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$\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$
$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | \begin{align*} \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ &= 1 \rightarrow \boxed{\text{B}}. \end{align*} | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sides of a triangle have lengths $6.5$, $10$, and $s$, where $s$ is a whole number. What is the smallest possible value of $s$?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$ | By [Triangle Inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_Inequality), $6.5 + s >10$ and therefore $s>3.5$. The smallest whole number that satisfies this is $\boxed{\text{(B)}\ 4}$. | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain?
$\text{... | 1)Model the amount left in the container as follows:
After the first pour $\frac12$ remains, after the second $\frac12 \times \frac23$ remains, etc.
This becomes the product $\frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{9}{10}$.
Note that the terms cancel out leaving $\frac{1}{10}$.
Now all that rem... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose that
means $a+b-c$.
For example,
is $5+4-6 = 3$.
Then the sum
is
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | The first triangle represents $1+3-4$
The 2nd triangle represents $2+5-6$
Solving the first triangle, we get $0$
Solving the 2nd triangle, we get $1$
Since we have to add the 2 triangles the final answer is $1$, which is $\boxed{D}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The digit-sum of $998$ is $9+9+8=26$. How many 3-digit whole numbers, whose digit-sum is $26$, are even?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | The highest digit sum for three-digit numbers is $9+9+9=27$. Therefore, the only possible digit combination is $9, 9, 8$. Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\boxed{\text{(A)}\ 1}$. | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$. Then $A+B=$
\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \... | The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The last square in the top row cannot be either 1 or 3, so it must be 2.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\... | 4 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$ | A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 \times 2 =\boxed{\textbf{(B)}\ 6}$ adults are fed. | 6 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately
$\text{(A)}\ 1\text{ month} \qquad \text{(B)}\ 4\text{ months} \qquad \text{(C)}\ 6\text{ months} \qquad \text{(D)}\ 8\text{ mo... | If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \times 4=240$ days which is about $\boxed{\text{(D)}\ 8\text{ months}}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$. Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \te... | Using the chart, $(2*4)=3$ and $(1*3)=3$. Therefore, $(2*4)*(1*3)=3*3=\boxed{\text{(D)}\ 4}$. | 4 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$ | We should list all the positive divisors of $36$ and count them. By trial and error, the divisors of $36$ are found to be $1,2,3,4,6,9,12,18,36$, for a total of $9$. However, $1$ and $2$ can't be equal to $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{\text{(A)}\ 7}$. | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$ | Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$. If the perimeter of one is $3$ times the other's, then $a=3b$. The area of the larger square over the area of the smaller square is
\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{\text{(E)}\ 9}\] | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number o... | Her driveway has $(4)(10)(3)=120$ cubic yards of snow. After the first hour she would have $120-20=100$ cubic yards, then $100-19=81$, $81-18=63$, $63-17=46$, $46-16=30$, $30-15=15$, and $15-14=1$ cubic yard after the seventh hour. It will take her a little more than seven hours to shovel it clean, which is closest to ... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
$\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$
$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ | $1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45$
$\frac{45+55}{10} = \dfrac{100}{10} = \boxed{\text{(D)}\ 10}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$ | If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{\text{th}}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\boxed{\text{(C)}\ 10}$ gumballs. | 10 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green... | If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.
Case 1: There are no green squares. This can be done in $1$ way.
Case 2: There is one green square and three red squares. This can... | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$. Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{\text{(A)}\ 0}$. | 0 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
For how many three-digit whole numbers does the sum of the digits equal $25$?
$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$ | Because $8+8+8=24$, it follows that one of the digits must be a $9$. The other two digits them have a sum of $25-9=16$. The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows
\[799,979,997,889,898,988\]
The number of configurations is $\boxed{\text{(C)}\ 6}$. | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The graph shows the distribution of the number of children in the families of the students in Ms. Jordan's English class. The median number of children in the family for this distribution is
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | Counting, there are thirteen total families. The middle number is 7th in either direction, and it is easy to see from the right side that this number is 4 $\text{(D)}$ | 4 | Other | MCQ | Yes | Yes | amc_aime | false |
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?
$\... | Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - $a:b$ - $a$ is for hrs. and $b$ is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minu... | 10 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$.
The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$.
Only $4, 12$ and $36$ appear on both lists, so the answer is $3$, which is option $\boxed{B}$. | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
$1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$ | Put the numbers in groups of $4$:
$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$
The first group has a sum of $0$.
The second group increases the two positive numbers on the end by $1$, and decreases the two negative numbers in the middle by $1$. Thus, the second group also has a sum of $0$.
Cont... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many subsets containing three different numbers can be selected from the set
\[\{ 89,95,99,132, 166,173 \}\]
so that the sum of the three numbers is even?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$ | To have an even sum with three numbers, we must add either $E+O+O$, or $E + E + E$, where $O$ represents an odd number, and $E$ represents an even number.
Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.
There are $2$ choices for... | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qqua... | Call this month "Month 0". Make a table of the fish that Brent and Gretel have each month.
$\text{Month / Brent / Gretel}$
$\text{0 / 4 / 128}$
$\text{1 / 16 / 256}$
$\text{2 / 64 / 512}$
$\text{3 / 256 / 1024}$
$\text{4 / 1024 / 2048}$
$\text{5 / 4096 / 4096}$
You could create a similar table without doing all of the... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is
$\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$ | If $AB = 10$ and $BC=4$, then $(10 - 4) \le AC \le (10 + 4)$ by the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality). In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.
Simplifying, this... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $\boxed{N}$ mean the number of whole number divisors of $N$. For example, $\boxed{3}=2$ because 3 has two divisors, 1 and 3. Find the value of
\[\boxed{\boxed{11}\times\boxed{20}}.\]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 24$ | $\boxed{\boxed{11}\times\boxed{20}}$
$11$ has $2$ factors, and $20$ has $\{1, 2, 4, 5, 10, 20\}$ as factors, for a total of $6$ factors. Plugging $\boxed{11} = 2$ and $\boxed {20}= 6$:
$\boxed{2 \times 6}$
$\boxed{12}$
$12$ has factors of $\{1, 2, 3, 4, 6, 12\}$, so $\boxed{12} = 6$, and the answer is $\boxed{A}$ | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$
There will be $10$ groups of $4$ numbers. The number now can be rewritten as $(2\cdot ... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Each of the letters $\text{W}$, $\text{X}$, $\text{Y}$, and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$, but not necessarily in that order. If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$, then the sum of $\text{W}$ and $\text{Y}$ is
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \te... | There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{ 1,2,3,4\}$.
$1$ has factor $1$.
$2$ has factors $1$ and $2$
$3$ has factors $1$ and $3$
$4$ has factors $1$, $2$, and $4$.
From here, we note that even though all numbers have the factor $1$, only $... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Harry has 3 sisters and 5 brothers. His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers. What is the product of $\text{S}$ and $\text{B}$?
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$ | Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.
$S = 3-1=2$
$B = 5+1=6$
$S\cdot B = 2\times6=12=\boxed{C}$ | 12 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
If $\begin{tabular}{r|l}a&b \\ \hline c&d\end{tabular} = \text{a}\cdot \text{d} - \text{b}\cdot \text{c}$, what is the value of $\begin{tabular}{r|l}3&4 \\ \hline 1&2\end{tabular}$?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | Plugging in values for $a$, $b$, $c$, and $d$, we get
$a=3$,
$b=4$,
$c=1$,
$d=2$,
$a\times d=3\times2=6$
$b\times c=4\times1=4$
$6-4=2$
$\boxed{E}$ | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$. Later the price is lowered again, this time by one-half the reduced price. The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qqu... | Solution 1
$100\%-20\%=80\%$
$10\times80\%=10\times0.8$
$10\times0.8=8$
$\frac{8}{2}=4=\boxed{C}$
Solution 2
The first discount has percentage 20, which is then discounted again for half of the already discounted price.
$100-20=80$
$\frac{80}{2}=40$
$40\%\times10=10\times0.4=4=\boxed{C}$ | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have ... | First, calculate how many of each type of problem she got right:
Arithmetic: $70\% \cdot 10 = 0.70 \cdot 10 = 7$
Algebra: $40\% \cdot 30 = 0.40 \cdot 30 = 12$
Geometry: $60\% \cdot 35 = 0.60 \cdot 35 = 21$
Altogether, Tori answered $7 + 12 + 21 = 40$ questions correct.
To get a $60\%$ on her test overall, she neede... | 5 | Other | MCQ | Yes | Yes | amc_aime | false |
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: $1\frac{1}{2}$ cups of flour, $2$ eggs, $3$... | If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\lceil \frac{216}{15}\rceil = 15$ pans.
$1$ pan requires $2$ ... | 5 | Other | MCQ | Yes | Yes | amc_aime | false |
At Central Middle School, the 108 students who take the AMC 8 meet in the evening to talk about food and eat an average of two cookies apiece. Hansel and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists these items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tabl... | For $216$ cookies, you need to make $\frac{216}{15} = 14.4$ pans. Since fractional pans are forbidden, round up to make $\lceil \frac{216}{15} \rceil = 15$ pans.
There are $3$ tablespoons of butter per pan, meaning $3 \cdot 15 = 45$ tablespoons of butter are required for $15$ pans.
Each stick of butter has $8$ tablesp... | 6 | Other | MCQ | Yes | Yes | amc_aime | false |
When $1999^{2000}$ is divided by $5$, the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | Note that the units digits of the powers of 9 have a pattern: $9^1 = {\bf 9}$, $9^2 = 8{\bf 1}$, $9^3 = 72{\bf 9}$, $9^4 = 656{\bf 1}$, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of $9$, ... | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $$1.02$, with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.
You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and q... | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
There is a list of seven numbers. The average of the first four numbers is $5$, and the average of the last four numbers is $8$. If the average of all seven numbers is $6\frac{4}{7}$, then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\... | Remember that if a list of $n$ numbers has an average of $k$, then the sum $S$ of all the numbers on the list is $S = nk$.
So if the average of the first $4$ numbers is $5$, then the first four numbers total $4 \cdot 5 = 20$.
If the average of the last $4$ numbers is $8$, then the last four numbers total $4 \cdot 8 = 3... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$ | The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$. The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$. There are five whole numbers in the interval. They are $2$, $3$, $4$, $5$, and $6$, so the answer is $\boxed{\text{(D)}\ 5... | 5 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Each principal of Lincoln High School serves exactly one $3$-year term. What is the maximum number of principals this school could have during an $8$-year period?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$ | If the first year of the $8$-year period was the final year of a principal's
term, then in the next six years two more principals would serve, and the last year of the
period would be the first year of the fourth principal's term. Therefore, the maximum
number of principals who can serve during an $8$-year period is $4... | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$-shaped region is
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$ | The side of the large square is $1 + 3 + 1 = 5$, so the area of the large square is $5^2 = 25$.
The area of the middle square is $3^2$, and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$.
Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$. This is the area of the two congruent ... | 7 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square?
\[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 ... | The $3$-digit powers of $5$ are $125$ and $625$, so space $2$ is filled with a $2$.
The only $3$-digit power of $2$ beginning with $2$ is $256$, so the outlined block is filled with
a $\boxed{\text{(D) 6}}$. | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If $a\otimes b = \dfrac{a + b}{a - b}$, then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$ | $6\otimes4=\frac{6+4}{6-4}=5$.
$5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}}$ | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$ | Let the numbers be $x$ and $y$. Then we have $x+y=11$ and $xy=24$. Solving for $x$ in the first equation yields $x=11-y$, and substituting this into the second equation gives $(11-y)(y)=24$. Simplifying this gives $-y^2+11y=24$, or $y^2-11y+24=0$. This factors as $(y-3)(y-8)=0$, so $y=3$ or $y=8$, and the corresponding... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Points $R$, $S$ and $T$ are vertices of an equilateral triangle, and points $X$, $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$ | There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to $\triangle RST$ There is obviously only $... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.
$\text{(A)}\ 1 \qquad \text{(B)}\ 1\frac{1}{2} \... | During the $10$ seconds, the sound traveled $1088\times10=10880$ feet from the lightning to Snoopy. This is equivalent to $\frac{10880}{5280}\approx2$ miles, $\boxed{\text{C}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A [circle](https://artofproblemsolving.com/wiki/index.php/Circle) and two distinct [lines](https://artofproblemsolving.com/wiki/index.php/Line) are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)... | The two [lines](https://artofproblemsolving.com/wiki/index.php/Line) can both [intersect](https://artofproblemsolving.com/wiki/index.php/Intersection) the [circle](https://artofproblemsolving.com/wiki/index.php/Circle) twice, and can intersect each other once, so $2+2+1= \boxed {\text {(D)}\ 5}.$ | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$. If she got six right her score would be $(6*5)-(2*4)=22$. That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$. Thus, our answer is $\boxed{\text{(C)}\ 7}$. ~avamarora | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. The area (in square inches) of the shaded region is
$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\q... | The shaded region is a right trapezoid. Assume WLOG that $YZ=8$. Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$. Because the line $AB$ is a midsegment, the top base of the trapezoid is $\frac12 AB = \frac14 YZ = 2$. Also, $AB$ divides $XC$ in two, so the height of the trapezoi... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$ | The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{\text{(B)}\ 4}$. | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by $6$?
$\textbf{(A)}\ 1/3\qquad\textbf{(B)}\ 1/2\qquad\textbf{(C)}\ 2/3\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 1$ | We have six cases: each different case, every one where a different number cannot be seen. The rolls that omit numbers one through five are all something times six: an example would be where the number you cannot see is one, so the product should be 2 x 3 x 4 x 5 x 6, and so product should be divisible by six. The roll... | 1 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
In this addition problem, each letter stands for a different digit.
$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$
If T = 7 and the letter O represents an even number, what is the only possible value for W?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(... | Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other possibili... | 3 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has $4$ seats: $1$ Driver seat, $1$ front passenger seat, and $2$ back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
$\textbf{(A... | There are only $2$ people who can go in the driver's seat--Bonnie and Carlo. Any of the $3$ remaining people can go in the front passenger seat. There are $2$ people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2\cdot3\cdot2$ or $12$ ways. The answer i... | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many class... | There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\boxed{\textbf{(D)}\ 6}$. | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$, the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$... | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The area of trapezoid $ABCD$ is $164\text{ cm}^2$. The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$, in centimeters?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Using the formula for the area of a trapezoid, we have $164=8(\frac{BC+AD}{2})$. Thus $BC+AD=41$. Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$. From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=... | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$ | Solution 1
If all the children were riding bicycles, there would be $2 \times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\boxed{\mathrm{(C)}\ 5}$ tricycles.
Solution 2
Setting up an equation, we have $a+b=7$ children and $3a+2b=19$. Solving for the variables, w... | 5 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Blake and Jenny each took four $100$-point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's aver... | Solution 1
Blake scored a total of $4 \times 78=312$ points. Jenny scored $10-10+20+20=40$ points higher than Blake, so her average is $\frac{312+40}{4}=88$.
the difference is $88-78=\boxed{\mathrm{(A)}\ 10}$.
Solution 2
The total point difference between Blake's and Jenny's scores is $10-10+20+20=40$. The average of... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes.... | When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its batt... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use [Ball-and-urn](https://artofproblemsolving.com/wiki/index.php/Ball-and-urn) to find the number of possibilities is $\binom{3+3-1}... | 10 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%$. How many of the last $5$ shots did she make?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Sally made $0.55*20=11$ shots originally. Letting $x$ be the number of shots she made, we have $\frac{11+x}{25}=0.56$. Solving for $x$ gives us $x=\boxed{\textbf{(C)}\ 3}$ | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\... | We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$, where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walked. Bec... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$.
\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]
Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\boxed... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many different isosceles triangles have integer side lengths and perimeter 23?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | Let $b$ be the base of the isosceles triangles, and let $a$ be the lengths of the other legs. From this, $2a+b=23$ and $b=23-2a$. From triangle inequality, $2a>b$, then plug in the value from the previous equation to get $2a>23-2a$ or $a>5.75$. The maximum value of $a$ occurs when $b=1$, in which from the first equatio... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.
In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise... | Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," w... | We can start at $200$ and work our way down to $1$. We want to press the button that multiplies by $2$ the most, but since we are going down instead of up, we divide by $2$ instead. If we come across an odd number, then we will subtract that number by $1$. Notice
$200 \div 2 = 100$,
$100 \div 2 = 50$,
$50 \div 2 =... | 9 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$ | Start by buying the largest packs first. After three $24$-packs, $90-3(24)=18$ cans are left. After one $12$-pack, $18-12=6$ cans are left. Then buy one more $6$-pack. The total number of packs is $3+1+1=\boxed{\textbf{(B)}\ 5}$. | 5 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10$ | We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$, which is $\boxed{\textbf{(C)}\ 5}$ values. | 5 | Inequalities | MCQ | Yes | Yes | amc_aime | false |
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven p... | Solution 1
The counting numbers that leave a remainder of $4$ when divided by $6$ are
$4, 10, 16, 22, 28, 34, \cdots$ The counting numbers that leave a remainder of $3$ when
divided by $5$ are $3,8,13,18,23,28,33, \cdots$ So $28$ is the smallest possible number
of coins that meets both conditions. Because 28 is divisib... | 0 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the multiplication problem below $A$, $B$, $C$, $D$ are different digits. What is $A+B$?
\[\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}\]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9$ | $CDCD = CD \cdot 101$, so $ABA = 101$. Therefore, $A = 1$ and $B = 0$, so $A+B=1+0=\boxed{\textbf{(A)}\ 1}$. | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$, $11$, $7$, $12$ and $10$ hours. How many hours must she work for the final week to earn the ticket... | Let $x$ be the number of hours she must work for the final week. We are looking for the average, so
\[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\]
Solving gives:
\[\frac{48 + x}{6} = 10\]
\[48 + x = 60\]
\[x = 12\]
So, the answer is $\boxed{\textbf{(D)}\ 12}$ | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The product of the two $99$-digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$ | We can first make a small example to find out $A$ and $B$. So,
$303\times505=153015$
The ones digit plus thousands digit is $5+3=8$.
Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\boxed{\textbf{(D)}\ 8}$
This is a direct multiplication way. | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the aver... | The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four ... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the sum of the two smallest prime factors of $250$?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12$ | The prime factorization of $250$ is $2 \cdot 5^3$. The smallest two are $2$ and $5$. $2+5 = \boxed{\text{(C) }7}$. | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ ... | The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$. Therefore the number in the lower right-hand square is $\boxed{\textbf{(B)}\ 2}... | 2 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf... | The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{\textbf{(A)}\ 7}$. | 7 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C... | Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$, the third $8/9$, the fourth $16/27$, and the fifth $32/81$. Half of $81$ is $40.5$, so the ball does not reach the required height on bounce $\boxed{\textbf{(C)}\ 5}$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Three $\text{A's}$, three $\text{B's}$, and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qqu... | There are $2$ ways to place the remaining $\text{As}$, $2$ ways to place the remaining $\text{Bs}$, and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$. | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$ | Instead of finding n, we find $x=\frac{n}{3}$. We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number is $100$, so that is our minimum value for $x$, since if $x \in \mathbb{Z^+}$, then $9x \in \mathbb{Z^+}$. The largest three-digit whole number divisible by $9$ is $999$, so our max... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the first year
her investment suffered a $15\%$ loss, but during the second year the remaining
investment showed a $20\%$ gain. Over the two-year period, what was the change
in Tammy's investment?
$\textbf{(A)}\ 5\%\text{ loss}\qquad \textbf{(B)}\ 2\%... | After the $15 \%$ loss, Tammy has $100 \cdot 0.85 = 85$ dollars. After the $20 \%$ gain, she has $85 \cdot 1.2 = 102$ dollars. This is an increase in $2$ dollars from her original $100$ dollars, a $\boxed{\textbf{(D)}\ 2 \%\ \text{gain}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?
$\... | Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$. They can be factored into $11\cdot13$ and $3\cdot5\cdot13$. The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth ... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The letters $A$, $B$, $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$,what digit does $D$ represent?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Because $B+A=A$, $B$ must be $0$.
Next, because $B-A=A\implies0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case.
Now we can rewrite $\begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular}$ as $\begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular}$. Therefore, $D=5+C$
Finally, $A-1-... | 9 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labe... | The areas of the tops of $A$, $B$, $C$, and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$. Now, one of the sides has an area of one, since it combines all of the heights of $A$, $B$, $C$, and $D$, which is $1$. The other side is also the same. Thus the total area now is... | 11 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Let $n$, $n+1$, and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$. Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that:
$n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$. The l... | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of (4,5), but we do lcm of 4,5 = 20. Since we are trying to find the minimum $x$, we must use the smallest possible value for the... | 3 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad... | Let the hundreds, tens, and units digits of the original three-digit number be $a$, $b$, and $c$, respectively. We are given that $a=c+2$. The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$. The hundreds, tens, and units digits of the reversed three-digit number are $c$, $b$, and $... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$. We find the average of this list to get $\boxed{\textbf{(A)}\ 6}$. ( In case you were wondering, the way to calculate the average is $\frac{(10+(-10)+20+30+(-20))}{5} = \frac{ 30}{5} = 6$. So the answer is indeed $\boxed{\textbf{(A... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many digits are in the product $4^5 \cdot 5^{10}$?
$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$ | \[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]
That is one $1$ followed by ten $0$'s, which is $\boxed{\textbf{(D)}\ 11}$ digits. | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the tens digit of $7^{2011}$?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$ | Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic.
\[7^1\equiv 07 \pmod{100}.\]
\[7^2\equiv 49 \pmod{100}.\]
\[7^3\equiv 43 \pmod{100}.\]
\[7^4\equiv 01 \pmod{100}.\]
We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$. Using this, we can say:
\[7^{... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$ | By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{\textbf{(B)}\ 5}$. | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf... | Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}$. | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$ | For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$, since all of the valid 4-digit number will always be greater than $1000$. The best way to solve this problem is by using casework.
There can be only two leading digits, namely $1$ or $2$.
When the... | 9 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$?
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$ | We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$, because of the above statement. Now we need to test the remaining answer choices.
Case 1: $x = 6$
Mode: $6$
Median: $6$
Mean: $\frac{37}{7}$
Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.
Case 2: ... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the units digit of $13^{2012}$?
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9$ | The problem wants us to find the units digit of $13^{2012}$, therefore, we can eliminate the tens digit of $13$, because the tens digit will not affect the final result. So our new expression is $3^{2012}$. Now we need to look for a pattern in the units digit.
$3^1 \implies 3$
$3^2 \implies 9$
$3^3 \implies 7$
$3^4 \im... | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$. Sharona bought some of the same pencils and paid $\textdollar 1.87$. How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05i... | We assume that the price of the pencils remains constant. Convert $\textdollar 1.43$ and $\textdollar 1.87$ to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$, which is $11$. Therefore, Jamar bought $\... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(... | The first answer choice ${\textbf{(A)}\ 3}$, can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$. The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$, which satisfies all the conditions... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | $6$ are blue and green - $b+g=6$
$8$ are red and blue - $r+b=8$
$4$ are red and green- $r+g=4$
We can do trial and error. Let's make blue $5$. That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$. To check this, let's plug $1$ and $3$ into $r+g=4$, which works. Now count the number of marbles - $5+3+1=9$. So t... | 9 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Let $R$ be a set of nine distinct integers. Six of the elements are $2$, $3$, $4$, $6$, $9$, and $14$. What is the number of possible values of the median of $R$?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05... | First, we find that the minimum value of the median of $R$ will be $3$.
We then experiment with sequences of numbers to determine other possible medians.
Median: $3$
Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14$
Median: $4$
Sequence: $-1, 0, 2, 3, 4, 6, 9, 10, 14$
Median: $5$
Sequence: $0, 2, 3, 4, 5, 6, 9, 10, 14$
Median:... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$ | Let the perimeter of the equilateral triangle be $3s$. The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$.
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$, sin... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textb... | The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{\textbf{(A)}\ 1}$ more model car.
~avamarora | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the tre... | We use that fact that $d=rt$. Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.
On Monday, he was at a rate of $5 \text{ m.p.h}$. So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$.
For Wednesday, he walked at a rate of $3 \text{ m.p.h}$. Therefore, $3x = 2 \text{ ... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A sign at the fish market says, "50$\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | 50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{\textbf{(D)}\ 12}$ dollars. | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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