AMC 8 · Test Mode

2022 AMC 8

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Problem 1 · 2022 AMC 8 Easy
Geometry & Measurement area-decomposition grid-counting
amc8-2022-01
Show hint (soft nudge)
There's a grid behind the logo — let it count for you instead of trying to eyeball the area.
Show hint (sharpest)
Lay the logo on the grid and add area piece by piece: count the whole gray squares, then the gray half-squares (triangles).
Show solution
  1. The logo splits into whole unit squares plus triangular half-squares along the grid lines.
  2. There are 4 whole gray squares in the middle, and the surrounding gray triangles add up to 6 more square inches (twelve half-squares).
  3. Total area: 4 + 6 = 10 square inches.
A 10 square inches.
Another way: spot five identical tilted unit-squares (MAA)
  1. The logo is made of five small tilted squares whose sides go diagonally across one grid cell, so each side is √(12 + 12) = √2.
  2. Each such square has area (√2)2 = 2.
  3. Five of them: 5 × 2 = 10 square inches.
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Problem 2 · 2022 AMC 8 Easy
Algebra & Patterns custom-operation substitution

Consider these two operations:

ab = a2b2
ab = (ab)2

What is the value of (5 ◆ 3) ★ 6?

Show hint
Work the inside (the ◆) first, then apply the ★ to that result.
Show solution
  1. Inside first: 5 ◆ 3 = 52 − 32 = 25 − 9 = 16.
  2. Now apply ★: 16 ★ 6 = (16 − 6)2 = 102.
  3. = 100.
D 100.
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Problem 3 · 2022 AMC 8 Medium
Number Theory factorization factor-triples casework

When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?

Show hint (soft nudge)
100 has only a handful of factor groupings. Be systematic so you don't miss any or double-count.
Show hint (sharpest)
List the ways to write 100 as a product of three different whole numbers, smallest to largest. (100 = 22 × 52.)
Show solution
  1. Try the smallest factor as 1: 1×2×50, 1×4×25, 1×5×20 — all have a < b < c.
  2. Now without a 1: 2×5×10 works (2 < 5 < 10).
  3. Others repeat a factor (like 1×10×10 or 2×2×25), so they don't count.
  4. That gives 4 valid choices.
E 4 ways.
Another way: bound the smallest factor first (MAA)
  1. Since a < b < c and abc = 100, we get a3 < 100, so a ≤ 4. And a must be a factor of 100, so a ∈ {1, 2, 4}.
  2. For each: a = 1 gives (1,2,50), (1,4,25), (1,5,20); a = 2 gives (2,5,10); a = 4 gives nothing (would need bc = 25 with 4 < b < c).
  3. Total: 4 ways.
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Problem 4 · 2022 AMC 8 Medium
Geometry & Measurement reflection transformations composition
amc8-2022-04
Show hint (soft nudge)
Two flips in a row often equal a single, simpler motion. Try to see the final position, not each step.
Show hint (sharpest)
Two reflections in a row, over two lines that cross, equal a single rotation about the crossing point — by twice the angle between the lines.
Show solution
  1. Reflect M over line q, then over line p.
  2. Doing two reflections about lines through one point is the same as rotating M about that point by twice the angle between p and q.
  3. Carrying M through both flips lands it in the position shown in figure (E).
E It matches figure (E).
Another way: just do the two reflections in order (MAA)
  1. Reflect M over line q: the M flips across that line, landing in its mirror image position.
  2. Reflect that result over line p: a second flip across the other line.
  3. The final position matches choice (E).
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Problem 5 · 2022 AMC 8 Stretch
Logic & Word Problems ages work-backward sum-constraint

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 30. How many years older than Bella is Anna?

Show hint (soft nudge)
Don't solve for ages directly — figure out who is how old today first.
Show hint (sharpest)
Work out each one's age today first. Bella was 6 five years ago; the kitten was newborn five years ago. The leftover part of 30 is Anna's age.
Show solution
  1. Bella today: 6 + 5 = 11. Kitten today: 0 + 5 = 5.
  2. The three ages sum to 30, so Anna = 30 − 11 − 5 = 14.
  3. Anna − Bella = 14 − 11 = 3 years.
C 3 years older.
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Problem 6 · 2022 AMC 8 Easy
Algebra & Patterns substitution arithmetic-sequence

Three positive integers are equally spaced on a number line. The middle number is 15 and the largest number is 4 times the smallest number. What is the smallest of these three numbers?

Show hint (soft nudge)
Equally spaced means the middle equals the average of the outer two.
Show hint (sharpest)
Let smallest = x, largest = 4x. Their average = 5x/2 = 15.
Show solution
  1. Let smallest = x, so largest = 4x. Equally spaced ⇒ middle = (x + 4x)/2 = 5x/2.
  2. Set 5x/2 = 15 ⇒ x = 6.
C 6.
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Problem 7 · 2022 AMC 8 Easy
Ratios, Rates & Proportions distance-speed-time unit-rate

When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about 56 kilobits per second. Approximately how many minutes would the download of a 4.2-megabyte song have taken at that speed? (Note that there are 8000 kilobits in a megabyte.)

Show hint (soft nudge)
Convert the song to kilobits first (units of the speed), then time = size ÷ speed.
Show hint (sharpest)
4.2 MB × 8000 kbits/MB = 33,600 kbits. Divide by 56 kbits/sec.
Show solution
  1. Song size: 4.2 × 8000 = 33,600 kilobits.
  2. Time = 33,600 ÷ 56 = 600 seconds = 10 minutes.
B 10 minutes.
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Problem 8 · 2022 AMC 8 Medium
Fractions, Decimals & Percents fraction-to-decimal

What is the value of

13 · 24 · 35 · … · 1820 · 1921 · 2022 ?
Show hint (soft nudge)
Every number 3 through 20 appears once as a numerator and once as a denominator. What cancels?
Show hint (sharpest)
After cancellation: only 1 · 2 (from numerators) and 21 · 22 (from denominators) survive.
Show solution
  1. Numerators run 1, 2, 3, …, 20. Denominators run 3, 4, 5, …, 22. Every number from 3 to 20 appears in both lists and cancels.
  2. Left over: 1 · 221 · 22 = 2462 = 1231.
B 1/231.
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Problem 9 · 2022 AMC 8 Easy
Fractions, Decimals & Percents percent-multiplier

A cup of boiling water (212°F) is placed to cool in a room whose temperature remains constant at 68°F. Suppose the difference between the water temperature and the room temperature is halved every 5 minutes. What is the water temperature, in degrees Fahrenheit, after 15 minutes?

Show hint (soft nudge)
Don't track the temperature directly — track the difference from room temp.
Show hint (sharpest)
Initial gap: 212 − 68 = 144. Halved three times in 15 minutes: 144 / 23 = 18. Add to room temp.
Show solution
  1. Initial gap above room: 212 − 68 = 144°F.
  2. 15 minutes = three 5-minute halvings: 144 → 72 → 36 → 18.
  3. Final temperature = 68 + 18 = 86°F.
B 86&deg;F.
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Problem 10 · 2022 AMC 8 Easy
Ratios, Rates & Proportions distance-speed-time graph-reading
amc8-2022-10
Show hint (soft nudge)
Compute the peak distance (when she reaches the trail) and the timing of the three phases.
Show hint (sharpest)
Drive 2 hr at 45 mph = 90 mi. Hike 3 hr at the trail (distance flat). Drive back at 60 mph → 90/60 = 1.5 hr.
Show solution
  1. Outbound: 2 hr × 45 mph = 90 miles, reaching the trail at 10am.
  2. Hike 3 hr (distance stays at 90 miles), so she leaves the trail at 1pm.
  3. Return: 90 miles ÷ 60 mph = 1.5 hr, so she's home at 2:30pm.
  4. The graph that peaks at 90 miles between 10am and 1pm and comes back to 0 at 2:30pm is choice E.
E Graph (E).
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Problem 11 · 2022 AMC 8 Easy
Arithmetic & Operations off-by-one

Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating 3 inches of pasta from the middle of one piece. In the end, he has 10 pieces of pasta whose total length is 17 inches. How long, in inches, was the piece of pasta he started with?

Show hint (soft nudge)
Each bite turns one piece into two. How many bites does it take to end with 10 pieces?
Show hint (sharpest)
10 pieces = 9 bites × 3 inches removed = 27 inches gone. Original = remaining + removed.
Show solution
  1. Each bite from the middle of a piece adds one new piece (one piece → two). Starting from 1 piece, reaching 10 pieces took 9 bites.
  2. 9 bites × 3 inches = 27 inches removed.
  3. Original length = 17 (remaining) + 27 (eaten) = 44 inches.
D 44 inches.
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Problem 12 · 2022 AMC 8 Medium
Counting & Probability careful-counting perfect-square
amc8-2022-12
Show hint (soft nudge)
N is a 2-digit number with tens digit from Spinner A and ones digit from Spinner B. List the perfect squares in range.
Show hint (sharpest)
Spinner A ∈ {5, 6, 7, 8}, so N is in the 50s, 60s, 70s, or 80s. Perfect squares there: 64 and 81.
Show solution
  1. N = 10A + B with A ∈ {5, 6, 7, 8} and B ∈ {1, 2, 3, 4}: N is between 51 and 84.
  2. Perfect squares in [51, 84]: 64 (= 82) and 81 (= 92). 64 = (A=6, B=4); 81 = (A=8, B=1). Both pairs are reachable.
  3. Probability = 24 × 4 = 18.
B 1/8.
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Problem 13 · 2022 AMC 8 Medium
Algebra & Patterns substitution sum-constraint

How many positive integers can fill the blank in the sentence below?

"One positive integer is ___ more than twice another, and the sum of the two numbers is 28."

Show hint (soft nudge)
Set up: smaller = a, larger = 2a + c. The blank is c.
Show hint (sharpest)
a + (2a + c) = 28 ⇒ c = 28 − 3a. For both to be positive integers, count valid a.
Show solution
  1. Let smaller = a, larger = 2a + c (with c ≥ 1 the blank). Sum: 3a + c = 28, so c = 28 − 3a.
  2. a ≥ 1 and c ≥ 1 ⇒ 28 − 3a ≥ 1 ⇒ a ≤ 9.
  3. a ∈ {1, 2, …, 9} ⇒ 9 values for the blank.
D 9 values.
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Problem 14 · 2022 AMC 8 Medium
Counting & Probability careful-counting

In how many ways can the letters in BEEKEEPER be rearranged so that two or more E's do not appear together?

Show hint (soft nudge)
BEEKEEPER has 5 E's and 4 non-E letters (B, K, P, R) in 9 positions. Where can the 5 E's go without two being adjacent?
Show hint (sharpest)
5 non-adjacent positions in a row of 9 force the E's into positions 1, 3, 5, 7, 9. Then the 4 non-E letters fill positions 2, 4, 6, 8 in some order.
Show solution
  1. BEEKEEPER = 5 E's + {B, K, P, R} in 9 slots. To keep no two E's adjacent, the 5 E's must occupy all 5 odd positions (1, 3, 5, 7, 9) — the only way to fit 5 non-adjacent positions in 9.
  2. The other 4 letters fill positions 2, 4, 6, 8 in any order: 4! = 24 arrangements.
D 24 ways.
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Problem 15 · 2022 AMC 8 Medium
Fractions, Decimals & Percents unit-rate estimate-and-pick
amc8-2022-15
Show hint (soft nudge)
For each weight, find the lowest price dot in that column. Then compute price ÷ weight.
Show hint (sharpest)
Lowest-price-per-ounce will favor a weight where the cheapest available pepper drops well below the dollar-per-ounce line.
Show solution
  1. Lowest price at each weight (reading off the scatter): 1 oz ≈ $1.25 (rate ≈ 1.25), 2 oz ≈ $2 (1.00), 3 oz ≈ $2.5 (≈ 0.83), 4 oz ≈ $3.9 (≈ 0.97), 5 oz ≈ $4.5 (≈ 0.90).
  2. The 3-ounce option has the lowest rate (~$0.83/oz). Answer: 3 ounces.
C 3 ounces.
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Problem 16 · 2022 AMC 8 Medium
Arithmetic & Operations sum-constraint

Four numbers are written in a row. The average of the first two is 21, the average of the middle two is 26, and the average of the last two is 30. What is the average of the first and last of the numbers?

Show hint (soft nudge)
Convert each average into a sum (multiply by 2). You don't need the four numbers individually — just a + d.
Show hint (sharpest)
(a + b) + (c + d) = (a+b+c+d), and you can subtract (b+c) to leave a+d.
Show solution
  1. a + b = 42, b + c = 52, c + d = 60.
  2. (a+b) + (c+d) − (b+c) = a + d = 42 + 60 − 52 = 50.
  3. Average = 50/2 = 25.
B 25.
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Problem 17 · 2022 AMC 8 Medium
Number Theory last-digit mod-10

If n is an even positive integer, the double-factorial notation n!! represents the product of all the even integers from 2 to n. For example: 8!! = 2 × 4 × 6 × 8. What is the units digit of the following sum?

2!! + 4!! + 6!! + … + 2022!!
Show hint (soft nudge)
Once a double-factorial includes 10 as a factor, it ends in 0. Which terms in the sum still affect the ones digit?
Show hint (sharpest)
Only 2!!, 4!!, 6!!, 8!! contribute — everything from 10!! onward ends in 0.
Show solution
  1. n!! for n ≥ 10 contains the factor 10, so its units digit is 0.
  2. Compute only the survivors: 2!! = 2, 4!! = 8, 6!! = 48, 8!! = 384. Their units digits: 2, 8, 8, 4.
  3. Sum of units digits: 2 + 8 + 8 + 4 = 22. Units digit: 2.
B Units digit 2.
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Problem 18 · 2022 AMC 8 Hard
Geometry & Measurement area area-decomposition

The midpoints of the four sides of a rectangle are (−3, 0), (2, 0), (5, 4), and (0, 4). What is the area of the rectangle?

Show hint (soft nudge)
The midpoints of any quadrilateral form a parallelogram — with area exactly half the original.
Show hint (sharpest)
Compute the midpoint-parallelogram's area, then double.
Show solution
  1. The four midpoints form a parallelogram. Its base ((-3,0) to (2,0)) is length 5; its height (0 to 4) is 4. Area = 5 × 4 = 20.
  2. Theorem: the midpoint quadrilateral has half the area of the original. So the rectangle has area 2 × 20 = 40.
C 40.
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Problem 19 · 2022 AMC 8 Medium
Arithmetic & Operations careful-counting
amc8-2022-19
Show hint (soft nudge)
With 20 students the median is the average of the 10th and 11th scores. For the median to be 85, both the 10th and 11th need to be (at least) 85.
Show hint (sharpest)
Currently 7 students scored ≥ 85. To make the 11th-highest hit 85, you need 4 more at ≥ 85.
Show solution
  1. Median of 20 scores = average of the 10th and 11th when sorted. To pin the median at 85, the 10th and 11th positions must both equal 85.
  2. From the dot plot, 7 students originally scored ≥ 85. Adding 5 to a student scoring 80 raises them to 85.
  3. Need at least 11 students at ≥ 85, so at least 11 − 7 = 4 students must be raised.
C 4 students.
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Problem 20 · 2022 AMC 8 Hard
Algebra & Patterns substitution sum-constraint
amc8-2022-20
Show hint (soft nudge)
Find the common row/column sum from the known top row, then express each missing cell in terms of x.
Show hint (sharpest)
Top row sums to 12 ⇒ magic sum is 12. The other three missing cells are 14−x, 4−x, x−1. x must beat all of them.
Show solution
  1. Top row: −2 + 9 + 5 = 12. So every row and column sums to 12.
  2. First column: −2 + (above x) + x = 12 ⇒ cell above x = 14 − x.
  3. Bottom row: x + (middle bottom) + 8 = 12 ⇒ middle bottom = 4 − x.
  4. Middle row: (14−x) + (center) + (−1) = 12 ⇒ center = x − 1.
  5. x must be the largest: x > 14−xx > 7; x > 4−xx > 2; x > x−1 always.
  6. Smallest integer satisfying all: x = 8.
D x = 8.
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Problem 21 · 2022 AMC 8 Hard
Fractions, Decimals & Percents percent-multiplier casework

Steph scored 15 baskets out of 20 attempts in the first half of a game, and 10 baskets out of 10 attempts in the second half. Candace took 12 attempts in the first half and 18 attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?

Show hint (soft nudge)
Steph took 20 + 10 = 30 attempts; Candace took 12 + 18 = 30. Same total, same overall % ⇒ same total makes.
Show hint (sharpest)
Steph: 15 + 10 = 25 makes. Candace must also make 25. With her per-half percentages strictly below Steph's, only one (f, s) split works.
Show solution
  1. Same total attempts (30 each) + same overall percentage ⇒ same total makes. Steph made 15 + 10 = 25, so Candace also made 25.
  2. Let Candace's makes be f (first half, out of 12) and s (second, out of 18). Per-half percentages strictly below Steph: f/12 < 15/20 = 3/4 ⇒ f ≤ 8. And s/18 < 1 ⇒ s ≤ 17.
  3. f + s = 25, f ≤ 8, s ≤ 17 ⇒ only f = 8, s = 17 fits.
  4. sf = 17 − 8 = 9.
C 9 more baskets.
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Problem 22 · 2022 AMC 8 Hard
Ratios, Rates & Proportions distance-speed-time casework

A bus takes 2 minutes to drive from one stop to the next, and waits 1 minute at each stop to let passengers board. Zia takes 5 minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus 3 stops behind. After how many minutes will Zia board the bus?

Show hint (soft nudge)
Zia only checks at multiples of 5 minutes (when she reaches a stop). At each check, see where the bus is and whether to wait.
Show hint (sharpest)
Bus cycle: 2 min driving + 1 min stopped = 3 min per stop. Track stop indexes at t = 0, 5, 10, 15.
Show solution
  1. Index stops by the bus's start (stop 0). At t = 0, Zia is at stop 3, bus at stop 0.
  2. t = 5: Zia at stop 4. Bus took 5 min → finished stop 1 (arrived at 2 min, left at 3 min, arrived at stop 2 at 5 min). Bus is at stop 2 — not yet at the previous stop (3), so Zia walks on.
  3. t = 10: Zia at stop 5. Bus: from t = 5 (at stop 2) waits 1 min (leaves at 6), drives 2 min to stop 3 (arrives at 8), waits till 9, drives to stop 4 (arrives at 11). So at t = 10, bus is mid-drive between stops 3 and 4 — not at the previous stop (4), so Zia walks on.
  4. t = 15: Zia at stop 6. Bus: arrives at stop 4 at 11, waits till 12, drives to stop 5 (arrives 14, waits till 15). At t = 15, bus is at stop 5 — the previous stop. Zia waits.
  5. Bus leaves stop 5 at t = 15 and drives 2 min to stop 6: arrives at t = 17.
A 17 minutes.
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Problem 23 · 2022 AMC 8 Hard
Counting & Probability careful-counting casework symmetry
amc8-2022-23
Show hint (soft nudge)
If ▵'s form a horizontal line, the ○'s line must also be horizontal (otherwise a cell would need to be both shapes). Same for vertical. So count vertical-line configurations and double.
Show hint (sharpest)
Casework on number of vertical lines: 3 (one for each column) or exactly 2.
Show solution
  1. A ▵ line and a ○ line can't be perpendicular (a cell would be both shapes). So both lines are horizontal, OR both are vertical. Count vertical, then multiply by 2.
  2. Vertical: case 3 lines — each of the 3 columns is monochromatic. 2³ = 8 colorings, minus the 2 all-same colorings (only one shape gets a line) → 6.
  3. Vertical: case 2 lines — one ▵ column, one ○ column, one mixed. 3 ways to pick the ▵ column × 2 remaining for ○ × 6 mixed configurations for the leftover column (2³ minus the two all-same = 6) = 36.
  4. Vertical total: 6 + 36 = 42. By symmetry, horizontal also = 42. Total: 42 + 42 = 84.
D 84 configurations.
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Problem 24 · 2022 AMC 8 Hard
Geometry & Measurement area spatial-reasoning
amc8-2022-24
Show hint (soft nudge)
Identify the prism: a triangular base of legs 6 and 8 (right triangle), with prism length 8.
Show hint (sharpest)
Match edges in the net: AH = EF = 8 forces the rectangle sides; GH = 14 with HJ = 8 gives GJ = 6 — the other leg of the triangular base.
Show solution
  1. Tracing the edges as the net folds: AB = BC = HJ = GF = 8 (the prism length). Also BJ = AH = 8.
  2. GJ = GH − HJ = 14 − 8 = 6.
  3. Triangular base BJG is a right triangle with legs 6 and 8: area = (6 × 8)/2 = 24.
  4. Volume = base × length = 24 × 8 = 192.
C Volume 192.
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Problem 25 · 2022 AMC 8 Hard
Counting & Probability careful-counting casework

A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops, what is the probability that the cricket has returned to the leaf where it started?

Show hint (soft nudge)
Track only one number: pn = probability the cricket is on the starting leaf after n hops.
Show hint (sharpest)
From start, the cricket must leave (contributes 0 to pn+1). From any other leaf, prob 1/3 of returning. So pn+1 = (1 − pn)/3.
Show solution
  1. Let pn = P(on starting leaf after n hops). From the start, the cricket can't stay; from any non-start, it returns with probability 1/3.
  2. pn+1 = (1 − pn) · 13.
  3. Iterate from p0 = 1: p1 = 0, p2 = 1/3, p3 = 2/9, p4 = (1 − 2/9)/3 = (7/9)/3 = 7/27.
E 7/27.
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