}=1$ , so $y=x+b_2$ . Plugging in the point $(2,0)$ gives us $b_2=-2$ , so $l_2$ is the line $y=x-2$ .At $E$ , the intersection point, both of the equations must be true, so $begin{align*} y=x-2, y=frac{-1}{2}x+3 &Rightarrow x-2=frac{-1}{2}x+3 \ &Rightarrow x=frac{10}{3} \ &Rightarrow y=frac{4}{3} \ end{align*}$ We have the coordinates of $A$ and $E$ , so we can use the distance formula here: $[sqrt{left(frac{10}{3}-0right)^2+left(frac{4}{3}-3right)^2}=frac{5sqrt{5}}{3}]$ which is answer choice $boxed{text{B}}$
$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2.5,.5),E); draw((6,0)--(4,2)); draw((0,3)--(2.5,.5)); [/asy]$

Draw the perpendiculars from $A$ and $B$ to $CD$ , respectively. As it turns out, $BC perp CD$ . Let $F$ be the point on $CD$ for which $AFperp CD$ .

$mangle AFE=mangle BCE=90^circ$ , and $mangle AEF=mangle CEB$ , so by AA similarity, $[triangle AFEsim triangle BCE Rightarrow frac{AF}{AE}=frac{BC}{BE}]$

By the Pythagorean Theorem, we have $AB=sqrt{3^2+6^2}=3sqrt{5}$ , $AF=sqrt{2.5^2+2.5^2}=2.5sqrt{2}$ , and $BC=sqrt{2^2+2^2}=2sqrt{2}$ . Let $AE=x$ , so $BE=3sqrt{5}-x$ , then $[frac{2.5sqrt{2}}{x}=frac{2sqrt{2}}{3sqrt{5}-x}]$ $[x=frac{5sqrt{5}}{3}]$

This is answer choice $boxed{text{B}}$

Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.

$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$

Drawing line $overline{BD}$ and parallel line $overline{CF}$ , we see that $triangle FCE sim triangle BDE$ by AA similarity. Thus $frac{FE}{EB} = frac{FC}{DB} = frac{2}{4} = frac{1}{2}$ . Reciprocating, we know that $frac{EB}{FE} = 2$ so $frac{EB+FE}{FE} = 2+1 Rightarrow frac{FB}{FE} = 3$ . Reciprocating again, we have $frac{FE}{FB} = frac{1}{3} Rightarrow FE = frac{1}{3}FB$ . We know that $FD = 2$ , so by the Pythagorean Theorem, $FB = sqrt{2^{2} + 4^{2}} = 2sqrt{5}$ . Thus $FE = frac{1}{3}FB = frac{2sqrt{5}}{3}$ . Applying the Pythagorean Theorem again, we have $AF = sqrt{1^{2}+2^{2}} = sqrt{5}$ . We finally have $AE = AF + FE = sqrt{5} + frac{2sqrt{5}}{3} = frac{5sqrt{5}}{3} Rightarrow boxed{text{B}}$
Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by $124$ cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of $124$ . Of the choices given, the only one is $boxed{text{D}}$
The area she sees looks at follows: $[asy] unitsize(0.8cm); path p1 = (0,0)--(5,0)--(5,5)--(0,5)--cycle; path p2 = (1,1)--(4,1)--(4,4)--(1,4)--cycle; path p3 = arc((0,0),1,180,270) -- arc((5,0),1,270,360) -- arc((5,5),1,0,90) -- arc((0,5),1,90,180) -- cycle; fill(p3,lightgray); unfill(p2); draw(p1,linewidth(bp)); draw(p2); draw(p3); draw( (0,0)--(-1,0), dashed ); draw( (0,0)--(0,-1), dashed ); draw( (5,0)--(6,0), dashed ); draw( (5,0)--(5,-1), dashed ); draw( (5,5)--(6,5), dashed ); draw( (5,5)--(5,6), dashed ); draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed ); draw( (0,-1)--(5,-1), Arrows ); label( "$5$", (2.5,-1), S ); draw( (1,1)--(4,1), Arrows ); label( "$3$", (2.5,1), N ); draw( (4,3.5)--(5,3.5), Arrows ); label( "$1$", (4.5,3.5), N ); draw( (5,3.5)--(6,3.5), Arrows ); label( "$1$", (5.5,3.5), N ); [/asy]$ The part inside the walk has area $5cdot 5 - 3cdot 3 = 16$ . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area $5cdot 1=5$ . The four arcs together form a circle with radius $1$ .Therefore the total area she can see is $16 + 4cdot 5 + picdot 1^2 = 36+pi simeq 39.14$ , which rounded to the nearest integer is $39$ . $boxed{C}$
$[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); [/asy]$ WLOG, let a side of the square be $1$ . Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = frac{1}{2}bh = frac{h}{2}$ , the height of the triangle with area $m$ is $2m$ . Therefore $frac{2m}{1} = frac{1}{x}$ where $x$ is the base of the other triangle. $x = frac{1}{2m}$ , and the area of that triangle is $frac{1}{2} cdot 1 cdot frac{1}{2m} = frac{1}{4m} text{(D)}$ .

$[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W); [/asy]$ From the diagram from the previous solution, we have $a$ , $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$ be $m$ times the area of square $C$ .

Since triangle $A$ is similar to the large triangle, it has $h_A = a(frac{c}{b}) = frac{ac}{b}$ , $b_A = c$ and $[[A] = frac{bh}{2} = frac{ac^2}{2b} = m[C] = mc^2]$ Thus $frac{a}{2b} = m$

Now since triangle $B$ is similar to the large triangle, it has $h_B = c$ , $b_B = bfrac{c}{a} = frac{bc}{a}$ and $[[B] = frac{bh}{2} = frac{bc^2}{2a} = nc^2 = n[C]]$

Thus $n = frac{b}{2a} = frac{1}{4(frac{a}{2b})} = frac{1}{4m}$ . $text{(D)}$ .
The trick is to realize that the sum $AMC+AM+MC+CA$ is similar to the product $(A+1)(M+1)(C+1)$ . If we multiply $(A+1)(M+1)(C+1)$ , we get $[(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.]$ We know that $A+M+C=10$ , therefore $(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11$ and $[AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.]$ Now consider the maximal value of this expression. Suppose that some two of $A$ , $M$ , and $C$ differ by at least $2$ . Then this triple $(A,M,C)$ is not optimal. (To see this, WLOG let $Ageq C+2.$ We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A to A-1$ and $C to C+1$ .)Therefore the maximum is achieved when $(A,M,C)$ is a rotation of $(3,3,4)$ . The value of $(A+1)(M+1)(C+1)$ in this case is $4cdot 4cdot 5=80,$ and thus the maximum of $AMC + AM + MC + CA$ is $80-11 = boxed{textbf{(C)} 69}.$ Notice that if we want to maximize $AMC + AM + MC + AC$ , we want A, M, and C to be as close as possible. For example, if $A = 7, B = 2,$ and $C=1,$ then the expression would have a much smaller value than if we were to substitute $A = 4, B = 5$ , and $C = 1$ . So to make A, B, and C as close together as possible, we divide $frac{10}{3}$ to get $3$ . Therefore, A must be 3, B must be 3, and C must be 4. $AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69$ . So the answer is $boxed{textbf{(C)} 69.}$
We interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:
- If $x$ is an $A$ , then $x$ is an $F$ .
- There is some $x$ that is a $C$ and at the same time an $A$ .
We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are $A$ s, but only Johnny is a $C$ " meets both conditions, but the first statement is false.

We CAN conclude that the second statement is true. We know that there is some $x$ that is a $C$ and at the same time an $A$ . Pick one such $x$ and call it Bobby. Additionally, we know that if $x$ is an $A$ , then $x$ is an $F$ . Bobby is an $A$ , therefore Bobby is an $F$ . And this is enough to prove the second statement -- Bobby is an $F$ that is also a $C$ .

We CAN NOT conclude that the third statement is true. For example, consider the situation when $A$ , $C$ and $F$ are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.

Therefore the answer is $boxed{text{(E) , II only}}$ .
Let be the total amount of coffee, of milk, and the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), soRegrouping, we get . Since both are positive, it follows that and are also positive, which is only possible when .One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has ounces. The "fair share" is So,Which requires that be since is a whole number.Again, let and be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is and also Thus, so and We also know that the amount Angela drank, which is is equal to ounces, thus Rearranging givesNow notice that (by the problem statement). In addition, so Therefore, and so We know that soFrom the leftmost inequality, we get and from the rightmost inequality, we get The only possible value of is .
Let $c,$ $m,$ and $p$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. $c$ and $m$ obviously can't be $0$ . We know $frac{c}{6} + frac{m}{4} = 8$ or $3c + 2m = 96$ and $c + m = 8p$ or $2c + 2m = 16p$ . Then, $[(2c + 2m) + c = 16p + c = 96]$ Because $16p$ and $96$ are both divisible by $16$ , $c$ must also be divisible by $16$ . Let $c = 16k$ . Now, $[3(16k) + 2m = 48k + 2m = 96]$ $k$ can't be $0$ , otherwise $c$ is $0$ , and $k$ can't be $2$ , otherwise $m$ is $0$ . Therefore $k$ must be $1$ , $c = 16$ and $m = 24$ . $c + m = 16 + 24 = 40 = 8p$ . Therefore, $p = 5 mathrm{(C)}$ .

Let $m$ , $c$ be the total amounts of milk and coffee, respectively. In order to know the number of people, we first need to find the total amount of mixture $t = m + c$ . We are given that $[frac{m}{4} + frac{c}{6} = 8]$ Multiplying the equation by $4$ yields $[m + frac{2}{3}c = (m + c) - frac{1}{3}c = t - frac{1}{3}c = 32]$ Since $frac{1}{3}c > 0$ , we have $t > 32$ . Now multiplying the equation by $6$ yields $[frac{3}{2}m + c = (m+c) + frac{1}{3}m = t + frac{1}{3}m = 48]$ Since $frac{1}{3}m > 0$ , we have $t < 48$ . Thus, $32 < t < 48$ .

Since $t$ is a multiple of $8$ , the only possible value for $t$ in that range is $40$ . Therefore, there are $frac{40}{8} = 5$ people in Angela's family. $mathrm{(C)}$ .
- he mean is $frac{10+2+5+2+4+2+x}{7} = frac{25+x}{7}$ .
- Arranged in increasing order, the list is $2,2,2,4,5,10$ , so the median is either $2,4$ or $x$ depending upon the value of $x$ .
- The mode is $2$ , since it appears three times.
We apply casework upon the median:
- If the median is $2$ ( $x le 2$ ), then the arithmetic progression must be constant.
- If the median is $4$ ( $x ge 4$ ), because the mode is $2$ , the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$ .
- If the median is $x$ ( $2 le x le 4$ ), we must have the arithmetic progression $2, x, frac{25+x}{7}$ . Thus, we find that $2x=2+frac{25+x}{7}$ so $x=3$ .
The answer is $3 + 17 = 20 mathrm{(E)}$ .
Let $y = frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -frac{1}{3}, frac{2}{9}$ . These sum up to $boxed{textbf{(B) }-frac19}$ . (We can also use Vieta's formulas to find the sum more quickly.)Set $f(frac{x}{3}) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-frac{1}{9}$ or $boxed{textbf{(B) }-frac19}$ .Since we have $f(x/3)$ , $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7: $81z^2 + 9z +1 = 7 Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-frac{b}{a}$ . Thus, the sum of the roots of this equation is $-frac{9}{81} = boxed{textbf{(B) }-frac19}$ .
There are either $[65 + 200 = 265]$ or $[66 + 200 = 266]$ days between the first two dates depending upon whether or not year $N$ is a leap year. Since $7$ divides into $266$ but not $265$ , for both days to be a Tuesday, year $N$ must be a leap year.Hence, year $N-1$ is not a leap year, and so since there are $[265 + 300 = 565]$ days between the date in years $N,text{ }N-1$ , this leaves a remainder of $5$ upon division by $7$ . Since we are subtracting days, we count 5 days before Tuesday, which gives us $boxed{mathbf{(A)} text{Thursday}.}$