2023 AMC 8 Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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Tiempo restante:

40:00

1.

What is the value of (8×4+2)(8+4×2)? (8 \times 4 + 2) - (8 + 4 \times 2)?

00

66

1010

1818

2424

Answer: D
Video solution:
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Written solution:

We can simplify this as follows.

(32+2)(8+8)=3416=18\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}

Thus, D is the correct answer.

2.

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?

Answer: E
Video solution:
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Written solution:

We can unfold the cut up paper to achieve the following figure.

Thus, E is the correct answer.

3.

Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation: W=T0.7SW=T-0.7\cdot S Where WW represents the wind chill, TT respresents air temperature measured in degrees Fahrenheit (F),(^{\circ}F), and SS represents wind speed measured in miles per hour (mph).

Suppose the air temperature is 36F36^{\circ}F and the wind speed is 1818 mph. Which of the following is closest to the approximate wind chill?

1818

2323

2828

3232

3535

Answer: B
Video solution:
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Written solution:

Using the formula, we can calculate the wind chill to be 360.718=3612.6=23.4.\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}

Thus, B is the correct answer.

4.

The numbers from 11 to 4949 are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number 7.7. How many of these four numbers are prime?

00

11

22

33

44

Answer: D
Video solution:
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Written solution:

We can fill in the other numbers to get the complete grid.

From this, we can see that the only prime numbers in the shaded boxes are 19,23,19, 23, and 47.47.

Thus, D is the correct answer.

5.

A lake contains 250250 trout, along with a variety of other fish. When a marine biologist catches and releases a sample of 180180 fish from the lake, 3030 are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

12501250

15001500

17501750

18001800

20002000

Answer: B
Video solution:
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Written solution:

Note that 30180=16.\dfrac{30}{180} = \dfrac{1}{6}. This means that a sixth of the fish in the lake are trout, or in other words, the total number of fish is 66 times the number of trout.

Therefore, there are 6250=15006 \cdot 250 = 1500 fish in the lake.

Thus, B is the correct answer.

6.

The digits 2,0,2,2, 0, 2, and 33 are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

00

88

99

1616

1818

Answer: C
Video solution:
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Written solution:

Note that we do not want 00 as a base, since that would make the expression equal to 0.0.

This means that 00 must be an exponent. The number whose exponent is 00 will automatically evaluate to 1.1.

Therefore, we should minimize this number, so we can put 20.2^0.

Now the other term is 232^3 or 32.3^2. Clearly, 323^2 is larger, so we should use that.

This gives us a final value of 20×32=1×9=9. 2^0 \times 3^2 = 1 \times 9 = 9.

Thus, C is the correct answer.

7.

A rectangle, with sides parallel to the xx-axis and yy-axis, has opposite vertices located at (15,3)(15, 3) and (16,5).(16, 5). A line is drawn through points A(0,0)A(0, 0) and B(3,1).B(3, 1). Another line is drawn through points C(0,10)C(0, 10) and D(2,9).D(2, 9). How many points on the rectangle lie on at least one of the two lines?

00

11

22

33

44

Answer: B
Video solution:
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Written solution:

We can graph the two lines.

From this, we see that only the top left corner of the rectangle intersects either line.

Thus, B is the correct answer.

8.

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers 11 and 00 represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo's win-loss record?

000101000101

001001001001

010000010000

010101010101

011000011000

Answer: A
Video solution:
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Written solution:

Note that for each round, there are going to 22 winners and 22 losers (two matches each with a winner and loser).

This means that for each match, there should be 22 ones and 22 zeros.

Using this fact, we can deduce that Tiyo's win-loss record is 000101.000101.

Thus, A is the correct answer.

9.

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between 44 and 77 meters?

66

88

1010

1212

1414

Answer: B
Video solution:
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Written solution:

The first time that she hits an elevation of 77 meters is at 22 seconds.

She then dips below 44 meters after 44 seconds. This adds 42=24 - 2 = 2 seconds to the total answer.

Malaika then goes above 44 meters at 66 seconds. She hits 77 meters again at 1010 seconds.

This adds 106=410 - 6 = 4 more seconds to the total. She finally dips below 77 meters for the last time at 1212 seconds.

She then falls below 44 meters at 1414 seconds, finally adding 1412=214 - 12 = 2 seconds to the total time.

The desired answer is therefore 2+4+2=8 seconds.  2 + 4 + 2 = 8 \text{ seconds. } Thus, B is the correct answer.

10.

Harold made a plum pie to take on a picnic. He was able to eat only 14\frac{1}{4} of the pie, and he left the rest for his friends. A moose came by and 13\frac{1}{3} of what Harold left behind. After that, a porcupine ate 13\frac{1}{3} of what the moose left behind. How much of the original pie still remained after the porcupine left?

112\dfrac{1}{12}

16\dfrac{1}{6}

14\dfrac{1}{4}

13\dfrac{1}{3}

512\dfrac{5}{12}

Answer: D
Video solution:
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Written solution:

Harold left 114=341 - \frac{1}{4} = \frac{3}{4} of the pie for his friends.

The moose ate 1334=14\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} of the pie, leaving 3414=12\frac{3}{4} - \frac{1}{4} = \frac{1}{2} of the pie.

Finally, the porcupine ate 1312=16\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} of the pie. This leaves 1216=13\frac{1}{2} - \frac{1}{6} = \frac{1}{3} of the pie.

Thus, D is the correct answer.

11.

NASA's Perseverance Rover was launched on July 30,2020.30, 2020. After traveling 292,526,838292,526,838 miles, it landed on Mars in Jezero Crater about 6.56.5 months later. Which of the following is closest to the Rover's average interplanetary speed in miles per hour?

6,0006,000

12,00012,000

60,00060,000

120,000120,000

600,000600,000

Answer: C
Video solution:
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Written solution:

We can round the distance up to 300,000,000300,000,000 miles. We can also approximate 6.56.5 months as 30×6.5=195200 30 \times 6.5 = 195 \approx 200 days. This means that the Rover traveled 300,000,000÷200=1,500,000 300,000,000 \div 200 = 1,500,000 miles per day. Dividing this by 2424 to get the speed in miles per hour yields 62,500,62,500, which is close to 60,000.60,000.

Thus, C is the correct answer.

12.

The figure below shows a large unshaded circle with a number of smaller unshaded and shaded circles in its interior. What fraction of the interior of the large unshaded circle is shaded?

14\dfrac{1}{4}

1136\dfrac{11}{36}

13\dfrac{1}{3}

1936\dfrac{19}{36}

59\dfrac{5}{9}

Answer: B
Video solution:
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Written solution:

WLOG, assume that each square has a side length of 22 units.

This means that there are 33 shaded unit circles, which total to 312π=3π3 \cdot 1^2 \pi = 3 \pi area.

There is also a shaded circle with radius 44 with two white circles of radius 22 inside.

This gives us an extra shaded area of 42π222π=16π8π=8π.\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}

Adding these values together yields 8π+3π=11π.8\pi + 3\pi = 11\pi. The area of the large white circle is 62π=36π.6^2\pi = 36\pi. Therefore, the desired fraction is 11π36π=1136.\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.

Thus, B is the correct answer.

13.

Along the route of a bicycle race, 77 water stations are evenly spaced between the start and finish lines, as shown in the figure below. There are also 22 repair stations evenly spaced between the start and finish lines. The 33rd water station is located 22 miles after the 11st repair station. How long is the race in miles?

88

1616

2424

4848

9696

Answer: D
Video solution:
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Written solution:

The 33rd water station is located 38\frac{3}{8} of the way along the race (the water stations split the race up into 88 equal spaces).

The first repair station is located 13\frac{1}{3} of the way along the race. The distance between the two is 3813=124 \dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24} the length of the race. We know that this equals 2,2, which means the race is 242=4824 \cdot 2 = 48 miles long.

Thus, D is the correct answer.

14.

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 55-cent, 1010-cent, and 2525-cent stamps, with exactly 2020 of each type. What is the greatest number of stamps Nicolas can use to make exactly $7.10 in postage?

(Note: The amount $7.10 corresponds to 77 dollars and 1010 cents. One dollar is worth 100100 cents.)

4545

4646

5151

5454

5555

Answer: E
Video solution:
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Written solution:

Note that we want to get 7100+10=7107 \cdot 100 + 10 = 710 cents. Let us try to see if it is possible to use up all of the 55-cent and 1010-cent stamps.

All of these two types of stamps combined would be worth 20(5+10)=2015=300 20(5 + 10) = 20 \cdot 15 = 300 cents. We then would need 710300=410710 - 300 = 410 cents, which cannot be created with just 2525-cent stamps.

We can, however, make 425425 cents with 2525-cent stamps. Using only 1919 each of 55 and 1010-cent stamps would total 300510=285 300 - 5 - 10 = 285 cents. This means we would then need 710285=425710 - 285 = 425 cents. This can be achieved with 425÷25=17425 \div 25 = 17 2525-cent stamps.

This lets us use 219+17=38+17=55\begin{align*} 2 \cdot 19 + 17 &= 38 + 17 \\ &= 55 \end{align*} stamps.

Thus, E is the correct answer.

15.

Visvam walks half a mile to get to school each day. His route consists of 1010 city blocks of equal length and he takes one minute to walk each block. Today, after walking 55 blocks, Visvam discovers that he has to make a detour, walking 33 blocks of equal length instead of 11 block to reach the next corner. From the time he starts his detour, at what speed, in miles per hour, must Visvam walk in order to arrive at school at his usual time?

44

4.24.2

4.54.5

4.84.8

55

Answer: B
Video solution:
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Written solution:

If half a mile is the same as 1010 blocks, then one block is 12÷10=120\dfrac{1}{2} \div 10 = \dfrac{1}{20} miles.

Starting from the detour, Visvam has to walk 7120=7207 \cdot \dfrac{1}{20} = \dfrac{7}{20} miles.

Normally, from this spot Visvam would take 55 minutes to walk to school. Now he has to travel 720\dfrac{7}{20} miles in 55 minutes.

Note that 55 minutes is 560=112\dfrac{5}{60} = \dfrac{1}{12} miles. This means his speed must be 720112=12720=215=4.2\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*} mph.

Thus, B is the correct answer.

16.

The letters P, Q, and R are entered into a 20×2020 \times 20 table according to the pattern shown below. How many Ps, Qs, and Rs will appear in the completed table.

132132 Ps, 134134 Qs, 134134 Rs

133133 Ps, 133133 Qs, 134134 Rs

133133 Ps, 134134 Qs, 133133 Rs

134134 Ps, 132132 Qs, 134134 Rs

134134 Ps, 133133 Qs, 133133 Rs

Answer: C
Video solution:
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Written solution:

Since 20=36+2,20 = 3 \cdot 6 + 2, the bottom 22 letters in each column will occur one more time than the third letter.

This means that the third letter in each column will occur 66 times, whereas the other 22 will appear 77 times.

Using the same analysis, we can see that RR and PP appear in the third row 77 times, whereas QQ only appears 66 times.

Therefore, in 207=1320 - 7 = 13 columns, PP and RR appear 77 times, for a total of 137=9113 \cdot 7 = 91 times.

They also appear 66 times in 77 columns, adding 67=426 \cdot 7 = 42 appearances to the total. This gives us a total of 91+42=13391 + 42 = 133 appearances for PP and R.R.

QQ will therefore appear 20202133=400266=134\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*} times.

Thus, C is the correct answer.

17.

A regular octahedron has eight equilateral triangle face with four faces meeting at each vertex. Jun will make the regular octahedron shown in the figure by folding the piece of paper below. Which number face will end up to the right of the shaded region Q?Q?

11

22

33

44

55

Answer: A
Video solution:
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Written solution:

Begin by observing that when folded, the sides labelled 2,2, 3,3, 4,4, and 55 form the bottom half of the octahedron. As such, the remaining four faces must make up the top half of the octahedron.

From here, we have narrowed down our possibilities to 1,1, 6,6, and 7.7. We can see that 66 will be the number to the left of the shaded region Q.Q. This also gives us that 77 is to the left of 6.6.

Therefore, we know that the only remaining face, 1,1, must be to the right of the shaded region Q.Q.

Thus, A is the correct answer.

18.

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 55 pads to the right or 33 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 20232023 pads to the right of her starting position?

405405

407407

409409

411411

413413

Answer: D
Video solution:
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Written solution:

An optimal strategy would be jumping as close as possible with the right jumps and then fine tuning with the left jumps.

It will take at least 404404 jumps to get to 2020.2020. Clearly, we cannot get to 20232023 in one more jump, so A cannot be right.

With 33 jumps, the only way to move forward is with 22 jumps right and the one jump left, but that puts us at 2207.2207.

This shows that B cannot happen.

Using 55 jumps, we can jump right twice and jump left thrice, but that puts us at 2021.2021. Jumping right rice and left twice would put us at 2029.2029.

This shows that C cannot happen either.

Finally, with 77 jumps, we can jump right thrice and left four times, putting us at 2023.2023.

Thus, D is the correct answer.

19.

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is 23\frac{2}{3} the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle.

1:31 : 3

3:83 : 8

5:125 : 12

7:167 : 16

4:94 : 9

Answer: C
Video solution:
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Written solution:

Since the inner triangle's side length is 23\frac{2}{3} the side length of the outer triangle, its area is 232=49\frac{2}{3}^2 = \frac{4}{9} the area of the outer triangle.

This means that the three trapezoids are 149=591 - \frac{4}{9} = \frac{5}{9} the area of the outer triangle.

Therefore, one trapezoid is 59÷3=527\frac{5}{9} \div 3 = \frac{5}{27} the area of the outer triangle.

This makes the ratio of the areas of one trapezoid and the inner triangle 52749=52794=512. \dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}.

Thus, C is the correct answer.

20.

Two integers are inserted into the list 3,3,8,11,283, 3, 8, 11, 28 to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?

5656

5757

5858

6060

6161

Answer: D
Video solution:
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Written solution:

Since the median remains unchanged. One number less than 88 and one number greater than 88 is added.

The inequalities are strict since the mode doesn't change, which means that only 33 can appear twice.

The current range is 283=25.28 - 3 = 25. The new range is therefore 225=50.2 \cdot 25 = 50.

To maximize the smaller number, we can set it equal to 7.7. The larger number is forced to be 50+3=53.50 + 3 = 53.

Their sum is 53+7=60.53 + 7 = 60.

Thus, D is the correct answer.

21.

Alina writes the numbers 1,2,,91,2,\cdots,9 on separate cards, one number per card. She wishes to divide the cards into 33 groups of 33 cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

00

11

22

33

44

Answer: C
Video solution:
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Written solution:

The sum of all the numbers is 9102=45.\dfrac{9 \cdot 10}{2} = 45. This means that the sum of each group is 45÷3=15.45 \div 3 = 15.

Consider the group with 99 in it. The other two numbers must add to 6.6. Therefore, the other cards in this group are 11 and 55 or 22 and 4.4.

Case 1:1: One group is 1,5,91, 5, 9

Consider the group with 88 in it. The other numbers must add to 7.7. The only option is 33 and 44 with the remaining cards.

The other group is then 2,6,2,6, and 7.7. This adds to 15,15, so this case contributes one possibility.

Case 2:2: One group is 2,4,92, 4, 9

Consider the group with 88 in it. As above, the other numbers have to add to 7.7. The only option is 11 and 6.6.

The final group is 3,5,3, 5, and 7,7, which adds to 15.15. This is another configuration.

We have gone through all the cases, which revealed that there are only 22 possible groupings.

Thus, C is the correct answer.

22.

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is 4000.4000. What is the first term?

11

22

44

55

1010

Answer: D
Video solution:
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Written solution:

Let xx be the first term and yy be the second.

Then we get the following sequence. x,y,xy,xy2,x2y3,x3y5, x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots From this, we get that x3y5=4000.x^3y^5 = 4000.

Factoring 4000,4000, we get 4000=2553. 4000 = 2^5 \cdot 5^3. We need xx and yy to be integers. The only fifth powers that divides 40004000 are 11 and 32.32.

If y=1,y = 1, then x3=4000,x^3 = 4000, which doesn't work since 40004000 is not a perfect cube. Therefore, y5=32 y^5 = 32 y=2. y = 2.

This forces xx to equal 5.5.

Thus, D is the correct answer.

23.

Each square in a 3×33 \times 3 grid is randomly filled with one of the 44 gray-and-white tiles shown below on the right.

What is the probability that the tiling will contain a large gray diamond in one of the smaller 2×22 \times 2 grids? Below is an example of such a tiling.

11024\dfrac{1}{1024}

1256\dfrac{1}{256}

164\dfrac{1}{64}

116\dfrac{1}{16}

14\dfrac{1}{4}

Answer: C
Video solution:
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Written solution:

There are a total of 494^9 configurations. The 2×22 \times 2 square that contains the large gray diamond can occur in 44 spots (as there are four 2×22 \times 2 squares in the big square).

The four tiles in the square have a fixed tile. The other 55 squares can be any tiles, which allows for 454^5 configurations for the other tiles.

The total number of configurations that include the large gray diamond is therefore 445=46.4 \cdot 4^5 = 4^6.

The desired probability is then 4649=143=164. \dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.

Thus, C is the correct answer.

24.

Isosceles triangle ABCABC has equal side lengths ABAB and BC.BC. In the figures below, segments are drawn parallel to AC\overline{AC} so that the shaded portions of ABC\triangle ABC have the same area. The heights of the two unshaded portions are 1111 and 55 units, respectively. What is the height hh of ABC?\triangle ABC?

14.614.6

14.814.8

1515

15.215.2

15.415.4

Answer: A
Video solution:
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Written solution:

Let aa be the area of ABC.\triangle ABC.

Note that the smaller triangles are similar to ABC.\triangle ABC. This means that the ratio of their areas is the ratio of their side lengths squared.

Then we get that aa(11h)2 a - a \cdot \left(\dfrac{11}{h}\right)^2 =a(h5h)2.= a \cdot \left(\dfrac{h - 5}{h}\right)^2. The left side is the area of the whole triangle minus the area of the unshaded region of the left triangle.

The right hand side is the area of the shaded triangle. We get this by finding the ratio of their side lengths (which is the same as the ratio of their heights) and squaring it.

Simplifying yields h2121=h210h+25. h^2 - 121 = h^2 - 10h + 25. This simplifies to 10h=146 10h = 146 h=14.6. h = 14.6.

Thus, A is the correct answer.

25.

Fifteen integers a1,a2,a3,,a15a_1, a_2, a_3, \cdots, a_{15} are arranged in order on a number line. The integers are equally spaced and have the property that 1a110, 1 \leq a_1 \leq 10,13a220, 13 \leq a_2 \leq 20, and 241a15250. 241 \leq a_{15} \leq 250.

What is the sum of the digits of a14?a_{14}?

88

99

1010

1111

1212

Answer: A
Video solution:
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Written solution:

Let dd be the common difference. If we let a1=10a_1 = 10 and a15=241,a_{15} = 241, we see that d2411014=6.5. d \geq \dfrac{241 - 10}{14} = 6.5. Since all the numbers are integers, dd must be at least 17.17.

Also, if a1=1a_1 = 1 and a15=250,a_{15} = 250, we get that d25011417.8. d \leq \dfrac{250 - 1}{14} \approx 17.8. Once again since all the numbers are integers, dd is at most 17.17. This tells us that dd is 17.17.

Note that 1714=238.17 \cdot 14 = 238. This means that a1a_1 must be at least 33 for a15a_{15} to be within the desired range.

If a1a_1 is greater than 3,3, however, a2a_2 becomes greater than 20,20, which is not allowed.

Now we know that a1=3a_1 = 3 and d=17.d = 17. This tell us that a14=3+1317=224. a_{14} = 3 + 13 \cdot 17 = 224.

Therefore, sum of the digits is 2+2+4=8.2 + 2 + 4 = 8.

Thus, A is the correct answer.