Download AIME15`s Tournament: Round 1 Solutions

AIME15’s Tournament: Round 1 Solutions
Written by: AIME15
AIME15’s Tournament: Round 1 Solutions
Problem 1
What day of the week will it be one million seconds after 1:00 AM on Tuesday?
Converting 1000000 seconds to days gives 1000000 ×
1
1
1
×
×
≈ 11.57 days. 11.57 days
60 60 24
Problem 1
from Tuesday, it will be Saturday .
Problem 2
If the area of trapezoid ABCD is 72 square centimeters, what is the number of centimeters in the length of
BC? Express your answer as a decimal to the nearest tenth.
A
8 cm
B
45◦
D
16 cm
C
The area of a trapezoid is equal to the average of the lengths of the bases times the perpendicular height. Therefore, to find the height of the trapezoid, realize that the height is a leg of
an isosceles right triangle. Using this formula and substituting values for the two bases gives:
8 + 16
× h = 72
2
12h = 72
h=6
Since a leg of the isosceles right triangle is 6 and DC = 16, then 16 − 6 − 8 = 2 is the length
of the remaining part of DC. Finally, by the Pythagorean Theorem:
62 + 22 = BC 2
36 + 4 = BC 2
40 = BC 2
So BC =
√
40 ≈ 6.3 .
Problem 2
AIME15’s Tournament: Round 1 Solutions
Problem 3
7 15
5
1
3
Five stocks has prices of 7 , 7 , 23 , 19 , and 28 dollars. What is the number of dollars in the average
8 16
8
2
8
price for these five stocks? Express your answer to the nearest tenth.
First, find the sum of all the stock prices.
7
15
5
1
3
5
7 + 7 + 23 + 19 + 28 = 87
8
16
8
2
8
16
87
1
Since there are five stocks, the average price is
+
≈ 17.5 .
5
16
Problem 3
Problem 4
How many distinct values can be obtained for the expression 1 ÷ 2 ÷ 3 ÷ 5 ÷ 7 ÷ 11 ÷ 13 ÷ 17 if an unlimited
number of parentheses may be placed in the expression?
Let’s first consider the expression 1 ÷ 2 ÷ 3, which is much less cumbersome. One pos1
1 1
·
= ; another placement gives
sible placement of parentheses gives (1 ÷ 2) ÷ 3 =
2 3
6
1 3
3
1 ÷ (2 ÷ 3) = · = . Of course, we could use fewer or more parentheses, but any other
1 2
2
result would be equivalent to one of these two results.
Similarly, when the expression contains four numbers - 1 ÷ 2 ÷ 3 ÷ 5 - there are four
15
1 3 5
,
, , and
. When 7 is added to the expression, there are eight
possible results:
30 10 6
2
possible results. It is probably becoming apparent that the number of possibilities doubles
each time a number is added to the expression. The reason for this is the movement of each
number because of parentheses. The 1 and the 2 are fixed; that is, in the final result, the 1
will be a factor of the numerator and the 2 will be a factor of the denominator. However, the
other numbers may move to either the numerator or the denominator. For each number, there
are two possibilities. Because there are six numbers beyond the 1 and 2 in the expression,
then, there are 26 = 64 possible values.
Problem 4
We must check to see if we have accounted for all possibilities. Notice that all the
numbers in the expression are prime numbers. Consequently, each number is relatively prime
to every other number in the expression. If the numbers shared a common factor, our method
of counting would allow for overlap. But since all numbers are relatively prime, there is no
overlap, and we can be certain that our method counted all possible values.
Problem 5
There are two prime numbers between 100 and 199 such that the tens digit is a prime number, the units
digit is a prime number, and the tens and units digit taken together are a two-digit prime number. Find the
sum of these two prime numbers.
The tens and the units digits can only come from the set {2, 3, 5, 7}. When paired, the only
two-digit prime numbers from this set are 23, 37, 53, and 73. Hence, the only numbers
between 100 and 199 that need to be considered are 123, 137, 153, and 173. But 123 = 41 × 3
and 153 = 9 × 17, so we must exclude those two, and the sum of the remaining numbers is
137 + 173 = 310 .
Problem 5
AIME15’s Tournament: Round 1 Solutions
Problem 6
As Jenna goes from home to school, she passes Mary’s house and then Jack’s house. The distance from
Jenna’s house to Mary’s is ten times the distance from Mary’s house to Jack’s. The distance from Mary’s
house to Jack’s is ten times the distance from Jack’s to school. The total distance from Jenna’s house to
school is 13.32 km. How many kilometers is it from Jenna’s house to Mary’s, assuming the houses and school
are collinear?
If we let x be the distance from school to Jack’s house, the distance from Jack’s house to
Mary’s house must be 10x and the distance from Mary’s house to Jenna’s house is 100x.
Therefore, the distance from Jack’s house to the school can be determined from:
100x + 10x + x = 13320
111x = 13320
Problem 6
x = 120
So this distance is 120 meters. The distance from Jenna’s house to Mary’s house is 100x, so
it is 100(120) = 12000 meters or 12 km.
Problem 7
Aida bought a jacket for $146.30 which included $6.30 in sales tax. What was the percent of tax that she
paid? Express your answer to the nearest tenth.
The pre-tax price is 146.30 − 6.30 = 140.00. Thus, the percent tax that Aida paid was
6.30
= 0.045, or 4.5% .
140.00
Problem 7
Problem 8
∠C is the complement of ∠A and the supplement of ∠B. Additionally, m∠A + m∠B = 292◦ . What is the
number of degrees in m∠C?
Let m∠C = x. Then, m∠A = 90 − x and m∠B = 180 − x. Plugging this information into the
equation for m∠A and m∠B gives:
m∠A + m∠B = 202
(90 − x) + (180 − x) = 202
270 − 2x = 202
−2x = −68
x = 34
So m∠C = 34◦ .
Problem 8
AIME15’s Tournament: Round 1 Solutions
Problem 9
A, B, and C are three arithmetic sequences. What is the median of the means of the three sequences?
A : 1, 3, 5, . . . , 91
B : 2, 4, 6, . . . , 92
C : 3, 6, 9, . . . , 93
Since each of these is an arithmetic sequence, the mean for each sequence is the mean of the
first and last numbers in the sequence. The means can be found very quickly:
91 + 1
Mean for A:
= 46
2
92 + 2
Mean for B:
= 47
2
93 + 3
Mean for C:
= 48
2
The median of these three numbers is 47 .
Problem 9
Problem 10
What is the least common multiple of 1537 and 1363?
Divisibility tests for 2, 3, 5, and 7 show that neither of these numbers has a one-digit factor.
Thus, the Euclidean Algorithm should be used. Once the factors are identified, the LCM
can be found easily:
1537 = 1363 · 1 + 174
1363 = 174 · 7 + 145
174 = 145 · 1 + 29
145 = 29 · 5 + 0
Thus, the greatest common factor is 29. Division reveals that 1537 = 29× and 1363 = 29×47.
The least common multiple is thus 29 × 47 × 53 = 72239 .
Problem 10
AIME15’s Tournament: Round 1 Solutions
Problem 11
Point B is a vertex of rectangle ABCD and the midpoint of a side of the larger rectangle. What is the
number of square inches in the area of ABCD?
3
A
D
45◦
B
C
Near point D there are three angles, one of which is a right angle, and the other two measure
45◦ . Consequently, the triangle in the upper-left corner of the
√ figure is a 45 − 45 − 90 triangle
- its leg lengths are 3 - so the hypotenuse AD has length 3 2.
The large triangle in the lower-left corner is also a 45 − 45 − 90 triangle. As a result, both angles in the lower-right corner of the figure measure 45◦ , and the triangle in √
the
lower-right
corner
is
a
45
−
45
−
90
triangle.
Because
ABCD
is
a
rectanlge
and
AD
=
3
2,
√
BC = 3 2 also. Hence, the hpotenuse of the lower-right triangle is length 6. Because B is
a midpoint of a side of the larger rectangle, both segments above and below B have length
6. The triangle in the upper-right
corner is a 45 − 45 − 90 triangle as well,
the length of
√
√ and√
its hypotenuse AB is thus 6 2. Then the area of rectangle ABCD is 6 2 × 3 2 = 36 .
Problem 11
Problem 12
The operation a ? b is defined to be (a + b)(a − b) and a4b is defined to be a2 + b2 . What is the value of
(5 ? 4)(544)?
Note that the operation a ? b can be simplified to a ? b = (a + b)(a − b) = a2 − b2 . Then,
(5 ? 4)(544) = (52 − 42 )(52 + 42 )
= (52 )2 − (42 )2
Problem 12
= 54 − 44
= 369
Problem 13
Compute: 2 − 4 + 6 − 8 + 10 − 12 + 14 − . . . + 210
Group the numbers as shown, notice that each group simplifies to −2, and compute the
partial sums (there are 52 groups with sum −2 plus 210 more at the end).
(2 − 4) + (6 − 8) + (10 − 12) + . . . + (206 − 208) + 210 = (−2 × 52) + 210 = 106 .
Problem 13
AIME15’s Tournament: Round 1 Solutions
Problem 14
A used car salesperson has a vehicle she can’t sell. The original price was $1024. She dropped the price
initially to $640. When the car still wouldn’t sell, she made an equivalent percent decrease. How many
dollars are in the next reduced price of the car?
The percent decrease can be found by dividing the reduced price by the original price.
640
= 0.625 = 62.5%. The sale price is 62.5% of the original price, so the next price will
1024
be 640 × 0.625 = $400 .
Problem 14
Problem 15
The sum of all positive integer factors of a number is 91. What is the number?
We can factor 91 as 7 × 13. The sum of the factors of 22 = 4 is 7 and the sum of the factors
of 32 = 9 is 13. Hence, the number we want is 22 × 32 = 36 . To check, we see that the sum
of the factors of 36 is 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91.
Problem 15