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Expand 10.2^3 and 10.3^3 and computing the difference, we find that there are 60 integers between 10.2^3 and 10.3^3.

Expand 10.2^3 and 10.3^3 and computing the difference, we find that there are 60 integers between 10.2^3 and 10.3^3.

Let m=37. We can show that n is divisible by 37 if and only if the result of this process is divisible by 37 as follows:

If n is divisible by 37, then the alternating sum of the two-digit chunks is divisible by 37.

This is because the alternating sum of the two-digit chunks is just the sum of the digits of n, modulo 37. Since n is divisible by 37, the sum of its digits is also divisible by 37, so the alternating sum of the two-digit chunks is also divisible by 37.

If the alternating sum of the two-digit chunks is divisible by 37, then n is divisible by 37.

This is because the sum of the digits of n is congruent to the alternating sum of the two-digit chunks, modulo 37.

Therefore, if the alternating sum of the two-digit chunks is divisible by 37, then the sum of the digits of n is also divisible by 37, so n is divisible by 37.

Therefore, the alternating sum of the two-digit chunks is a divisibility test for 37.

To see this in action, let's consider the number 354764. The alternating sum of the two-digit chunks of 354764 is 64−47+35=52. Since 52 is divisible by 37, we know that 354764 is also divisible by 37.

We can solve this problem by considering the two ways to win and summing their probabilities:

Matching at least two white balls:

Favorable outcomes: We need to calculate the number of winning tickets where we match at least two of the chosen white balls.

Overcounting: We initially consider all possibilities where we pick three white balls (220 ways) and any red ball (8 ways). However, this overcounts tickets where we match all three white balls (counted three times for each red ball).

Correcting the overcount: There are 220 ways to pick three white balls and only 1 way to pick the red ball (doesn't matter which) for a total of 220 tickets where we match all three white balls. Subtracting this from the overcount gives us the actual number of winning tickets with at least two white ball matches.

Matching the red SuperBall:

Favorable outcomes: Here, we can pick any three white balls (220 ways) as long as they don't match the chosen red SuperBall (8 ways).

Calculations:

Matching at least two white balls:

Total outcomes (overcounted): 220 (white balls) * 8 (red balls) = 1760

Overcounted matches (all three white balls): 220 (white balls) * 1 (red ball) = 220

Corrected favorable outcomes: 1760 (total) - 220 (overcounted) = 1540

Matching the red SuperBall:

Favorable outcomes: 220 (white balls) * 8 (red balls that don't match) = 1760

Total Probability:

We win if either of these scenarios occurs. So, the total probability is the sum of the probabilities of each scenario:

Probability (matching at least two white balls): 1540 favorable outcomes / (12C3 * 8) total outcomes = 1540 / (220 * 8) = 44 / 248

Probability (matching the red SuperBall): 1760 favorable outcomes / (12C3 * 8) total outcomes = 1760 / (220 * 8) = 11 / 31

Total Probability (winning the SuperLottery):

(44 / 248) + (11 / 31) = 33/62.

Therefore, the probability that you win a super prize is 33/62.

To solve this problem, let's break it down step by step:

1. **Finding the probability of drawing an even number on the first draw:**

   There are 7 numbers on each color, and 3 of them are even (2, 4, 6). So, the probability of drawing an even number on the first draw is \(\frac{3}{7}\).

2. **Finding the probability of drawing a multiple of 3 on the second draw:**

   There are still 7 numbers on each color, and 2 of them are multiples of 3 (3, 6). After the first draw, there are 13 cards remaining. Among these, there are 4 cards labeled with multiples of 3. So, the probability of drawing a multiple of 3 on the second draw is \(\frac{4}{13}\).

3. **Multiplying the probabilities:**

   The probability of drawing an even number on the first draw and a multiple of 3 on the second draw is the product of the individual probabilities:
   \(\frac{3}{7} \times \frac{4}{13}\)

4. **Calculating the final probability:**

   Multiply the fractions:

   \(\frac{3}{7} \times \frac{4}{13} = \frac{12}{91}\)

So, the probability that the first card Grok draws is labeled with an even number and the second card Grok draws is labeled with a multiple of 3 is \(\frac{12}{91}\).

Since M is the midpoint of QR​, then QM=MR=2QR​=26​=3.

Triangle PQR is a right triangle since ∠PQR=90∘ (by the Pythagorean Theorem, PQ2+QR2=PR2, so 52+62=72).

Since M is the midpoint of the hypotenuse of this right triangle, then segment PM is half the length of the hypotenuse. Therefore, PM = PR/2 = 7/2.

There are two events we need to consider:

Event 1: Drawing an even number first.

There are 4 colors in the deck.

Each color has one even number (2, 4, or 6).

So, there are 4 * 3 = 12 cards that satisfy this condition (even number, any color).

Event 2: Drawing a multiple of 3 after drawing an even number (without replacement).

After drawing the first card, there are only 27 cards remaining.

There are 3 multiples of 3 left (3 and 6, since one even number - 6 - is already drawn).

However, since Grok isn't replacing the first card, there are only 2 colors left that have multiples of 3 (red and blue, as the even number 6 was likely green or yellow).

Therefore, there are 2 * 1 = 2 cards that satisfy this condition (multiple of 3, remaining colors).

Total Favorable Cases:

To get the probability, we need the number of favorable cases (both events happening) divided by the total number of possible cases (drawing any two cards).

Favorable cases: 12 (Event 1) * 2 (Event 2) = 24

Total possible cases: There are 28 cards total (7 numbers * 4 colors), and we draw 2 without replacement. So, the total number of possible choices is 28C2 (28 choose 2) which is 28 * 27 / (2 * 1) = 378

Probability:

Therefore, the probability that Professor Grok draws an even number first, followed by a multiple of 3 (without replacement), is:

Probability = Favorable Cases / Total Possible Cases Probability = 24 / 378

Simplifying the fraction:

Both the numerator (24) and denominator (378) have a common divisor of 6. We can simplify:

Probability = (24 / 6) / (378 / 6) Probability = 4 / 63

So, the probability is 4/63.

The problem has been posted here: https://web2.0calc.com/questions/pre-calc-problem