AMC 10 A
AMC 10 B
AMC 12 A(x2)
AMC 12 B(x2)
100
AMC 10A 2021 Problem 1
What is the value of
D=8
This corresponds to 
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1
100
AMC 10B 2017 Problem 3
Real numbers 
, 
, and 
satisfy the inequalities 
, 
, and 
. Which of the following numbers is necessarily positive?
(E)y+z
Notice that 
must be positive because 
. Therefore the answer is 
.
The other choices:
As grows closer to 
, 
decreases and thus becomes less than .
can be as small as possible (
), so 
grows close to as approaches .
For all , 
, and thus it is always negative.
The same logic as above, but when 
this time.
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_3
100
AMC 12A 2021 Problem 1
What is the value of
(B)50
We evaluate the given expression to get that
100
AMC 12B 2004 Problem 1
At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
(A)3
Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made 
free throws, on the third 
, on the second 
, and on the first 
.
Because there are five days, or four transformations between days (day 1 
day 2 day 3 day 4 day 5), she makes 
200
AMC 10A 2021 Problem 8
When a student multiplied the number 
by the repeating decimal, 
where 
and 
are digits, he did not notice the notation and just multiplied times 
Later he found that his answer is 
less than the correct answer. What is the 
-digit number 
(E) 75
We are given that 
from which
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_5
200
AMC 10B 2017 Problem 7
Samia set off on her bicycle to visit her friend, traveling at an average speed of 
kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at 
kilometers per hour. In all it took her 
minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
(C)2.8Let's call the distance that Samia had to travel in total as 
, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both 
, or .
She bikes at a rate of kph, so she travels the distance she bikes in 
hours. She walks at a rate of kph, so she travels the distance she walks in 
hours.The total time is 
. This is equal to 
of an hour. Solving for , we have:
Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about 
.https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7
, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both , or .She bikes at a rate of kph, so she travels the distance she bikes in hours. She walks at a rate of kph, so she travels the distance she walks in hours.The total time is . This is equal to of an hour. Solving for , we have:Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about .https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7200
AMC 12B 2021 Problem 6
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is 
. When 
black cards are added to the deck, the probability of choosing red becomes 
. How many cards were in the deck originally?
(C)12
If the probability of choosing a red card is 
, the red and black cards are in ratio 
. This means at the beginning there are red cards and black cards.
After black cards are added, there are 
black cards. This time, the probability of choosing a red card is 
so the ratio of red to black cards is 
. This means in the new deck the number of black cards is also 
for the same red cards.
So, 
and 
meaning there are red cards in the deck at the start and 
black cards.
So, the answer is 
.
200
AMC 12B 2004 Problem 5
On a trip from the United States to Canada, Isabella took 
U.S. dollars. At the border she exchanged them all, receiving 
Canadian dollars for every 
U.S. dollars. After spending 
Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ?
(A)5
Isabella had 
Canadian dollars. Setting up an equation we get 
, which solves to 
, and the sum of digits of is 
.
300
AMC 10A 2011 Problem 15
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 
miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 
gallons per mile. On the whole trip he averaged 
miles per gallon. How long was the trip in miles?
(C) 440
We know that 
. Let be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is 
. The total distance traveled is 
, so we get 
. Solving this equation, we get 
, so the total distance is 
.
300
AMC 10B 2015 Problem 12
For how many integers is the point 
inside or on the circle of radius centered at 
?
(A)11
The equation of the circle is 
. Plugging in the given conditions we have 
. Expanding gives: 
, which simplifies to 
and therefore 
and 
. So ranges from 
to , for a total of 
integer values.
Note by Williamgolly: Alternatively, draw out the circle and see that these points must be on the line 
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_12
300
AMC 12A 2021 ProblemOf the following complex numbers , which one has the property that 
has the greatest real part?
, which one has the property that has the greatest real part?
First, is 
, is 
, is 
.
Taking the real part of the th power of each we have:
, whose real part is
Thus, the answer is .
300
5613 - X9 = 5211
The subscript means base. e.g (56 base 13).
What is X?
37.
56 base 13 = 5(13) + 6(1) = 91.
52 base 11 = 5(11) + 2(1) = 57
X = 34 base 10.
37 base 9 = 3(9) + 7(1) = 34 base 10.
400
For some positive integer n, the number 110n3 has 110 positive integer divisors, including 1 and the number 110n3. How many positive integer divisors does the number 81n4 have? (2016 AMC 10A Problem 22)
325.
Since the prime factorization of 
is 
, we have that the number is equal to 
. This has 
factors when 
. This needs a multiple of 11 factors, which we can achieve by setting 
, so we have 
has factors. To achieve the desired factors, we need the number of factors to also be divisible by , so we can set 
, so 
has factors. Therefore, . In order to find the number of factors of 
, we raise this to the fourth power and multiply it by 
, and find the factors of that number. We have 
, and this has 
factors.
400
AMC 10B 2006 Problem 19
Let 
be a sequence for which 
, 
, and 
for each positive integer 
. What is 
?
(E)3
Looking at the first few terms of the sequence:
Clearly, the sequence repeats every 6 terms.
Since 
,
400
AMC 12A 2009 Problem 20
Convex quadrilateral 
has 
and 
. Diagonals 
and 
intersect at 
, 
, and 
and 
have equal areas. What is 
?
(E)6
Let 
denote the area of triangle 
. 
, so 
. Since triangles 
and share a base, they also have the same height and thus 
and 
with a ratio of 
. 
, so 
400
AMC 12B 2004 Problem 12
In the sequence 
, 
, 
, 
, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 
. What is the 
term in this sequence?
(C)0
We already know that 
, 
, 
, and 
. Let's compute the next few terms to get the idea how the sequence behaves. We get 
, 
, 
, and so on.
We can now discover the following pattern: 
and 
. This is easily proved by induction. It follows that 
.
600
AMC 10A 2007 Problem 25
For each positive integer 
, let 
denote the sum of the digits of 
For how many values of is 
(D)4
For the sake of notation let 
. Obviously 
. Then the maximum value of 
is when 
, and the sum becomes 
. So the minimum bound is 
. We do casework upon the tens digit:
Case 1: 
. Easy to directly disprove.
Case 2: 
. 
, and 
if 
and 
otherwise.
Subcase a: 
. This exceeds our bounds, so no solution here.Subcase b: 
. First solution.
Case 3: 
. 
, and 
if 
and 
otherwise.
Subcase a: 
. Second solution.Subcase b: 
. Third solution.
Case 4: 
. But 
, and 
clearly sum to 
.
Case 5: 
. So 
and 
(recall that 
), and 
. Fourth solution.
In total we have 
solutions, which are 
and .
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22
600
How many ways can the statement fall math madness be arranged. The arrangements do not how to be actual words. The first letter must be 4 letters and have a F,A,L,L and same for the next two words.
725760.
We can arrange a word in n!/a1! + a2! +...
where n is the length of the word and a1 and so on are the number of times a letter is repeated for a letters.
600
AMC 12A 2009 Problem 25
The first two terms of a sequence are 
and 
. For 
,
What is 
?
(A)0
Consider another sequence 
such that 
, and 
.
The given recurrence becomes
It follows that 
. Since 
, all terms in the sequence will be a multiple of 
.
Now consider another sequence 
such that 
, and 
. The sequence 
satisfies 
.
As the number of possible consecutive two terms is finite, we know that the sequence is periodic. Write out the first few terms of the sequence until it starts to repeat.
Note that 
and 
. Thus 
has a period of 
: 
.
It follows that 
and 
. Thus 
Our answer is 
.
600
AMC 12B 2003 Problem 21
An object moves 
cm in a straight line from 
to 
, turns at an angle 
, measured in radians and chosen at random from the interval 
, and moves cm in a straight line to 
. What is the probability that 
?
(D)1/3
It follows that 
, and the probability is 
.