2022 AMC 12 A
MATH & SCIENCE ACADEMY
2022 AMC12 A
Just Calculating
Answer : D
Let three numbers be 𝑎, 𝑏, 𝑐.
𝑎 + 𝑏 + 𝑐 = 96, 𝑎 = 3𝑐, 𝑐 = 𝑏 − 40.
Answer : E
Sum of Areas of the five rectangles : 6 + 8 + 30 + 14 + 6 = 64.
So the length of a side of the square is 8.
Sum of all the sides of the outside 4 rectangles is 8 × 4 × 2 = 64.
Sum of all the sides of the given 5 rectangles is (7 + 6 + 11 + 9 + 5) × 2 = 76.
So, the length of all the sides of the rectangle in the middle is 12.
That is, the rectangle in the middle has dimensions 2 × 4.
Answer : B
2
MATH & SCIENCE ACADEMY
2022 AMC12 A
180 = 22 × 32 × 5, 18 = 2 × 32 .
So, 𝑛 should be at least 22 × 5 and at most 22 × 32 × 5.
45 = 32 × 5. So n should have factors 3 and 5, but the power of 3 should be 1.
Therefore, 𝑛 = 22 × 3 × 5 = 60.
Answer : B
Taxicab Distance (Manhattan Distance)
is the shortest length of paths between two points when
you move along horizontal and vertical segments only,
without ever going back, like those described by a car
moving in a lattice-like street pattern.
So, the distance between points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) in the
coordinate plane is given by 𝑥2 − 𝑥1 + 𝑦2 − 𝑦1 .
This problem wants to count the number of points with
integer coordinates (𝑥, 𝑦) such that 𝑥 + 𝑦 ≤ 20.
Let’s look at the quadrant I only.
There are 1, 2, 3, ... , 19 points in each horizontal line
where 𝑦 = 19, 18, ⋯ , 1 respectively.
So, the number of points on quadrant I is
1 + 2 + ⋯ + 19 = 190.
The number of points on all 4 quadrants is 760.
And there are some points on the 𝑥- and 𝑦-axis.
The number of points on 𝑥-axis is 41.
Of course there are 41 points on 𝑦-axis.
But we counted the origin twice.
So the answer is 760 + 41 + 41 − 1 = 841.
3
Answer : C
MATH & SCIENCE ACADEMY
2022 AMC12 A
20 + 𝑋 = 6 × 1 : impossible
20 + 𝑋 = 6 × 2 : impossible
20 + 𝑋 = 6 × 5 : 𝑋 = 10
20 + 𝑋 = 6 × 7 : 𝑋 = 22
20 + 𝑋 = 6 × 𝑋 : 𝑋 = 4
So the sum of all positive values of X is 36.
Answer : D
Let’s name the regions as the right figure.
A
Theoretically, we can assign colors in any order.
But if you proceed C-A-B-D-E, you will get into a
C
D
complicated situation when allocate color to D.
If B and C were painted the same color, D has 3 choices.
But B and C had different colors, D has only 2 choices.
Of course we can count the possible ways in this case, but so annoying.
B
E
When we solve the coloring problem like this, start with the region with the most
contacts. So, for this problem, proceeding D-A-B-C-E order will be the best solution.
So the answer is 5 × 4 × 3 × 3 × 3 = 540.
Answer : D
4
MATH & SCIENCE ACADEMY
2022 AMC12 A
Infinite Geometric Series
For geometric sequence 𝑎𝑛 = 𝑎𝑟 𝑛−1
the partial sum 𝑆𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 =
𝑎 1−𝑟 𝑛
1−𝑟
.
If the absolute value of the common ratio of the geometric sequence is more than
or equal to 1, the sum of the infinite geometric series does not exist.
When 𝑟 < 1, the infinite geometric series exists, and the value is
𝑎
𝑆=
1−𝑟
3
10 ×
3 3
10 ×
3 3
3
1
1 1 1
1
+ +
+⋯= 3 = .
1
3 9 27
1−3 2
1
So the answer is 102 = 10.
5
1
1
1
1 1
1
10 × ⋯ = 103 × 109 × 1027 × ⋯ = 103+9+27+⋯
Answer : A
MATH & SCIENCE ACADEMY
2022 AMC12 A
Let the number of truth-teller be T, lier L, alternater choosing the truth as their first
response AT, choosing a lie AL.
22 children who answered yes for the principal’s first question are truth-tellers, liars
and alternaters choosing a lie as their first response. So T+L+AL=22.
15 children who answered yes for the principal’s second question are liars and
alternaters choosing a lie as their first response. So L+AL=15.
9 children who answered yes for the principal’s third question are alternaters
choosing the truth as their first response. So AT=9.
T=7. (The problem is not so good.)
Answer : A
Let the greater number be G, and the lesser number L.
The numbers 8 through 14 cannot be a L.
So, the numbers 1 through 7 are L’s and the numbers 8 through 14 are G’s.
Let L’s pick their G pair.
7 can pick only 14 as its G : 1 choice.
6 can pick among 12, 13 : 2 choices.
5 can pick among 10, 11, 12, 13 except 6’ G : 3 choices.
4 can pick among 8, 9, 10, 11, 12, 13 except 6’, and 5’ Gs : 4 choices.
3 can pick among 8, 9, 10, 11, 12, 13 except 6’, 5’, and 4’ Gs : 3 choices.
2 can pick among 8, 9, 10, 11, 12, 13 except 6’, 5’, 4’ and 3’ Gs : 2 choices.
1 can pick among 8, 9, 10, 11, 12, 13 except 6’, 5’, 4’, 3’ and 2’ Gs : 1 choices.
So, the answer is 1 × 2 × 3 × 4 × 3 × 2 × 1 = 144
Answer : E
6
MATH & SCIENCE ACADEMY
2022 AMC12 A
log 6 𝑥 − log 6 9 = 2 log 6 10 − 1
2
𝑥
5
25
log 6 = ± log 6
,
𝑥 = 9⋅ ,
9
3
9
So, the product is 81.
9⋅
9
25
Answer : E
Law of Cosines
For a triangle ABC
𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
𝑏 2 = 𝑐 2 + 𝑎2 − 2𝑐𝑎 cos 𝐵
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶
Let the length of a edge be 2𝑎 and ∠𝐶𝑀𝐷 be 𝜃.
𝐶𝑀 = 𝑀𝐷 = 3𝑎.
𝐶𝐷 = 2𝑎
By the Law of Cosines,
𝐶𝑀2 + 𝑀𝐷2 + 𝐶𝐷2 3𝑎2 + 3𝑎2 − 4𝑎2 1
cos 𝜃 =
=
=
3
2 ⋅ 𝐶𝑀 ⋅ 𝑀𝐷
2 ⋅ 3𝑎 ⋅ 3𝑎
Answer : B
7
MATH & SCIENCE ACADEMY
2022 AMC12 A
Complex Plane
is the plane formed by the complex numbers, with a
Cartesian coordinate system, such that the 𝑥-axis (real axis)
is formed by the real numbers, and the 𝑦-axis (imaginary
axis) is formed by the imaginary numbers.
A complex number 𝑎 + 𝑏𝑖 can be represented by a point
at (𝑎, 𝑏) in the complex plane.
All the 𝑧1 form a line segment which ends at two points (3,0)
and (0,4) in the Cartesian Coordinate system.
All the 𝑧2 form a circle with radius 1 and inside the circle.
So, 𝑧 = 𝑧1 + 𝑧2 form the shape in the right figure.
The area is 𝜋 + 5 × 2 ≈ 13.14.
Answer : A
Let log 2 = 𝑥. Then
1 − 𝑥 3 + 1 + 𝑥 3 + 3𝑥 × −2𝑥
= 1 − 3𝑥 + 3𝑥 2 − 𝑥 3 + 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 − 6𝑥 2 = 2
Answer : C
8
MATH & SCIENCE ACADEMY
2022 AMC12 A
Let 𝑃 𝑥 = 10𝑥 3 − 39𝑥 2 + 29𝑥 − 6.
If ℎ, 𝑙, 𝑤 are three roots of the polynomial 𝑃(𝑥), then
𝑃 𝑥 = 10 𝑥 − ℎ 𝑥 − 𝑙 𝑥 − 𝑤 .
The volume of the new rectangular box V is
1
1
300
𝑉 = ℎ + 2 𝑙 + 2 𝑤 + 2 = − 𝑃 −2 =
80 + 156 + 58 + 6 =
= 30
10
10
10
Answer : D
Triangular Number
is simply the number of dots in each equilateral
triangular pattern.
Triangular Number 𝑡𝑛 can be obtained by
𝑛 𝑛+1
𝑡𝑛 = 1 + 2 + 3 + ⋯ + 𝑛 =
2
If 𝑡𝑛 is a perfect square, one of two things must occur. Either
be squares or 𝑛 and
𝑛+1
2
𝑛
2
and 𝑛 + 1 must both
must both be squares(*). That is 𝑛 or 𝑛 + 1 must be squares.
Also if 𝑛 is a square, 𝑛 must be an odd number because
𝑛+1
2
must be a square too.
𝑛
And if 𝑛 + 1 is a square, 𝑛 must be an even number because 2 must be a square too.
So, we only need to check odd and square numbers.
𝑛
𝑛+1
For 𝑛 > 49, the first odd and square number is 81. But both 2 and 2 are not squares.
𝑛
112 = 121, 132 = 169, 152 = 225 are all invalid. But for 172 = 289, 𝑛 = 288, 2 = 122 .
So, the fourth smallest triangular number that is also a perfect square is 𝑡288 .
288⋅289
𝑡288 = 2 = 41 616. The answer is 4 + 1 + 6 + 1 + 6 = 18.
Answer : D
𝑛
𝑛
(*) If 2 and 𝑛 + 1 are not both be squares, the two numbers must be of the form 2 =
𝑎𝑏 2 , 𝑛 + 1 = 𝑎𝑐 2 . Then 𝑛 = 2𝑎𝑏 2 , 2𝑎𝑏 2 + 1 = 𝑎𝑐 2 , 𝑎 𝑐 2 − 2𝑏 2 = 1. The only
solution is 𝑎 = 1, 𝑐 2 − 2𝑏 2 = 1.
9
MATH & SCIENCE ACADEMY
2022 AMC12 A
Sum and Double Angle Formulas of Trigonometry
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 ,
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 ,
tan 𝐴 + tan 𝐵
tan 𝐴 + 𝐵 =
,
1 − tan 𝐴 tan 𝐵
sin 2𝐴 = 2 sin 𝐴 cos 𝐴
cos 2𝐴 = cos2 𝐴 − sin2 𝐴
2 tan 𝐴
tan 2𝐴 =
1 − tan2 𝐴
LHS : 𝑎 sin 𝑥 + sin 2𝑥 = 𝑎 sin 𝑥 + 2 sin 𝑥 cos 𝑥 = 𝑎 sin 𝑥 (1 + 2 cos 𝑥)
RHS : sin 3𝑥 = sin 𝑥 + 2𝑥 = sin 𝑥 cos 2𝑥 + cos 𝑥 sin 2𝑥
= sin 𝑥 cos2 𝑥 − sin2 𝑥 + 2 sin 𝑥 cos 2 𝑥 = sin 𝑥 3 cos 2 𝑥 − sin2 𝑥
= sin 𝑥 4 cos2 𝑥 − 1 = sin 𝑥 (2 cos 𝑥 + 1)(2 cos 𝑥 − 1)
𝑎 sin 𝑥 (1 + 2 cos 𝑥) = sin 𝑥 (2 cos 𝑥 + 1)(2 cos 𝑥 − 1)
Since sin 𝑥 ≠ 0 on the interval (0, 𝜋), we can divide out sin 𝑥 from both sides, and get
𝑎 1 + 2 cos 𝑥 = 4 cos2 𝑥 − 1.
4 cos2 𝑥 − 2𝑎 cos 𝑥 − 𝑎 − 1 = 0
𝑎 ± 𝑎2 + 4𝑎 + 4 𝑎 ± (𝑎 + 2)
cos 𝑥 =
=
4
4
Now the equation is changed as the following :
𝑎+1
1
cos 𝑥 =
,
cos 𝑥 = −
2
2
Since the second equation has one solution in the interval, the first equation should
have more than one solution other than the solution of the second solution. So
𝑎+1
1
1 𝑎+1
−1 <
<− ,
− <
<1
2
2
2
2
−3 < 𝑎 < −2,
−2 < 𝑎 < 1
That is, the set of all 𝑎 is −3, −2 ∪ (−2,1).
So the answer is −3 − 2 + 1 = −4.
Answer : A
10
MATH & SCIENCE ACADEMY
2022 AMC12 A
Polar Coordinate System
In two dimensions, the Cartesian coordinates (𝑥, 𝑦) specify
the location of a point P in the plane. Another two-dimensional
coordinate system is polar coordinates. Instead of using the
signed distances along the two coordinate axes, polar
coordinates specifies the location of a point P in the plane
by its distance 𝑟 from the origin and the angle 𝜃 made between the line segment
from the origin to P and the positive 𝑥-axis. The polar coordinates (𝑟, 𝜃) of a point
P are illustrated in the above figure.
For this problem, we will use the polar coordinate system.
Let the initial point be 𝑃(1, 0°) and 𝑃𝑛 be the point rotated by the sequence of
transformations 𝑇1 , 𝑇2 , ⋯ , 𝑇𝑛 .
Then the distance 𝑟 of all the points is 1, unchanged. Only the angle 𝜃 will be changed.
Let 𝜃𝑛 be the angle of 𝑃𝑛 in degrees. Then the angles of the points will be as follows :
𝜃1 = 179, 𝜃2 = −1,
𝜃3 = 178, 𝜃4 = −2,
𝜃5 = 177, 𝜃6 = −3,
𝜃7 = 176, 𝜃8 = −4,
⋯
So, 𝜃2𝑘−1 = 180 − 𝑘, 𝜃2𝑘 = −𝑘.
We want to know when the angle will be 0. And the answer is when 𝑘 = 180.
So the answer is 2 × 180 − 1 = 359.
Answer : A
11
MATH & SCIENCE ACADEMY
2022 AMC12 A
Let number the spots that the cards are placed as 1, 2, 3, ..., 13 from left to right.
Suppose we picked 3 cards on the first pass. It means that the number 1,2,3 cards
were placed in ascending order any 3 spots except the 1,2,3 spots. And the number
4~13 cards was placed in ascending order at the remaining spots.
In this case, the number of possible ways to arrange the cards is 13𝐶3 − 1.
We can pick between 1 and 12 cards on the first pass. So, the number of possible
ways is
12
෍
𝑘=1
12
13𝐶𝑘
13
− 1 = ෍ 13𝐶𝑘 − 12 = ෍ 13𝐶𝑘 − 14 = 213 − 14 = 8192 − 14 = 8178
𝑘=1
𝑘=0
Answer : D
Let the coordinates of A and D be (-a,0) and (a,0).
Let the height of the trapezoid be h, and let the
coordinates of B and D be (-b,h) and (b,h).
Let P be located at point (c,d).
By the distance formula,
𝑐 + 𝑎 2 + 𝑑2 = 1, 𝑐 − 𝑎 2 + 𝑑2 = 16
𝑐 + 𝑏 2 + 𝑑 − ℎ 2 = 4, 𝑐 − 𝑏 2 + 𝑑 − ℎ 2 = 9
4𝑎𝑐 = −15, 4𝑏𝑐 = −5 ⇒ 𝑎 = 3𝑏
𝐵𝐶 2𝑏 1
=
=
𝐴𝐷 2𝑎 3
B
C
P
A
D
Answer : B
12
MATH & SCIENCE ACADEMY
2022 AMC12 A
(A)The remainder of P(x) divided by 𝑥 2 − 𝑥 + 1 is the same as the remainder of P(x)
divided by 𝑥 3 + 1. The remainder of P(x) divided by 𝑥 3 + 1 is obtained by
substitute −1 into 𝑥 3 . 𝑃 𝑥 = 𝑥 3 674 + 𝑥 3 337 + 1. So the remainder is 1.
(B)The remainder of P(x) divided by 𝑥 2 + 𝑥 + 1 is the same as the remainder of P(x)
divided by 𝑥 3 − 1. The remainder is 3.
(C) The remainder of P(x) divided by 𝑥 4 + 1 is obtained by substitute −1 into 𝑥 4 .
𝑃 𝑥 = 𝑥 4 505 𝑥 2 + 𝑥 4 252 𝑥 3 + 1. So the remainder is 𝑥 3 + 𝑥 2 + 1.
(D)The remainder of P(x) divided by 𝑥 6 − 𝑥 3 + 1 is the same as the remainder of P(x)
divided by 𝑥 9 + 1. The remainder of P(x) divided by 𝑥 9 + 1 is obtained by
substitute −1 into 𝑥 9 . 𝑃 𝑥 = 𝑥 9 224 𝑥 6 + 𝑥 9 112 𝑥 3 + 1 = 𝑥 6 + 𝑥 3 + 1. So the
remainder is 2𝑥 3 .
(E)The remainder of P(x) divided by 𝑥 6 + 𝑥 3 + 1 is the same as the remainder of P(x)
divided by 𝑥 9 − 1. The remainder of P(x) divided by 𝑥 9 − 1 is obtained by
substitute 1 into 𝑥 9 . 𝑃 𝑥 = 𝑥 9 224 𝑥 6 + 𝑥 9 112 𝑥 3 + 1 = 𝑥 6 + 𝑥 3 + 1. So the
remainder is 0.
Answer : E
13
MATH & SCIENCE ACADEMY
2022 AMC12 A
According to the Vieta’s Formula, 𝑧1 + 𝑧2 = 𝑐, 𝑧1 𝑧2 = 10.
According to the Conjugate Root Theorem, 𝑧1 = 𝑧2 .
𝑧1 = 10, 𝑅𝑒 𝑧1 =
𝑐
2
𝑧1 = 10 cos 𝜃 + 𝑖 sin 𝜃 , 𝑧2 = 10 cos 𝜃 − 𝑖 sin 𝜃
1
1
1
1
=
cos 𝜃 − 𝑖 sin 𝜃 , =
(cos 𝜃 + 𝑖 sin 𝜃)
𝑧1
𝑧2
10
10
1
1
Let the points of 𝑧1 , 𝑧2 , 𝑧 , 𝑧 be A, B, C, D, respectively.
1
2
2
𝐴𝐵 = 2 10 sin 𝜃 , 𝐶𝐷 =
10
sin 𝜃 , ℎ =
10 −
1
10
cos 𝜃
the Area of Q :
1
2
1
1
99
2 10 +
sin 𝜃 ⋅ 10 −
cos 𝜃 = 10 −
sin 𝜃 cos 𝜃 =
sin 2𝜃
2
10
10
10
10
𝜋
The area of Q has its maximum possible value when 𝜃 = 4 .
𝜋
𝑐 = 2 𝑅𝑒 𝑧1 = 2 × 10 cos = 20 ≈ 4.47
4
Answer : A
Let 𝑧1 = 10(𝑎 + 𝑏𝑖) where 𝑎2 + 𝑏 2 = 1. then
1
1
𝑧2 = 10 𝑎 − 𝑏𝑖 ,
=
𝑎 − 𝑏𝑖 ,
𝑧1
10
1
1
2
1
1
1
=
(𝑎 + 𝑏𝑖)
𝑧2
10
Let the points of 𝑧1 , 𝑧2 , 𝑧 , 𝑧 be A, B, C, D, respectively.
𝐴𝐵 = 2 10𝑏, 𝐶𝐷 =
the Area of Q :
1
2
2 10 +
𝑏⋅
2
10
2
10
𝑏, ℎ =
10 −
1
10 −
1
10
𝑎 = 10 −
𝑎
1
99
𝑎𝑏 =
𝑎𝑏
10
10
10
Since 𝑎 + 𝑏 = 1, 𝑎𝑏 has its maximum when 𝑎 = 𝑏 =
1
𝑐 = 2 𝑅𝑒 𝑧1 = 2 × 10 ×
= 20 ≈ 4.47
2
2
14
2
1
.
2
MATH & SCIENCE ACADEMY
2022 AMC12 A
1 1 3
+ = , 𝑘 = 2, 𝐿2 = 2
1 2 2 2
1 1 1 6 + 3 + 2 11
+ + =
=
, 𝑘 = 6, 𝐿3 = 6
1 2 3
6
6 3
1 1 1 1 12 + 6 + 4 + 3 25
+ + + =
=
, 𝑘 = 12, 𝐿4 = 12
1 2 3 4
12
12 4
1 1 1 1 1 60 + 30 + 20 + 15 + 12 137
+ + + + =
=
, 𝑘 = 60, 𝐿5 = 60
1 2 3 4 5
60
60 5
1 1 1 1 1 1 60 + 30 + 20 + 15 + 12 + 10 147 49
+ + + + + =
=
=
, 𝑘 = 20, 𝐿6 = 60
1 2 3 4 5 6
60
60
20 6
ℎ𝑛 ℎ𝑛−1 1
−
=
𝑘𝑛 𝑘𝑛−1 𝑛
According to the Vieta’s Formula, 𝑧1 + 𝑧2 = 𝑐, 𝑧1 𝑧2 = 10.
According to the Conjugate Root Theorem, 𝑧1 = 𝑧2 .
15
Answer : A
MATH & SCIENCE ACADEMY
2022 AMC12 A
Case 1 : The string has 1 different digit : only one possibility. 00000
1
Case 2 : The string has 2 different digits
00001, 00002, 00003, 00004
: 4 × 5 = 20
00011, 00022, 00033
: 3 × 5𝐶2 = 30
00111, 00222
: 2 × 5𝐶3 = 20
01111
:5
75
Case 3 : The string has 3 different digits
00012, 00013, 00014, 00023, 00024, 00034 : 10 × 5 × 4 = 200
00112, 00113, 00114, 00223, 00224
00122, 00133,
00233
: 10 × 5!/2! 2! = 300
01112, 01113, 01114
01122, 01133
01222
500
Case 4 : The string has 4 different digits
00123, 00124, 00134, 00234
01123, 01124, 01134
01223, 01224
01233
: 10 × 5!/2! = 600
600
Case 5 : The string has 5 different digits : 01234 : 5! = 120
120
So the answer is 1296.
Answer : E
16
MATH & SCIENCE ACADEMY
2022 AMC12 A
We can think the situation in two cases.
Case 1 : The circle is an incircle. (𝑎, 𝑏 > 2𝑟)
𝑐 = (𝑎 − 𝑟) + (𝑏 − 𝑟) = 𝑎 + 𝑏 − 2𝑟
𝑎2 + 𝑏 2 = 𝑐 2 = 𝑎 + 𝑏 − 2𝑟 2
4𝑟 2 − 4𝑎𝑟 − 4𝑏𝑟 + 2𝑎𝑏 = 0
𝑎 − 2𝑟 𝑏 − 2𝑟 = 2𝑟 2
Case 2 : The circle is an excircle. (0 < 𝑎, 𝑏 < 𝑟)
𝑐 = 𝑟 − 𝑎 + 𝑟 − 𝑏 = 2𝑟 − 𝑎 − 𝑏
𝑎2 + 𝑏 2 = 𝑐 2 = 2𝑟 − 𝑎 − 𝑏 2
4𝑟 2 − 4𝑎𝑟 − 4𝑏𝑟 + 2𝑎𝑏 = 0
𝑎 − 2𝑟 𝑏 − 2𝑟 = 2𝑟 2
Oh! The same equation is obtained.
We want the smallest 𝑟 such that this equation has at least 14 distinct pairs (𝑎, 𝑏),
where 𝑎 and 𝑏 are integers.
Let 𝑎 − 2𝑟 = 𝑢, 𝑏 − 2𝑟 = 𝑣, then the equation becomes 𝑢𝑣 = 2𝑟 2 .
Since 2𝑟 2 is not a perfect square, the number of positive (𝑢, 𝑣) is equal to the number
of positive divisor of 2𝑟 2 . All these solutions are the solutions in Case 1.
For each feasible positive solution (𝑢, 𝑣), its opposite (−𝑢, −𝑣) is also a solution.
However, since 𝑎 and 𝑏 are both positive integers, some of (−𝑢, −𝑣)’s are not feasible
solutions. And All these feasible solutions are the solutions in Case 2.
To get at least 14 distinct pairs (𝑎, 𝑏), 2𝑟 2 should have about 10~14 positive factors.
Since we want the smallest 𝑟, we can select 72 as 2𝑟 2 , where 𝑟 = 6.
72 has 12 positive factors : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Corresponding (𝑢, 𝑣) pairs are (1, 72), (2, 36), ...
and corresponding (𝑎, 𝑏) pairs are (13, 84), (14, 48), ...
For the case 2, only (-8, -9) and (-9, -8) pairs for (𝑢, 𝑣) can give feasible (𝑎, 𝑏) pairs,
which are (4, 3) and (3, 4).
So, we have got 14 distinct pairs (𝑎, 𝑏).
The maximum value of 𝑐 can be obtained when (𝑎, 𝑏) pair is (13, 84), and the
minimum value of 𝑐 can be obtained when (𝑎, 𝑏) pair is (3,4).
That is, 𝑐14 = 132 + 842 = 85, 𝑐1 = 32 + 42 = 5. So the answer is 17.
Answer : E
17
MATH & SCIENCE ACADEMY