Mock AMC 12 Solutions: Scrabbler94 L TEX Template By: Scrabbler94
Mock AMC 12 Solutions: Scrabbler94 L TEX Template By: Scrabbler94
Mock AMC 12 Solutions: Scrabbler94 L TEX Template By: Scrabbler94
scrabbler94
LATEX template by: scrabbler94
2. Answer: (A) Suppose there are 3x boys and 4x girls in the class.
Then 14 · 3x = 43 x boys have a pet, and 13 · 4x = 43 x girls have a pet.
Then there are 34 x + 34 x = 25
12 x sophomores who have a pet. The
percentage of boys is
3 3
4x 4 9
25 = 25 = = (A) 36%
12 x 12
25
1
2020 Mock AMC 12 A Solutions 2
..
.
... ...
..
.
8. Answer: (D) Extend the side lengths of the triangles to form square
AP QC of side length 7 as shown, where B is contained on P Q:
A
4 3
3
4
P
C
4
B 3
3 4
Q
9. Answer: (D) It is easy to check that Joanna can make all monetary
amounts of the form $5k, where k = 0, 1, . . . , 74, using denominations
which are multiples of 5. Using one or two $1 bills, she can also make
the amounts $5k +1 and $5k +2. Thus, there are 75 possible values for
k, each of which gives 3 different monetary amounts, giving 75 × 3 =
225 different amounts. However this includes $0 which is invalid (since
one or more bills is needed), so the answer is 225 − 1 = (D) 224 .
10. Answer: (B) If the √ ant crawls along the red path, as shown below,
the ant travels 2 + 3 units:
A better solution is for the ant to crawl along two square faces as
shown:
B
2020 Mock AMC 12 A Solutions 5
Unfolding the two square faces, we see that the length of the path
equals the hypotenuse of a right triangle with leg lengths
√ 2 and 3;√by
the Pythagorean theorem, the distance crawled is 22 + 32 = 13
units. However, the optimum solution is to crawl along one square
face and one triangular face:
2 A
r √ 2
3 2 3
We see that the length of the dotted path is 2 + 2+ 2 =
√
q
(B) 7 + 2 3 .
11. Answer: (D) By the change of base formula, we can rewrite log8 (a3 )
3
as log 2 (a )
log2 8 =
3 log2 a
3 = log2 a. Thus, the given equation simplifies to a
cubic in log2 a, namely 8(log2 a)3 − 8 log2 a − 1 = 0.
We observe that the cubic polynomial 8x3 − 8x − 1 has three distinct
real roots. This can be shown by looking at the graph of y = 8x3 −8x =
3
√−x), which
8(x √
has roots 0, ±1 as well as local minima/maxima at x =
3 16 3
± 3 , ∓ 9 , in which shifting the graph down by 1 will still result
in a graph which crosses the x-axis at exactly three distinct points
(note that the computation of the local extrema is not needed; we
can also observe that y ± 12 = ∓3, which yields the same conclusion).
Thus, each real root of the cubic 8x3 −8x−1 gives exactly one solution
2020 Mock AMC 12 A Solutions 6
x ≡ a (mod 2)
x ≡ b (mod 3)
x ≡ c (mod 5)
13. Answer: (D) Observe that the prime factorization of 2021 is 43 × 47;
this can be seen easily from the difference of squares factorization
2021 = 452 − 22 = (45 − 2)(45 + 2).
Using Legendre’s formula, we have 2021! = 4348 × 4743 × K, where K
is not divisible by 43 or 47. Also, we have 2020! = 4347 × 4742 × K.
In order for lcm(n, 2020!) = 2021!, the smallest possible n is n =
4348 × 4743 , and the number of factors of n is (48 + 1)(43 + 1) =
49 × 44 = (D) 2156 .
three such pairs sum to less than or equal to (3, 3) (where (x, y) ≤ (3, 3)
if and only if x ≤ 3 and y ≤ 3).
Suppose these three pairs are (a1 , b1 ), (a2 , b2 ), (a3 , b3 ). By stars and
bars, the number of non-negative integer solutions to a1 + a2 + a3 ≤ 3
equals the number of non-negative solutions to a1 + a2 + a3 + s = 3,
which is 63 = 20 (and similarly b1 + b2 + b3 ≤ 3). Any such 6-
y
ω2
`2
`1
ω1
x
We claim that P is the point (3, 4). Let A and B be the points where
`1 is tangent to circles ω1 and ω2 , respectively. By a simple AAA
argument, 4P AO1 ∼ 4P BO2 ; since the radius of ω2 is twice that of
ω1 , the ratio P O1 : P O2 is 1 : 2. This implies `1 intersects O1 O2 at
(3, 4). Similarly, `2 intersects O1 O2 at (3, 4), so P = (3, 4).
For simplicity, shift everything by (−3, −4) so that P is now the origin.
Thus, we have two lines passing through the origin which are tangent
to the circle (x + 3)2 + (y + 4)2 = 1, for which we can determine the
possible slopes. Let y = mx where m is the slope of `1 or `2 . Then
the following quadratic must have exactly one real solution x in order
for y = mx to be tangent:
2020 Mock AMC 12 A Solutions 8
Let m1 and m2 be the two real solutions of the above quadratic, cor-
responding to the slopes of `1 and `2 . By simple application of Vieta’s
formulas, we obtain m1 + m2 = 24 8 = 3.
Back in the original setting (before shifting by (−3, −4)), we have that
both lines intersect at P = (3, 4). Thus, if the equations of `1 and `2
are y = m1 x + b1 and y = m2 x + b2 , we have
4 = 3m1 + b1
4 = 3m2 + b2
16. Answer: (D) Let ak denote the largest real solution to the equation
f k (x) = 0, so that x1 = a2020 . We observe a pattern:
√
k=1: x2 − 20 = 0 =⇒ a1 = 20
√
q
2 2
k=2: (x − 20) − 20 = 0 =⇒ a2 = 20 + 20
r
√
q
k=3: ((x2 − 20)2 − 20)2 − 20 = 0 =⇒ a3 = 20 + 20 + 20
..
.
√
(provable inductively), and that 20 + ak > ak iff 20 + ak > a2k iff
(ak − 5)(ak + 4) < 0. As 4 <qak < 5, this inequality holds, so (ak )
p √
is strictly increasing. Let a = 20 + 20 + 20 + . . .; squaring both
sides gives a2 = 20 + a =⇒ a = −4, 5 by the quadratic formula. Since
a > 0, we take the positive root, or a = 5. However, a2020 is less than
5 (by an extremely small amount); that is, x1 = a2020 = 4.999 . . . < 5.
Here it is sufficient to bound 4.95 < x1 < 5, so that 24.5 < x21 < 25.
To find x0 , we observe that f k (x) is an even function for all k ≥ 1.
This can be shown inductively; f 1 (x) is even, and if f k−1 (x) is even,
then f k (−x) = f 1 (f k−1 (−x)) = f 1 (f k−1 (x)) = f k (x). Thus for real
x, we have f 2020 (x) = 0 iff f 2020 (−x) = 0, so x0 = −x1 ≈ −4.999 . . .;
that is, |x0 | = |x1 | ≈ 4.999 . . .. Then 49 < x20 + x21 < 50, so the largest
integer less than or equal to x20 + x21 is (D) 49 .
Case 1: (x1 , x2 ) = (0, 0) (all digits are 0 (mod 3)). This gives
23 − 1 = 7 subsets.
Case 2: (x1 , x2 ) = (1, 1). This gives 31 × 31 × 23 = 72 subsets.
subsets.
2020 Mock AMC 12 A Solutions 10
Since ω 2 + ω = −1:
Thus there are 168 ascending numbers which are 1 (mod 3), since the
RHS represents this quantity. Proceed as above.
18. Answer: (C) Label the persons 1, . . . , 6 with 1 the shortest and 6
the tallest. Note that 1 must stand next to 2 and 6 must stand next
to 5. Thus we may treat 1 and 2 as one “pair” (similarly with 5 and
6). The remaining constraint is that 3 must stand next to 2 or 4, and
4 must stand next to 3 or 5.
2020 Mock AMC 12 A Solutions 11
P A
Q
O1 1 O2
x2 = 2 − 2 cos(180◦ − 2α)
4x2 = 2 − 2 cos(2α − 60◦ )
the Pythagorean identity and using the fact that 2α ∈ (0, π) (so that
sin 2α is positive). Hence our system of equations implies
x2 = 2 + 2 cos 2α
√ p
4x2 = 2 − cos 2α − 3 1 − cos2 2α
3 − 3y 2 = 81y 2 + 108y + 36
84y 2 + 108y + 33 = 0
28y 2 + 36y + 11 = 0
√
√
3 21
11
= 7 , so P A = 721 and P Q = 3x = (B)
3
P A2 = 2 − 2 − 14 .
7
20. Answer: (E) For n ≥ 1, let pn denote the probability that Paige’s
first n rolls sum to 7. Note that p1 = 0 since each roll is at most
6, and pn = 0 for n ≥ 8 since each roll is at least 1. The answer is
p2 + p3 + . . . + p7 .
To find pn for 2 ≤ n ≤ 7, we can find the number of outcomes on n
dice rolls which sum to 7, then divide by 6n . The number of “good”
outcomes equals the number of positive integer solutions to the equa-
tion a1 + . . . + an = 7 where 1 ≤ ai ≤ 6. By stars and bars, this yields
6
n−1 outcomes.
2020 Mock AMC 12 A Solutions 13
√ 2
2
value varies continuously between 0 and 4−a 2 , when b = a4 and b = 1
respectively. Hence for a fixed a ∈ [0, 1], the set of non-real numbers
in R which are the root of some polynomial√ of the form z 2 + az + b
2
for some b is − a2 + mi where 0 ≤ |m| ≤ 4−a 2 .
√
2
Consider points of the form (− a2 , 4−a
2 ), where a ∈ [0, 1]. Notice
2
that these points lie on the unit circle; this is checked as − a2 +
√ 2 2 2
4−a2
2 = a4 + 4−a
4 = 1. This gives us enough information to plot
the region R. Note that R also contains the interval [−1, 0], but this
has area zero.
Im(z)
Re(z)
−1
Im(z)
1
A
√
3
2
1
2 O Re(z)
−1
B
2020 Mock AMC 12 A Solutions 15
√ √
3
The area of isosceles triangle AOB is 21 · 3· 12 = 4 ,
and the combined
√
◦ π 3 π
area of the two 30 sectors is 6 . The total area of R is + =
√ 4 6
3 3 + 2π
(A) .
12
22. Answer: (C) First, we observe that M N OP QR is an equilateral
hexagon; note that M N is the side opposite the 120◦ angle in 4BM N .
To find M N , we can either use the law of cosines on 4BM N , or
drop an altitude from N onto BM and use the Pythagorean theorem.
Though trigonometry is not required for this problem, we compute
M N using the law of cosines:
E P D
Q O
F C
Y
R
N
7
A M 14 B
X
2020 Mock AMC 12 A Solutions 16
R
β PX
PM β √
7 7
A
r α
α
X PR M
23. Answer: (E) We will consider the problem in binary. The binary
representation of 2020 is 111111001002 .
Let fP
(n) denote the number of tuples of the form (a0 , a1 , . . . , ak ) such
that ki=0 ai 2i = n and ai ∈ {0, 1, 2}. The desired answer is f (2020)
(note that 211 > 2020, so the maximum possible k is 10). To represent
the tuples, we will express them in a way similar to binary, except
that digits may be 0, 1, or 2 (for example, 102 corresponds to a2 = 1,
a1 = 0, a0 = 4):
f (63) = f (1111112 ) = 1
f (252) = f (111111002 ) = f (63) + 2f (62)
= 1 + 12 = 13
f (505) = f (1111110012 ) = f (252) = 13
f (2020) = f (111111001002 ) = f (505) + 2f (504)
= 13 + 2(f (252) + f (251))
= 13 + 2(13 + f (62)) ((2), f (251) = f (125) = f (62))
= 13 + 2(13 + 6) = (E) 51
24. Answer: (E) Note that points (Pi ), (Qi ), (Ri ), (Si ) are positioned on
the sides of the square roughly as follows, with a sample quadrilateral
P1 Q1 R3 S2 shown.
S R1 R2 R3 R
S3
S2
S1 Q1
Q2
Q3
P P3 P2 P1 Q
2020 Mock AMC 12 A Solutions 19
1 2019
E([4P Ss Pp ]) = E(P Ss )E(P Pp ) ≈ 2
2 2 · 2020
1 2019 1
= +ε −ε
2 2020 2020
2019
= − ε0
2 · 20202
for some small ε0 > 0 (note that ε and ε0 are on the order of 2−2020 ,
which can be treated as a negligibly small number). By symmetry,
the quantities E([4QPp Qq ]), E([4RQq Rr ]), and E([4SRr Ss ]) are also
equal. Thus, by linearity of expectation, we have
E([Pp Qq Rr Ss ]) = E(1 − [4P Ss Pp ] − [4QPp Qq ] − [4RQq Rr ] − [4SRr Ss ])
= 1 − E([4P Ss Pp ]) − E([4QPp Qq ])
− E([4RQq Rr ]) − E([4SRr Ss ])
1
= 1 − 4 · (E(P Ss )E(P Pp ))
2
2 · 2019
=1− + 4ε0
20202
Since K is the sum over all possible areas [Pp Qq Rr Ss ], and there are
20204 equally likely choices for (p, q, r, s), then we have
2 · 2019
K = 2020 4
1− + 4ε0
20202
It follows that bKc = 20204 − 2 · 2019 · 20202 ≡ −2 · 2019 · 20202 ≡
(E) 800 (mod 1000).
2020 Mock AMC 12 A Solutions 20
x1 + x2 + . . . + x6 = 5