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We can solve this problem by counting the number of rolls that don't score any points and then dividing by the total number of possible rolls. Here's how:

Rolls that score 0 points:

Six singles: There are 5 ways to roll a single 1, 5 ways to roll a single 2, etc., giving 6⋅5=30​ ways to roll six singles.

Three pairs: There are (26​)=15 ways to choose two dice for the first pair, and then 4 ways to choose a value for that pair (since the dice cannot be used in more than one combination).

Similarly, there are 4 choices for the second pair, giving a total of 15⋅4⋅4=240​ ways to roll three pairs.

One triple and one pair: There are (16​)=6 ways to choose the die for the triple, and then 5 choices for its value.

There are (25​)=10 ways to choose the two dice for the pair, and then 4 choices for its value, giving a total of 6⋅5⋅10⋅4=1200​ ways to roll one triple and one pair.

Four singles and one double: There are (16​)=6 ways to choose the die for the double, and then 5 choices for its value.

There are (45​)=5 ways to choose the four dice for the singles, giving a total of 6⋅5⋅5=150​ ways to roll four singles and one double.

Total number of rolls:

There are 6⋅6⋅6⋅6⋅6⋅6 = 6^6 possible outcomes when rolling six dice.

Probability of scoring 0 points:

The probability of scoring 0 points is the number of rolls with no points divided by the total number of rolls: (30 + 240 + 1200 + 150)/^6 = ​15/424.

Here's how to convert 2231_4 to base 2 without first converting to decimal:

Start with the largest power of 4 that fits into 2231:

4^3 = 64 is the largest power of 4 less than 2231.

Divide 2231 by 64 and get the quotient 34 and the remainder 3.

Write down the quotient 34 as the first digit in the base-2 representation.

Repeat for the remaining digits:

Repeat the process with the remainder 3.

The largest power of 4 that fits into 3 is 4^1 = 4.

Divide 3 by 4 and get the quotient 0 and the remainder 3.

Write down the quotient 0 as the second digit.

Repeat with the remaining 3.

The largest power of 4 that fits into 3 is 4^0 = 1.

Divide 3 by 1 and get the quotient 3 and the remainder 0.

Write down the quotient 3 as the third digit.

Combine the digits:

The digits obtained in the previous steps, in reverse order, give the base-2 representation.

Therefore, 2231_4 in base 2 is 100111_2.

This method directly uses the powers of 4 in the base-4 representation to express the number in the base-2 system, avoiding the need for intermediate decimal conversion.

Here's how to find the number of ordered pairs (a, b) of integers:

Multiply both sides by the common denominator (a + 5) in the first term:

a + 2 = b * (a + 5) / 4

Simplify and expand the expression on the right:

4 * (a + 2) = b * (a + 5)

4a + 8 = ab + 5b

Move all terms containing a to one side:

4a - ab = 5b - 8

a(4 - b) = 5b - 8

Factor out the greatest common factor (GCF) in both sides:

a(b - 4) = 5(b - 8/5)

Analyze the equation:

This equation implies that a and (b - 4) are factors of 5.

Since a and b are integers, the possible factors of 5 are 1, -1, 5, and -5.

Consider each factor combination:

Case 1: a = 1, (b - 4) = 5 - b = 9

Case 2: a = -1, (b - 4) = -5 - b = -1

Case 3: a = 5, (b - 4) = 1 - b = 5

Case 4: a = -5, (b - 4) = -1 - b = -3

Check for solutions:

All four cases lead to valid integer solutions for a and b: (1, 9), (-1, -1), (5, 5), and (-5, -3).

Therefore, there are 4 ordered pairs (a, b) of integers that satisfy the given equation.

Let's analyze the situation step-by-step:

Initial Weight:

Total weight = 200 kg

Water content = 98%

Non-water content = 2% (100% - 98%)

Weight of non-water content = 200 kg * 2% = 4 kg

Water Evaporation:

After 2 days, water content decreases to 95%.

Water loss = 3% (98% - 95%)

Weight of water lost = 200 kg * 3% = 6 kg

Final Weight:

All non-water content remains (it doesn't evaporate).

Final weight = Weight of non-water content + Remaining water

Final weight = 4 kg + (200 kg - 6 kg) = 4 kg + 194 kg = 198 kg

Therefore, the total weight of the berries after two days in the sun is 198 kg.

Since AN=3 and BN=4, we have AB=AN+BN=3+4=7. Let I be the incenter of triangle ABC, so IN is a radius of the incircle. Since AN=3, we have IN=3.

The incenter of a triangle is also the intersection of the angle bisectors. Therefore, ∠AIB=∠ACB/2. Also, since ∠INB=90∘, we have ∠ANB=90∘−∠AIB=90∘−∠ACB/2.

By the Law of Cosines on triangle ABC,

[\cos C = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{81 + BC^2 - 49}{18BC} = \frac{BC^2 + 32}{18BC}.]

By the Law of Cosines on triangle ANB,

[\cos (90^\circ - C/2) = \sin \frac{C}{2} = \frac{AN}{AB} = \frac{3}{7}.]

Therefore,

[\cos C = 1 - 2 \sin^2 \frac{C}{2} = 1 - 2 \cdot \left(\frac{3}{7}\right)^2 = \frac{47}{49}.]

We have [\frac{BC^2 + 32}{18BC} = \frac{47}{49}.]

Multiplying both sides by 882BC, we get 49BC2+1536=882BC. Then 49BC2−882BC+1536=0, which factors as (7BC−48)(7BC−32)=0. Therefore, BC=48/7 or BC=32/7.

Since AN=3<7/2, the incenter I is closer to A than to B. Therefore, I is inside the smaller of the two triangles formed by AC and AB.

This means that BC>AC=9. Therefore, BC = 48/7.

We can solve this problem by considering the number of moves right and up that Josh makes.

Josh needs to move 6 steps to the right and 7 steps up, for a total of 13 moves.

Each move can be either right or up, so there are 2 choices for each of the 13 moves.

Therefore, the total number of ways to reach his home is 2^13 = 8192.

However, this includes paths that go through the fire at (2,3). To exclude these paths, we need to consider how many paths go through the fire and subtract them from the total.

A path goes through the fire if it includes a move right at (1,2) and a move up at (2,2).

There are 11 ways to choose which of the remaining 11 moves will be the right move at (1,2).

For each of these 11 choices, there are 6 ways to choose which of the remaining 6 moves will be the up move at (2,2).

Therefore, there are 11 * 6 = 66 paths that go through the fire.

Finally, we subtract the number of paths that go through the fire from the total number of paths to find the number of paths that avoid the fire:

8192 total paths - 66 paths through the fire = 8126 paths that avoid the fire.

Therefore, there are 8126 ways for Josh to reach his home without going through the fire.

Converting the first equation to base 10, we have 6A+B+C=6C. Simplifying, we have 6A+B=5C.

Converting the second equation to base 10, we have 6A+B+6B+A=6C+C. Simplifying, we have 7A+7B=7C. Dividing both sides by 7, we have A+B=C.

Substituting C=A+B into the first equation, we have 6A+B+A+B=6(A+B). Simplifying, we have 8A+2B=6A+6B. Combining like terms, we have 2A=4B. Dividing both sides by 2, we have A=2B.

Since A, B, and C are distinct digits less than 6, the only possible values for (A,B,C) are (2,1,3) and (4,2,6). However, C must be less than 6, so the only solution is ABC = 213.

By the Triangle Inequality,

[|a_1 - a_6| = |a_1 - a_2 + a_2 - a_3 + a_3 - a_4 + a_4 - a_5 + a_5 - a_6|

\le |a_1 - a_2| + |a_2 - a_3| + |a_3 - a_4| + |a_4 - a_5| + |a_5 - a_6|.]

Equality occurs when a2​, a3​, a4​, and a5​ all lie on the line segment joining a1​ and a6​.

Also by the Triangle Inequality,

[|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_4| + |a_4 - a_5| + |a_5 - a_6| + |a_6 - a_{10}| + |a_{10} - a_1|

\ge |a_1 - a_{10}| + |a_{10} - a_1| = 2|a_1 - a_{10}|.]

Equality occurs when a2​, a3​, a4​, a5​, and a6​ all lie on the line segment joining a1​ and a10​.

Thus, if a2​, a3​, a4​, a5​, and a6​ all lie on the line segment joining a1​ and a10​, then

[|a_1 - a_6| = 500 - |a_6 - a_{10}| - |a_9 - a_{10}| - |a_8 - a_9| - |a_7 - a_8|.]

To maximize ∣a1​−a6​∣, we minimize ∣a6​−a10​∣+∣a9​−a10​∣+∣a8​−a9​∣+∣a7​−a8​∣.

The minimum value is 0, which occurs when a6​=a7​=a8​=a9​=a10​.

Thus, the largest possible value of ∣a1​−a6​∣ is 500​.