AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Mei can travel to her grandma's house by a direct route, or by a scenic route that is 5\ \text{km} longer. When she travels by the scenic route, and comes directly home, the round trip is 35\ \text{km}. How long is the direct route?

  • A.

    5\ \text{km}

  • B.

    12.5\ \text{km}

  • C.

    15\ \text{km}

  • D.

    20\ \text{km}

  • E.

    22.5\ \text{km}

Answer:C

Let the length of the direct route be x kilometers, so the length of the scenic route is x + 5 kilometers. When Mei chooses the scenic route to her grandmother's house and then directly returns home, the total round trip includes the scenic route on the way there and the direct route on the way back. Therefore, the total length can be represented as: (x + 5) + x = 35. Solve this equation, x = 15. So, the length of the direct route is 15 kilometers.

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Problem 2 Easy

A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?

  • A.

    15

  • B.

    30

  • C.

    40

  • D.

    60

  • E.

    70

Answer:E

Let the total amount of flowers be x. Thus, the number of pink flowers is 0.6x, and the number of red flowers is 0.4x. The number of pink carnations is \dfrac{2}{3}\left( 0.6x\right)=0.4x and the number of red carnations is \dfrac{3}{4}\left( 0.4x\right)=0.3x. Summing these, the total number of carnations is 0.4x+0.3x=0.7x. Dividing, we see that \dfrac{0.7x}{x}=0.7=\left( \text{E}\right)70\%.

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Problem 3 Easy

In isosceles \triangle ABC with AB = AC, point D lies on \overline{AC} such that \overline{BD} \perp \overline{AC}. Given that \triangle ABD is isosceles, what is the degree measure of \angle C?

  • A.

    45^{\circ}

  • B.

    60^{\circ}

  • C.

    67.5^{\circ}

  • D.

    72^{\circ}

  • E.

    75^{\circ}

Answer:C

Since \triangle ABD is isosceles with \overline{BD} \perp \overline{AC}, we know that \triangle ABD is an isosceles right triangle. This gives the measure of \angle C is 67.5^{\circ} in isosceles \triangle ABC.

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Problem 4 Easy

Ponce rolls a four-sided die (from 1 to 4), a six-sided die (from 1 to 6), and an eight-sided die (from 1 to 8), all of which are perfectly fair. What is the probability that all three dice show the same number?

  • A.

    \frac{1}{24}

  • B.

    \frac{1}{32}

  • C.

    \frac{1}{36}

  • D.

    \frac{1}{48}

  • E.

    \frac{1}{64}

Answer:D

The result of the 4-sided die does not matter. To match with the next two tosses, we have the probability be \frac 16 \times \frac 18=\frac{1}{48}.

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Problem 5 Easy

The lines with equations y=mx+7, y=2, x=0, and y=0 form a trapezoid with area 3. If m>0, what is the value of m?

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    5

  • E.

    8

Answer:C

Let A and B be the points at which the line with equation y=mx+7 intersects the lines with equations y=0 and y=2, respectively. Also, suppose C has coordinates (0,2) and D has coordinates (0,0). The trapezoid in the problem is ABCD, as shown.

We can find the coordinates of A and B in terms of m. To find the coordinates of A, we find the point of intersection of the line with equation y=mx+7 and the line with equation y=0. Setting 0=mx+7, we get x=-\frac7m. Note that the y-coordinate of A must be 0 since it is, by definition, on the line with equation y=0. Therefore, the coordinates of A are (−\frac7m,0). Similarly, the coordinates of B are (−\frac5m,2). Trapezoid ABCD has parallel bases AD and BC and height CD. The two bases are horizontal and have lengths AD=\frac7m and BC=\frac5m. The length of CD is 2. Therefore, the area of ABCD is 12⋅CD⋅(AD+BC)=12⋅2⋅(\frac7m+\frac5m)=\frac{12}{m}. It is given that the area of the trapezoid is 3, so we have \frac{12}{m}=3, or m=4.

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Problem 6 Easy

Exactly one of the five numbers shown below is not a divisor of Molly's favorite positive integer. Which one is it?

  • A.

    7

  • B.

    9

  • C.

    12

  • D.

    21

  • E.

    28

Answer:B

9=3\times 3, while for the other four answer choices, none of which contains 3^2. Therefore, 9 not necessarily divides Molly's favorite positive integer.

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Problem 7 Easy

When two ants work together they can build an anthill in 24 minutes. When the bigger ant works alone, an anthill can be built in 14 minutes less than when the smaller ant works alone. How many minutes does it take the smaller ant to build an anthill when working alone?

  • A.

    52

  • B.

    54

  • C.

    56

  • D.

    58

  • E.

    60

Answer:C

Let x be the number of minutes that it would take the bigger ant to build an anthill alone and let y be the number of minutes that it would take the smaller ant to build an anthill alone. Since the bigger ant can build an anthill in x minutes, the bigger ant builds 1/x anthills per minute. Likewise, the smaller ant can build 1/y anthills per minute. Thus, working together, the two ants build 1/x+1/y anthills per minute. It is also given that it takes the two ants 24 minutes to build an anthill together, so this means they build 1/24 anthills per minute working together. Hence, we get the equation 1/x+1/y=1/24. Multiplying this equation through by 24xy gives 24y+24x=xy. From the other given condition, we get x=y−14, so we can substitute to get 24y+24(y−14)=(y−14)y. Expanding and rearranging, this equation becomes y^2−62y+336=0, which can be factored as (y−56)(y−6)=0. If y=6, then x=−8, which does not make sense since x must be positive. Therefore, y=56.

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Problem 8 Easy

Call a positive integer \textit{good} if all its digits are nonzero and it is divisible by each of its digits. How many two-digit positive integers are \textit{good}?

  • A.

    9

  • B.

    14

  • C.

    15

  • D.

    16

  • E.

    18

Answer:B

Let \overline{ab} = 10a + b be a two-digit good integer, where a and b are nonzero digits. Then, a and b must both divide 10a + b. This implies a divides b and b divides 10a. Consider cases on the value of a.

Case 1: a = 1. Then, the divisibility becomes b divides 10. There are 3 such digits b, namely 1, 2, and 5.

Case 2: a = 2. Then, the divisibility becomes 2 divides b and b divides 20. There are 2 such digits b, namely 2 and 4.

Case 3: a = 3. Then, the divisibility becomes 3 divides b and b divides 30. There are 2 such digits b, namely 3 and 6.

Case 4: a = 4. Then, the divisibility becomes 4 divides b and b divides 40. There are 2 such digits b, namely 4 and 8.

Case 5: 5 ≤ a ≤ 9. Then, the only possible nonzero digit that is a multiple of a is a itself. Thus, b = a. Then, b is guaranteed to divide 10a. Thus, there are 5 good integers here. The total number of two-digit good integers is 3 + 2 + 2 + 2 + 5 =14.

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Problem 9 Easy

Sally arranges the numbers from 1 to 8 in a circle and then calculates the sum of each set of 3 adjacent numbers. She would like an arrangement in which the smallest of these sums is as large as possible. If she chooses such an arrangement, what will this smallest sum be?

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:C

Let the smallest of the sums of three adjacent numbers be x.

Then adding from the number 1 around the circle and back to 1 adds 9 numbers in a row, so this value is at least 3x.

However these 9 numbers add to 1+1+2+3+\cdots +8 = 37.

Hence 37 ⩾ 3x so that 12\dfrac{1}{3}\geqslant x  and since x is a whole number, x ⩽ 12.

Once it is known that 12 has a chance of being the largest minimum, it is straightforward to find one of the many arrangements where all sums of 3 adjacent numbers (shown in blue) are 12 or more, such as the arrangement shown. This confirms that the smallest possible sum of 3 adjacent numbers is 12.

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Problem 10 Easy

Six identical equilateral triangles of side length 2 are drawn outside a regular hexagon of side length 1, defining a larger hexagon as shown. What is the ratio of the area of the larger hexagon to the area of the smaller hexagon?

  • A.

    5:1

  • B.

    6:1

  • C.

    7:1

  • D.

    8:1

  • E.

    9:1

Answer:C

Each yelllow equilateral triangle can be divided in to 4 unit equilateral triangles, for a total of 24 unit equilateral triangles. The small regular hexagon counts 6 unit equilateral triangles. Last, each white obtuse triangle is half of the area of a yellow equilateral trianlges, for a total of 3 yellow equilateral traingles, or 12 unit equilateral triangles.

The entire figure consists of 24+6+12=42 unit equilateral triangles. Therefore, the required ratio is 6:42=1:7.

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Problem 11 Medium

Triangle ABC is equilateral with AB=1. Points E and G are on \overline{AC} and points D and F are on \overline{AB} such that both \overline{DE} and \overline{FG} are parallel to \overline{BC}. Furthermore, triangle ADE and trapezoids DFGE and FBCG all have the same perimeter. What is DE+FG?

  • A.

    1

  • B.

    \frac{3}{2}

  • C.

    \frac{21}{13}

  • D.

    \frac{13}{8}

  • E.

    \frac{5}{3}

Answer:C

Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y.

Based on the fact that ADE, DEFG, and BCFG have the same perimeters, we can say the following:

3x = x + 2(y-x) + y = y + 2(1-y) + 1

Simplifying, we can find that

3x = 3y-x = 3-y

Since 3-y = 3x, y = 3-3x.

After substitution, we find that 9-10x = 3x, and x = \frac{9}{13}.

Again substituting, we find y = \frac{12}{13}.

Therefore, x+y = \frac{21}{13}, which is C.

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Problem 12 Medium

A sequence of numbers t_1, t_2, t_3, \cdots has its terms defined by t_n=\frac{1}{n}−\frac{1}{n+2} for every integer n \geqslant 1. For example, t_4=\frac 14−\frac 16. What is the largest positive integer k for which the sum of the first k terms (that is, t_1+t_2+⋯+t_{k−1}+t_k) is less than 1.499?

  • A.

    2000

  • B.

    1999

  • C.

    2002

  • D.

    2001

  • E.

    1998

Answer:E

Note that

\begin{aligned} t_1+t_2+t_3+\ldots+t_{k-1}+t_k & =\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots \\ & +\left(\frac{1}{k-1}-\frac{1}{k+1}\right)+\left(\frac{1}{k}-\frac{1}{k+2}\right) \\ & =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k-1}+\frac{1}{k}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots-\frac{1}{k+1}-\frac{1}{k+2} \\ & =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\cdots+\frac{1}{k-1}-\frac{1}{k-1}+\frac{1}{k}-\frac{1}{k}-\frac{1}{k+1}-\frac{1}{k+2} \\ & =\frac{1}{1}+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2} \\ & =1.500-\frac{1}{k+1}-\frac{1}{k+2} \end{aligned}

This means that the sum of the first k terms is less than 1.499 exactly when \frac{1}{k+1}+\frac{1}{k+2} is greater than 0.001.

As k increases from 4 , each of \frac{1}{k+1} and \frac{1}{k+2} decreases, which means that their sum decreases as well. When k=1998, \frac{1}{k+1}+\frac{1}{k+2}=\frac{1}{1999}+\frac{1}{2000}>\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}=0.001.

When k=1999, \frac{1}{k+1}+\frac{1}{k+2}=\frac{1}{2000}+\frac{1}{2001}<\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}=0.001.

This means that \frac{1}{k+1}+\frac{1}{k+2} is greater than 0.001 exactly when k \leq 1998 and is less than 0.001 when k \geq 1999.

In other words, the sum of the first k terms is less than 1.499 for k=1,2,3,4 as well as for 5 \leq k \leq 1998, which is the same as saying that this is true for 1 \leq k \leq 1998.

Therefore, k=1998 is the largest positive integer for which the sum of the first k terms is less than 1.499.

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Problem 13 Medium

A gold coin is worth x\% more than a silver coin. The silver coin is worth y\% less than the gold coin. Both x and y are positive integers. How many possible values for x are there?

  • A.

    0

  • B.

    3

  • C.

    6

  • D.

    9

  • E.

    12

Answer:E

Let the values of a gold coin and a silver coin be g and s dollars, respectively. Since a gold coin is worth x \% more than a silver coin, g=\frac{100+x}{100} \times s. Hence \frac{g}{s}=\frac{100+x}{100}. Since a silver coin is worth y \% less than a gold coin, s=\frac{100-y}{100} \times g. Hence \frac{g}{s}=\frac{100}{100-y}. Therefore \frac{100+x}{100}=\frac{100}{100-y}. Hence (100+x)(100-y)=100 \times 100=10000. Because x is a positive integer, it follows that 100+x is a factor of 10000 with 100+x>0. Therefore we need to count the factors of 10000 which are greater than 100. There are 12 such factors.

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Problem 14 Medium

A network of pathways lead from a single opening to three bins, labelled A, B, C as shown. If a ball is dropped into the opening, it will follow a path and land in one of the bins. Every time a path splits, it is equally likely for the ball to follow either of the downward paths. Ellen drops two balls, one after the other, into the opening. What is the probability that the two balls land in different bins?

  • A.

    \frac { 1 7 } { 3 2 }

  • B.

    \frac { 2 7 } { 5 0 }

  • C.

    \frac { 2 5 } { 6 4 }

  • D.

    \frac { 1 } { 3 }

  • E.

    \frac { 1 5 } { 3 2 }

Answer:A

There are 6 different locations at which the path splits, and we label these splits 1 to 6, as shown.

We begin by determining the probability that a ball lands in the bin labelled A.

There is exactly one path that leads to bin A.

This path travels downward to the left at each of the three splits labelled 1, 2 and 3.

At each of these splits, the probability that a ball travels to the left is \frac{1}{2}, and so the probability that a ball lands in bin A is \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}.

Next, we determine the probability that a ball lands in the bin labelled C.

There are exactly three paths that lead to bin C.

One of these paths travels downward to the right at each of the three splits labelled 1, 4 and 6.

Thus, the probability that a ball lands in bin C by following this path is \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}.

A second path to bin C travels downward to the right at split 1, to the left at split 4, to the right at split 5, and to the right at split 6.

The probability that a ball follows this path is \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}.

The third and final path to bin C travels left at split 1, and to the right at each of the three splits 2, 5 and 6.

The probability that a ball follows this path is also \frac{1}{16}.

The probability that a ball lands in bin C is the sum of the probabilities of travelling each of these three paths or

\frac{1}{8}+\frac{1}{16}+\frac{1}{16}=\frac{2 + 1+1}{16}=\frac{4}{16}=\frac{1}{4}

Finally, we determine the probability that a ball lands in bin B.

There are six different paths that lead to bin B, and we could determine the probability that a ball follows each of these just as we did for bins A and C.

However, it is more efficient to recognize that a ball must land in one of the three bins, and thus the probability that it lands in bin B is 1 minus the probability that it lands in bin A minus the probability that it lands in bin C, or

1-\frac{1}{8}-\frac{1}{4}=\frac{8 - 1-2}{8}=\frac{5}{8}

The probability that the two balls land in different bins is equal to 1 minus the probability that the two balls land in the same bin.

The probability that a ball lands in bin A is \frac{1}{8}, and so the probability that two balls land in bin A is \frac{1}{8}\times\frac{1}{8}=\frac{1}{64}.

The probability that a ball lands in bin C is \frac{1}{4}, and so the probability that two balls land in bin C is \frac{1}{4}\times\frac{1}{4}=\frac{1}{16}.

The probability that a ball lands in bin B is \frac{5}{8}, and so the probability that two balls land in bin B is \frac{5}{8}\times\frac{5}{8}=\frac{25}{64}.

Therefore, the probability that the two balls land in different bins is equal to

1-\frac{1}{64}-\frac{1}{16}-\frac{25}{64}=\frac{64 - 1-4 - 25}{64}=\frac{34}{64}=\frac{17}{32}

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Problem 15 Medium

A rectangle with dimensions 100 \text{ cm} by 150 \text{ cm} is tilted so that one corner is 20 \text{ cm} above a horizontal line, as shown. To the nearest centimeter, determine the height of vertex Z above the horizontal line.

  • A.

    164

  • B.

    165

  • C.

    166

  • D.

    167

  • E.

    168

Answer:D

Since V W Y Z is a rectangle, then Y W=Z V=100 and Z Y=V W=150. Since \triangle Y C W is right-angled at C, by the Pythagorean Theorem,

C W^2=Y W^2-Y C^2=100^2-20^2=10000-400=9600

Since C W>0, then C W=\sqrt{9600}=\sqrt{1600 \cdot 6}=\sqrt{1600} \cdot \sqrt{6}=40 \sqrt{6}. The height of Z above the horizontal line is equal to the length of D C, which equals D Y+Y C which equals D Y+20. Now \triangle Z D Y is right-angled at D and \triangle Y C W is right-angled at C. Also, by Triple Perpendicular Model, \triangle Z D Y is similar to \triangle Y C W.

Therefore, \frac{D Y}{Z Y}=\frac{C W}{Y W} and so D Y=\frac{Z Y \cdot C W}{Y W}=\frac{150 \cdot 40 \sqrt{6}}{100}=60 \sqrt{6}. Finally, D C=D Y+20=60 \sqrt{6}+20 \approx 166.97. Rounded to the nearest integer, D C is 167.

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Problem 16 Hard

A 4\times4\times h rectangular box contains a sphere of radius 2 and eight smaller spheres of radius 1. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is h?

  • A.

    2+2 \sqrt{7}

  • B.

    3+2 \sqrt{5}

  • C.

    4+2 \sqrt{7}

  • D.

    4 \sqrt{5}

  • E.

    4 \sqrt{7}

Answer:A

Let one of the corners be (0, 0, 0). We can orient the box such that the center of the small sphere closest to the corner is (1,1,1), and the center of the large sphere is (2, 2, h/2).

Since the two spheres are tangent, the distance between their centers is 1+2 = 3, so \sqrt{(2-1)^2+(2-1)^2+(h/2-1)^2} = 3. Solving, h=2 + 2\sqrt{7}=\boxed{\textbf{(A)}}

~

Let A be the point in the same plane as the centers of the top spheres equidistant from said centers. Let B be the analogous point for the bottom spheres, and let C be the midpoint of \overline{AB} and the center of the large sphere. Let D and E be the points at which line AB intersects the top of the box and the bottom, respectively.

Let O be the center of any of the top spheres (you choose!). We have AO=1\cdot\sqrt{2}, and CO=3, so AC=\sqrt{3^2-\left(\sqrt2\right)^2}=\sqrt{7}. Similarly, BC=\sqrt{7}. \overline{AD} and \overline{BE} are clearly equal to the radius of the small spheres, 1. Thus the total height is AD+AC+BC+BE=2+2\sqrt7, or \boxed{\textbf{(A)}}.

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Problem 17 Hard

For how many ordered triples (A,B,C) of integers with 0\leqslant A\leqslant 9 and 0\leqslant B \leqslant 9 and 0\leqslant C \leqslant 9 is the sum of three six-digit positive integers \overline{7A6B5C}+\overline{2B9C5A}+\overline{7C1A6B} divisible by 36?

  • A.

    77

  • B.

    78

  • C.

    79

  • D.

    80

  • E.

    81

Answer:E

Set n=7A6B5C+2B9C5A+7C1A6B. We can rearrange as follows: n=(706050+A0B0C)+(209050+B0C0A)+(701060+C0A0B) =(706050+209050+701060)+(A0B0C+B0C0A+C0A0B) =1616160+(A+B+C)(10000)+(B+C+A)(100)+(C+A+B) =1616160+(A+B+C)(10101) =(10101)(160+A+B+C) and so n=(10101)(160+A+B+C).

We need to determine for which values of A, B, and C the integer n is divisible by 36.

Observe that 36=4\times9, and since 4 and 9 have no prime factors in common, n will be divisible by 36 exactly when it is divisible by both 4 and 9.

Since 10101 is odd, it is not a multiple of 4, and so n is a multiple of 4 exactly when 160+A+B+C is a multiple of 4.

Factoring 10101 gives 10101=3\times3367, and since 3367 is not a multiple of 3, we get that 10101 has a factor of 3 but not a factor of 9.

Therefore, n is a multiple of 9 exactly when 160+A+B+C is a multiple of 3.

We now have that n is a multiple of 36 exactly when 160+A+B+C is a multiple of 3 and a multiple of 4, or equivalently, when 160+A+B+C is a multiple of 12.

Since 156 is a multiple of 12, 160+A+B+C=156+4+A+B+C is a multiple of 12 exactly when 4+A+B+C is a multiple of 12.

The digits A, B, and C are each between 0 and 9 inclusive, so 4+A+B+C is at least 4 and at most 4+3\times9=31. The only multiples of 12 between 4 and 31 are 12 and 24, so we must have that 4+A+B+C=12 or 4+A+B+C=24.

To answer the question, we count the number of triples (A,B,C) of integers, each between 0 and 9 inclusive, for which A+B+C=8 or A+B+C=20.

First, suppose A+B+C=8. If A=0, then B+C=8. In this case, we can have B=0 and C=8, or B=1 and C=7, or B=2 and C=6, and so on to B=8 and C=0. This gives a total of 9 triples. If A=1, then B+C=7. In this case, we can have B=0 and C=7, B=1 and C=6, and so on to B=7 and C=0 for a total of 8 triples. Continuing in this way, there are 7 triples when A=2, 6 triples when A=3, 5 triples when A=4, 4 triples when A=5, 3 triples when A=6, 2 triples when A=7, and there is one triple when A=8. The number of triples (A,B,C) when A+B+C=8 is 9+8+7+6+5+4+3+2+1=45.

 

Now assume A+B+C=20. If A=0, then B+C=20, which is impossible given that B≤9 and C≤9. There are no triples with A=0. Similarly, if A=1, then there are no triples. If A=2, then B+C=18, which means B=9 and C=9, so there is only one triple. If A=3, then B+C=17, so either B=9 and C=8 or B=8 and C=9. There are 2 triples in this case. If A=4, then B+C=16, and there are 3 triples. Continuing in this way, when A=5, there are 4 triples, when A=6, there are 5 triples, when A=7, there are 6 triples, when A=8, there are 7 triples, and when A=9, there are 8 triples. The number of triples when A+B+C=20 is 1+2+3+4+5+6+7+8=36. The number of triples (A,B,C) is 45+36=81.

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Problem 18 Hard

Some integers m with 1<m<100000 have the property that the product of the digits of m is equal to 200. If N is the number of such integers m, what is the integer formed by the rightmost two digits of N?

  • A.

    17

  • B.

    27

  • C.

    37

  • D.

    47

  • E.

    57

Answer:B

There are no 2-digit integers with the given property. This is because the product of the digits of a 2-digit number is at most 9\times 9=81<200.

Observe that 200=2^3\times 5^2. We are essentially looking for products of 3, 4, or 5 digits that equal 200. The only digit that is a multiple of 5 is 5 itself, so no matter how many digits m has, exactly two of its digits must be 5. The number of 3-digit numbers m with the given property is 3, since if two of the digits are 5, the third must be 8. There are three 3-digit numbers with one digit equal to 8 and two digits equal to 5. They are 558, 585, and 855.

If m has four digits, then two digits are 5 and the other two digits have a product of 8. The only ways to express 8 as a product of two integers is 8=2\times 4 and 8=1\times 8. In either case, there are six ways to choose where the two digits equal to 5 go. They are

55_ _

5_5_

5__5

_55_

_5_5

__55

In each of these 6 cases, there are 2 ways to place the remaining digits. There are also two choices for the remaining two digits (2 and 4 or 1 and 8), so the number of 4-digit numbers with the given property is 6\times 2\times 2=24.

Using a similar method of counting, we conclude that if m has 5 digits, then there are 10 ways to place the two digits that are equal to 5. The remaining three digits must have a product of 8, so they must be 1, 1, and 8 or 1, 2, and 4 or 2, 2, and 2. If the remaining digits are 1, 1, and 8, then there are 3 choices of where to place the remaining digits. This is because once the 8 is placed, the last two digits must be 1 (there is no choices). If the remaining digits are 1, 2, and 4, then there are 6 ways to place the remaining digits. This is because there are 6 ways to order the digits 1, 2, and 4. Finally, if the remaining digits are all 2, then there is only one way to place the digits. Thus, there are 3\times 10+6\times 10+10=100 five-digit numbers with the given property.

The total number of integers m with 1<m<100000 with the property that the product of the digits is 200 is 3+24+100=127. The integer formed by the rightmost 2 digits of N is 27.

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Problem 19 Hard

The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD=3. Point E is chosen so that ED=5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of \triangle ABC?

  • A.

    \frac{120}{37}

  • B.

    \frac{140}{39}

  • C.

    \frac{145}{39}

  • D.

    \frac{140}{37}

  • E.

    \frac{120}{31}

Answer:D

We note that \triangle ACB\sim\triangle ADE by AA similarity. Also, since the area of \triangle ADE=\dfrac{7\cdot5}{2}=\dfrac{35}{2} and AE=\sqrt {74},\dfrac{A_{\triangle ABC}}{A_{\triangle ADE}}=\dfrac{A_{\triangle ABC}}{\dfrac{35}{2}}=\left( \dfrac{4}{\sqrt{74}}\right)^{2} , so the area of \triangle ABC=\frac {140}{37}.

 

Let's call the center of the circle that segment AB is the diameter of O. Note that \triangle ODE is an isosceles right triangle. Solving for side OE, using the Pythagorean theorem, we find it to be 5\sqrt 2. Calling the point where segment OE intersects circle O, the point I, segment IE would be 5\sqrt 2-2. Also, noting that \triangle ADE is a right triangle, we solve for side AE, using the Pythagorean Theorem, and get \sqrt {74}. Using Power of Point on point E, we can solve for CE. We can subtract CE from AE to find AC and then solve for CB using Pythagorean theorem once more: AE \cdot CE=\text{Diameter of circle} O+IE \cdot IE

\Rightarrow\sqrt{74}\left( CE\right)=\left( 5\sqrt{2}+2\right)\left( 5\sqrt{2}-2\right)\Rightarrow CE=\dfrac{23\sqrt{74}}{37},

AC=AE-CE\rightarrow AC=\sqrt{74}-\dfrac{23\sqrt{74}}{37}\Rightarrow AC=\dfrac{14\sqrt{74}}{37}.

Now to solve for CB: AB^{2}-AC^{2}=CB^{2}\rightarrow4^{2}+\dfrac{14\sqrt{74}^{2}}{37}=CB^{2}\Rightarrow CB=\dfrac{10\sqrt{74}}{37},

Note that \triangle ABC is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases AC and BC, we get the area of triangle ABC to be \frac {140}{37}.

Link Problem
Problem 20 Hard

Suppose that f\left( x \right) and g\left( x \right) are both quadratic functions with quadratic coefficient 2. If g\left( 10 \right)=25 and \frac{f\left( -2 \right)}{g\left( -2 \right)}=\frac{f\left( 2 \right)}{g\left( 2 \right)}=\frac{16}{15}, determine f\left( 10 \right).

  • A.

    -25

  • B.

    0

  • C.

    1

  • D.

    25

  • E.

    \frac{208}{15}

Answer:E

If f\left( x \right)={2{x}^{2}}+ax+b and g\left( x \right)={2{x}^{2}}+cx+d, by the given conditions,

15\left( 8-2a+b \right)=16\left( 8-2c+d \right)\cdots \cdots①;

15\left( 8+2a+b \right)=16\left( 8+2c+d \right)\cdots \cdots②.

①+② gives 240+30b=256+32d, so d=\frac{15}{16} b-\frac12;

②-① gives 60a=64c, c=\frac{15}{16} a.

Also, by g\left( 10 \right)=25, we have 200+10c+d=25.

∴ 200+10\times \frac{15}{16}a+\frac{15}{16}b-\frac12=25 and 10a+b=-\frac{2792}{15},

∴ f\left( 10 \right)=200+10a+b=\frac{208}{15}.

The answer is \text{E}

Link Problem
Problem 21 Hard

What is the number of positive integer values of n \leqslant 2025 such that

x+\lfloor\sqrt{x}\rfloor+\lfloor\sqrt[3]{x}\rfloor=n has no real solution for x?

  • A.

    54

  • B.

    55

  • C.

    56

  • D.

    57

  • E.

    58

Answer:C

Since the floor function only outputs integers and n is an integer, x must be an integer. Let f(x)=x+\lfloor\sqrt{x}\rfloor+\lfloor\sqrt[3]{x}\rfloor for integers x \geqslant 0 (to prevent the number under the square root from being negative). If x is incremented by 1, x will always increase. While \lfloor\sqrt{x}\rfloor or \lfloor\sqrt[3]{x}\rfloor may not increase, they will never decrease. Thus, f(x) is a strictly increasing function along the nonnegative integers. Clearly, f(0)=0 and f(1)=3. It should be checked if there is an integer a such that f(a)=2025 (if not, the integer a where f(a) is as closest to 2025 as possible). Clearly, f(2025)>2025 and f\left(44^2\right)=1936+44+12=1992. Since f(x) is strictly increasing, 44^2 \leqslant a<45^2 if a exists. In this range, \lfloor\sqrt{a}\rfloor can only be 44 and \lfloor\sqrt[3]{a}\rfloor can only be 12.

Then, a=2025-44-12=1969, which lies in the range. Hence, f(1969)=2025. By the strictly increasing behavior of f(x), the values f(1), f(2), f(3), \cdots, f(1969) are distinct and the only values of f that can be one of the first 2025 positive integers. Thus, exactly 1969 of those integers are covered by f, so the number of integers n not covered by f is 2025-1969=56.

Link Problem
Problem 22 Hard

Let S be the set of circles in the coordinate plane that are tangent to each of the three circles with equations x^2+y^2=4, x^2+y^2=64 and \left( x-5\right)^{2}+y^{2}=3. What is sum of the areas of all circles in S.

  • A.

    48\pi

  • B.

    68\pi

  • C.

    96\pi

  • D.

    102\pi

  • E.

    136\pi

Answer:E

There are two circles that are externally tangent to the two small circles and internally tangent to the large circle. Here r=3 and A=18\pi. (blue part)

There are two circles that are internally tangent to the concentric circles and externally tangent to the other circle. Here r=5 and A=50\pi. (red part)

There are two circles that are internally tangent to the big and non-center circles, but externally tangent to the small center circle. Here r=3 and A=18\pi. (green part)

There are two circles that are internally tangent to all three circles. Here r=5 and A=50\pi. (brown part)

The sum of area of all circles is 136\pi.

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Problem 23 Hard

How many strings of length 5 formed from the digits 0, 1, 2, 3, 4 are there such that for each j\in\left\{ 1,2,3,4\right\}, at least j of the digits are less than j? (For example, 02214 satisfies the condition because it contains at least 1 digit less than 1, at least 2 digits less than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string 23404 does not satisfy the condition because it does not contain at least 2 digits less than 2.)

  • A.

    500

  • B.

    625

  • C.

    1089

  • D.

    1199

  • E.

    1296

Answer:E

Suppose a<b<c<d<e. Then let's do the casework.

① If 5 numbers are all distinct, 5!=120.

② If one number is counted twice, 10\times \frac{5!}{2}=600.

③ If two numbers are counted twice, 10\times \frac{5!}{2 \cdot 2}=300.

④ If one number is counted 3 times, 10\times \frac{5!}{3!}=200.

⑤If one number is counted 3 times and the others are counted twice, 5\times \frac{5!}{2!\cdot3!}=50.

⑥ If one number is counted 4 times, 5 \times 5=25.

⑦ If one number is counted 5 times, only 1 possible way.

In total, 120+600+300+200+50+25+1=1296.

Link Problem
Problem 24 Hard

Four regular hexagons surround a square with side length 1, each one sharing an edge with the square, as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be written as m \sqrt{n} + p, where m, n, and p are integers and n is not divisible by the square of any prime. What is m+n+p?

  • A.

    -12

  • B.

    -4

  • C.

    4

  • D.

    24

  • E.

    32

Answer:B

Note that each of the green sections is a rectangle, so its interior angles are all 90^{\circ}. Since \angle{ABC}=120^{\circ}, every one of the orange sections is a 30-60-90 right triangle.

Define x to be the distance from the corner of the square with side length 1 to the corner of the larger blue square. Due to the sides of the two squares being parallel to each other, the large blue triangle is a 45-45-90 right triangle. By AAA similarity, the smaller blue triangles are also 45-45-90, and have side lengths of \frac{x\sqrt{2}}{2}, \frac{x\sqrt{2}}{2}, and x. By 30-60-90 triangle relations, the largest altitude of the orange triangle is \frac{x\sqrt{6}}{2}.

Now, we can find the height of the hexagon to obtain an equation in terms of x. Consider a hexagon with side length 1, where point P is the foot of the perpendicular dropped from B, bisecting \angle{ABC}:

Note that triangles \triangle{APB} and \triangle{CPB} are congruent 30-60-90 triangles, by SAS congruence. Since the side length of this hexagon is 1, the length of AP is \frac{1}{2}\cdot \sqrt{3} =\frac{\sqrt{3}}{2}, by 30-60-90 triangle relations. The height of the hexagon is twice this value, or \sqrt{3}.

The height is also equal to the sum of the values along the long blue line, in the first diagram. Therefore, 1+\frac{x\sqrt{2}}{2}+\frac{x\sqrt{6}}{2}=\sqrt{3}. Solving and rationalizing, x=2\sqrt{2}-\sqrt{6}.

The area of the dodecagon is equal to the sum of the areas of the four rectangles, eight orange triangles, and purple square. In terms of x, this is 8\cdot \frac{1}{2}\cdot \frac{x\sqrt{2}}{2}\cdot \frac{x\sqrt{6}}{{2}}+4\cdot 1\cdot\frac{x\sqrt{6}}{2}+(x\sqrt{2}+1)^2.

Plugging in x=2\sqrt{2}-\sqrt{6}, the area of the dodecagon is 16\sqrt{3}-23. Therefore, the answer is 16+3-23= \boxed{\textbf{(B) }-4}.

Link Problem
Problem 25 Hard

Let N be the number of triples (x,y,z) of positive integers such that x<y<z and xyz=2^2 \times 3^2 \times 5^2 \times 7^2 \times 11^2 \times 13^2 \times 17^2 \times 19^2. When N is divided by 100, what is the remainder?

  • A.

    8

  • B.

    28

  • C.

    48

  • D.

    68

  • E.

    88

Answer:A

Suppose that a, b and c are positive integers with a b c=2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 17^2 \cdot 19^2. We determine the number of triples (a, b, c) with this property. (We are temporarily ignoring the size ordering condition in the original question.) Since the product a b c has two factors of 2, then a, b and c have a total of two factors of 2. There are 6 ways in which this can happen: both factors in a, both factors in b, both in c, one each in a and b, one each in a and c, and one each in b and c. Similarly, there are 6 ways of distributing each of the other squares of prime factors. Since a b c includes exactly 8 squares of prime factors and each can be distributed in 6 ways, there are 6^8 ways of building triples (a, b, c) using the prime factors, and so there are 6^8 triples (a, b, c) with the required product.

Next, we include the condition that no pair of a, b and c should be equal. (We note that a, b and c cannot all be equal, since their product is not a perfect cube.) We count the number of triples with one pair equal, and subtract this number from 6^8. We do this by counting the number of these triples with a=b. By symmetry, the number of triples with a=c and with b=c will be equal to this total. In order to have a=b and a \neq c and b \neq c, for each of the squared prime factors p^2 of a b c, either p^2 is distributed as p and p in each of a and b, or p^2 is distributed to c. Thus, for each of the 8 squared prime factors p^2, there are 2 ways to distribute, and so 2^8 triples (a, b, c) with a=b and a \neq c and b \neq c. Similarly, there will be 2^8 triples with a=c and 2^8 triples with b=c. This means that there are 6^8-3 \cdot 2^8 triples (a, b, c) with the required product and with no two of a, b, c equal. The original problem asked us to the find the number of triples (x, y, z) with the given product and with x<y<z. To convert triples (a, b, c) with no size ordering to triples (x, y, z) with x<y<z, we divide by 6. (Each triple (x, y, z) corresponds to 6 triples (a, b, c) of distinct positive integers with no size ordering.) Therefore, the total number of triples (x, y, z) with the required properties is

N=\frac{1}{6}\left(6^8-3 \cdot 2^8\right)=6^7-2^7=279808

When N is divided by 100, the remainder is 8.

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AMC 10 Daily Practice - Exponent AMC 10 Daily Practice Round 4 AMC 10 Daily Practice Round 2 AMC 10 Daily Practice Round 1
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