AMC10 2017 B

Working backwards, switching the digits of the numbers 71, 72, 73, 74, and 75 and subtracting 11 gives, respectively, 6, 16, 26, 36, and 46. Only 6 and 36 are divisible by 3, and only $36 \div 3 = 12$ is a two-digit number.

Each lap took Sofia $\frac{100 \text{ m}}{4 \text{ m/s}} + \frac{300 \text{ m}}{5 \text{ m/s}} = 85$ seconds, so 5 laps took her $5 \cdot 85 = 425$ seconds, which is 7 minutes and 5 seconds.

Adding the inequalities $y > -1$ and $z > 1$ yields $y + z > 0$. The other four choices give negative values if, for example, $x = \frac{1}{8}$, $y = -\frac{1}{4}$, and $z = \frac{3}{2}$.

The given equation implies that $3x + y = -2(x - 3y)$, which is equivalent to $x = y$. Therefore $$\frac{x + 3y}{3x - y} = \frac{4y}{2y} = 2.$$

Suppose Camilla originally had $b$ blueberry jelly beans and $c$ cherry jelly beans. After eating 10 pieces of each kind, she now has $b - 10$ blueberry jelly beans and $c - 10$ cherry jelly beans. The conditions of the problem are equivalent to the equations $b = 2c$ and $b - 10 = 3(c - 10)$. Then $2c - 10 = 3c - 30$, which means that $c = 20$ and $b = 2 \cdot 20 = 40$.

A possible arrangement of 4 blocks is shown by the figure. Four blocks do not completely fill the box because the combined volume of the blocks is only 4(2 · 2 · 1) = 16 cubic inches, whereas the volume of the box is 3 · 2 · 3 = 18 cubic inches. Because the unused space, 18 − 16 = 2 cubic inches, is less than the volume of a block, 4 cubic inches, no more than 4 blocks can fit in the box.

Let 2d be the distance in kilometers to the friend’s house. Then Samia bicycled distance d at rate 17 and walked distance d at rate 5, for a total time of d/17 + d/5 = 44/60 hours. Solving this equation yields d = 17/6 = 2.833…. Therefore Samia walked about 2.8 kilometers.

The altitude AD lies on a line of symmetry for the isosceles triangle. Under reflection about this line, B will be sent to C. Because B is obtained from D by adding 3 to the x-coordinate and subtracting 6 from the y-coordinate, C is obtained from D by subtracting 3 from the x-coordinate and adding 6 to the y-coordinate. Thus the third vertex C has coordinates (−1 − 3, 3 + 6) = (−4, 9). \n\nOR \n\nTo find the coordinates of C(x, y), note that D is the midpoint of BC. Therefore (x + 2)/2 = −1 and (y − 3)/2 = 3. Solving these equations gives x = −4 and y = 9, so C = (−4, 9).

The probability of getting all 3 questions right is (1/3)³ = 1/27. Because there are 3 ways to get 2 of the questions right and 1 wrong, the probability of getting exactly 2 right is 3(1/3)²(2/3) = 6/27. Therefore the probability of winning is 1/27 + 6/27 = 7/27.

Because the lines are perpendicular, their slopes, a/2 and −2/b, are negative reciprocals, so a = b. Substituting b for a and using the point (1, −5) yields the equations b + 10 = c and 2 − 5b = −c. Adding the two equations yields 12 − 4b = 0, so b = 3. Thus c = 3 + 10 = 13.

The students who like dancing but say they dislike it constitute 60% · (100% − 80%) = 12% of the students. Similarly, the students who dislike dancing and say they dislike it constitute (100% − 60%) · 90% = 36% of the students. Therefore the requested fraction is 12 / (12 + 36) = 1/4 = 25%.

For Elmer’s old car, let M be the fuel efficiency in kilometers per liter, and let C be the cost of fuel in dollars per liter. Then for his new car, the fuel efficiency is 1.5M, and the cost of fuel is 1.2C. The cost in dollars per kilometer for the old car is C/M, and for the new car it is 1.2C / 1.5M = 0.8 (C/M). Therefore, fuel for the long trip will cost 20% less in Elmer’s new car.

Let x, y, and z be the number of people taking exactly one, two, and three classes, respectively. The condition that each student in the program takes at least one class is equivalent to the equation x + y + z = 20. The condition that there are 9 students taking at least two classes is equivalent to the equation y + z = 9. The sum 10 + 13 + 9 = 32 counts once the students taking one class, twice the students taking two classes, and three times the students taking three classes. Then x + 2y + 3z = 32, which is equivalent to z = 32 − (x + y + z) − (y + z) = 32 − 20 − 9 = 3.

An integer will have a remainder of 1 when divided by 5 if and only if the units digit is either 1 or 6. The randomly selected positive integer will itself have a units digit of each of the numbers from 0 through 9 with equal probability. This digit of N alone will determine the units digit of N¹⁶. Computing the 16th power of each of these 10 digits by squaring the units digit four times yields one 0, one 5, four 1s, and four 6s. The probability is therefore 8/10 = 4/5.

Triangles ADE and ABE have the same area because they share the base AE and, by symmetry, they have the same height. By the Pythagorean Theorem, AC = 5. Because $\triangle ABE \sim \triangle ACB$, the ratio of their areas is the square of the ratio of their corresponding sides. Their hypotenuses have lengths 3 and 5, respectively, so their areas are in the ratio 9 to 25. The area of $\triangle ACB$ is half that of the rectangle, so the area of $\triangle ABE$ is (9/25) · 6 = 54/25. Thus the area of $\triangle ADE$ is also 54/25.

It will be easier to count the complementary set. There are 9 one-digit numerals that do not contain the digit 0, 9·9 = 81 two-digit numerals that do not contain the digit 0, 9 · 9 · 9 = 729 three-digit numerals that do not contain the digit 0, and 1 · 9 · 9 · 9 = 729 four-digit numerals starting with 1 that do not contain the digit 0, a total of 1548. All four-digit numerals between 2000 and 2017, inclusive, contain the digit 0. Therefore 2017 − 1548 = 469 numerals in the required range do contain the digit 0.

The monotonous positive integers with one digit or increasing digits can be put into a one-to-one correspondence with the nonempty subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9}. The number of such subsets is 2⁹ − 1 = 511. The monotonous positive integers with one digit or decreasing digits can be put into a one-to-one correspondence with the subsets of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} other than ∅ and {0}. The number of these is 2¹⁰ − 2 = 1022. The single-digit numbers are included in both sets, so there are 511 + 1022 − 9 = 1524 monotonous positive integers.

By symmetry, there are just two cases for the position of the green disk: corner or non-corner. If a corner disk is painted green, then there is 1 case in which both red disks are adjacent to the green disk, there are 2 cases in which neither red disk is adjacent to the green disk, and there are 3 cases in which exactly one of the red disks is adjacent to the green disk. Similarly, if a non-corner disk is painted green, then there is 1 case in which neither red disk is in a corner, there are 2 cases in which both red disks are in a corner, and there are 3 cases in which exactly one of the red disks is in a corner. The total number of paintings is 1 + 2 + 3 + 1 + 2 + 3 = 12.

Draw segments CB′, AC′, and BA′. Let X be the area of △ABC. Because △BB′C has a base 3 times as long and the same altitude, its area is 3X. Similarly, the areas of △AA′B and △CC′A are also 3X. Furthermore, △AA′C′ has 3 times the base and the same height as △ACC′, so its area is 9X. The areas of △CC′B′ and △BB′A′ are also 9X by the same reasoning. Therefore the area of △A′B′C′ is X + 3(3X) + 3(9X) = 37X, and the requested ratio is 37 : 1.

There are \binom{21}{2} + \binom{21}{4} + \binom{21}{8} + \binom{21}{16} = 10 + 5 + 2 + 1 = 18 powers of 2 in the prime factorization of 21!. Thus 21! = 2¹⁸k, where k is odd. A divisor of 21! must be of the form 2ⁱb where 0 ≤ i ≤ 18 and b is a divisor of k. For each choice of b, there is one odd divisor of 21! and 18 even divisors. Therefore the probability that a randomly chosen divisor is odd is 1/19.

By the converse of the Pythagorean Theorem, ∠BAC is a right angle, so BD = CD = AD = 5, and the area of each of the small triangles is 1/2 (half the area of △ABC). The area of △ABD is equal to its semiperimeter, 1/2 · (5 + 5 + 6) = 8, multiplied by the radius of the inscribed circle, so the radius is 1/2 ÷ 8 = 3/2. Similarly, the radius of the inscribed circle of △ACD is 4/3. The requested sum is 3/2 + 4/3 = 17/6.

Because ∠ACB is inscribed in a semicircle, it is a right angle. Therefore △ABC is similar to △AED, so their areas are related as AB² is to AE². Because AB² = 4² = 16 and, by the Pythagorean Theorem, AE² = (4 + 3)² + 5² = 74, this ratio is 16/74 = 8/37. The area of △AED is 35/2, so the area of △ABC is (35/2) · (8/37) = 140/37.

The remainder when N is divided by 5 is clearly 4. A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of N is 4(0 + 1 + 2 + · · · + 9) + 10·1 + 10·2 + 10·3 + (4+0) + (4+1) + (4+2) + (4+3) + (4+4) = 270, so N must be a multiple of 9. Then N − 9 must also be a multiple of 9, and the last digit of N − 9 is 5, so it is also a multiple of 5. Thus N − 9 is a multiple of 45, and N leaves a remainder of 9 when divided by 45.

Assume without loss of generality that two of the vertices of the triangle are on the branch of the hyperbola in the first quadrant. This forces the centroid of the triangle to be the vertex (1, 1) of the hyperbola. Because the vertices of the triangle are equidistant from the centroid, the first-quadrant vertices must be (a, 1/a) and (1/a, a) for some positive number a. By symmetry, the third vertex must be (−1, −1). The distance between the vertex (−1, −1) and the centroid (1, 1) is 2√2, so the altitude of the triangle must be (3/2) · 2√2 = 3√2, which makes the side length of the triangle s = (2/√3) · 3√2 = 2√6. The required area is (√3/4) s² = 6√3. The requested value is (6√3)² = 108.

Let S be the sum of Isabella’s 7 scores. Then S is a multiple of 7, and 658 = 91 + 92 + 93 + · · · + 97 ≤ S ≤ 94 + 95 + 96 + · · · + 100 = 679, so S is one of 658, 665, 672, or 679. Because S − 95 is a multiple of 6, it follows that S = 665. Thus the sum of Isabella’s first 6 scores was 665 − 95 = 570, which is a multiple of 5, and the sum of her first 5 scores was also a multiple of 5. Therefore her sixth score must have been a multiple of 5. Because her seventh score was 95 and her scores were all different, her sixth score was 100.