💡To determine the number of distinct prime factors of (30!), I need to identify all prime numbers that are less than or equal to 30. First, I recall that a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. I start by listing the numbers from 2 up to 30 and check each one to see if it is prime:- 2 is prime.- 3 is prime.- 4 is not prime (divisible by 2).- 5 is prime.- 6 is not prime (divisible by 2 and 3).- 7 is prime.- 8 is not prime (divisible by 2).- 9 is not prime (divisible by 3).- 10 is not prime (divisible by 2 and 5).- 11 is prime.- 12 is not prime (divisible by 2 and 3).- 13 is prime.- 14 is not prime (divisible by 2 and 7).- 15 is not prime (divisible by 3 and 5).- 16 is not prime (divisible by 2).- 17 is prime.- 18 is not prime (divisible by 2 and 3).- 19 is prime.- 20 is not prime (divisible by 2 and 5).- 21 is not prime (divisible by 3 and 7).- 22 is not prime (divisible by 2 and 11).- 23 is prime.- 24 is not prime (divisible by 2 and 3).- 25 is not prime (divisible by 5).- 26 is not prime (divisible by 2 and 13).- 27 is not prime (divisible by 3).- 28 is not prime (divisible by 2 and 7).- 29 is prime.- 30 is not prime (divisible by 2, 3, and 5).After this process, I compile the list of prime numbers less than or equal to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Counting these primes, I find there are 10 distinct prime factors.

💡Okay, so I have this problem where I need to figure out how many ways I can replace the asterisks in the number 2 * 0 * 1 * 6 * 0 * 2 * with digits from the set {0, 2, 4, 5, 7, 9}. The catch is that the resulting 12-digit number has to be divisible by 75. Hmm, okay, let me break this down step by step.First, I remember that 75 is 25 multiplied by 3, so for a number to be divisible by 75, it must be divisible by both 25 and 3. That gives me two separate conditions to check.Starting with divisibility by 25. From what I recall, a number is divisible by 25 if its last two digits form a number that's divisible by 25. So, looking at the number 2 * 0 * 1 * 6 * 0 * 2 *, the last two digits are represented by the last asterisk and the digit before it. Wait, actually, let me count the positions. The number is 2, *, 0, *, 1, *, 6, *, 0, *, 2, *. So, the last two digits are the 11th and 12th positions, which are both asterisks. So, both of these need to form a number divisible by 25.Looking at the digits I can use: 0, 2, 4, 5, 7, 9. Let me list all possible two-digit numbers from these digits that are divisible by 25. The possible two-digit numbers divisible by 25 are 00, 25, 50, 75. But from the given digits, I can only form 00, 25, 50, 75. Wait, but 25 is 2 and 5, which are both in the set. 50 is 5 and 0, which are also in the set. 75 is 7 and 5, which are in the set. 00 is 0 and 0, which is also possible.But hold on, the number is 12 digits long, and the first digit is 2, so it can't start with 0. But the last two digits can be 00, 25, 50, or 75. So, the possible endings are 00, 25, 50, or 75. Each of these endings uses two digits, so I need to assign these to the last two asterisks.But wait, the problem says each asterisk is replaced by any of the digits, so digits can be repeated. So, for the last two digits, I have four possible options: 00, 25, 50, 75. Each of these is a two-digit number. So, how many ways can I assign these?Well, for each of these endings, the last two asterisks are determined. So, if I choose 00, both last asterisks are 0. If I choose 25, the 11th digit is 2 and the 12th is 5. Similarly for 50 and 75. So, that gives me 4 possible endings.But wait, the problem says that each asterisk is replaced by any of the digits 0, 2, 4, 5, 7, 9, and digits can be repeated. So, does that mean that for the last two digits, I have 4 choices? Or is there something else?Wait, actually, the last two digits must form a number divisible by 25, so the possible endings are 00, 25, 50, 75. Each of these is a specific two-digit number, so each ending is a specific combination of digits. Therefore, for the last two asterisks, I have 4 possible combinations.But hold on, the problem is asking for the number of ways to replace the asterisks. So, if I fix the last two digits as one of these four possibilities, then the remaining asterisks can be filled with any of the digits, right? But I also have to satisfy the divisibility by 3 condition.So, let me note that down: for divisibility by 25, the last two digits must be one of 00, 25, 50, 75. So, 4 possibilities for the last two digits.Now, moving on to divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. So, I need to calculate the sum of all the digits of the number and ensure that it's a multiple of 3.Given the number: 2 * 0 * 1 * 6 * 0 * 2 *. Let me write down the fixed digits and the positions of the asterisks.Positions: 1:2, 2:*, 3:0, 4:*, 5:1, 6:*, 7:6, 8:*, 9:0, 10:*, 11:2, 12:*.Wait, no, actually, the number is 2 * 0 * 1 * 6 * 0 * 2 *. So, let's count the positions correctly.Position 1: 2Position 2: *Position 3: 0Position 4: *Position 5: 1Position 6: *Position 7: 6Position 8: *Position 9: 0Position 10: *Position 11: 2Position 12: *So, the fixed digits are at positions 1,3,5,7,9,11: 2,0,1,6,0,2.The asterisks are at positions 2,4,6,8,10,12.So, the sum of the fixed digits is 2 + 0 + 1 + 6 + 0 + 2 = 11.Now, the sum of the digits at the asterisk positions (positions 2,4,6,8,10,12) must be such that the total sum (11 + sum of asterisks) is divisible by 3.But wait, the last two digits (positions 11 and 12) are fixed as part of the divisibility by 25 condition. Wait, no, position 11 is fixed as 2, and position 12 is an asterisk. So, actually, position 12 is part of the asterisks.Wait, no, hold on. Let me clarify. The number is 2 * 0 * 1 * 6 * 0 * 2 *. So, position 12 is an asterisk, and position 11 is 2. So, the last two digits are position 11 (2) and position 12 (*). So, the last two digits are 2 followed by an asterisk. Wait, but that contradicts my earlier thought.Wait, no, hold on. Let me write the number structure again:Position 1: 2Position 2: *Position 3: 0Position 4: *Position 5: 1Position 6: *Position 7: 6Position 8: *Position 9: 0Position 10: *Position 11: 2Position 12: *So, the last two digits are position 11:2 and position 12:*. So, the last two digits are 2 followed by an asterisk. So, the last two digits are 2x, where x is the digit replacing the last asterisk.But for divisibility by 25, the last two digits must form a number divisible by 25. So, 2x must be divisible by 25. But 2x is a two-digit number where the first digit is 2 and the second digit is x.Looking at the possible x digits: 0,2,4,5,7,9.So, 20, 22, 24, 25, 27, 29.Which of these are divisible by 25? 20 is divisible by 25 (20 ÷ 25 = 0.8, which is not an integer). Wait, 20 is 20, which is not divisible by 25. 25 is 25, which is divisible by 25. 25 ÷ 25 = 1. So, 25 is good. 20 is not, 22 is not, 24 is not, 27 is not, 29 is not.Wait, so the only possible two-digit number ending with 2x that is divisible by 25 is 25. So, x must be 5. Therefore, the last digit must be 5.So, that means the last asterisk (position 12) must be 5. So, that's fixed. Therefore, the last two digits are 2 and 5, making 25, which is divisible by 25.So, that's one condition. So, position 12 is fixed as 5. Therefore, we only have 5 asterisks left to replace: positions 2,4,6,8,10.Each of these can be replaced by any of the digits 0,2,4,5,7,9, with repetition allowed.Now, moving on to divisibility by 3. The sum of all digits must be divisible by 3. The fixed digits are 2,0,1,6,0,2,5 (since position 12 is fixed as 5). Wait, position 11 is 2, and position 12 is 5. So, the fixed digits are 2,0,1,6,0,2,5. Wait, but position 12 was an asterisk, so it's now fixed as 5. So, the fixed digits are 2,0,1,6,0,2,5.Wait, let me recount the fixed digits:Position 1:2Position 3:0Position 5:1Position 7:6Position 9:0Position 11:2Position 12:5 (fixed)So, fixed digits: 2,0,1,6,0,2,5.Sum of fixed digits: 2 + 0 + 1 + 6 + 0 + 2 + 5 = 16.Now, the asterisks are at positions 2,4,6,8,10. Each of these can be 0,2,4,5,7,9. So, five asterisks, each with 6 choices, but with the condition that the total sum of all digits is divisible by 3.So, the total sum is fixed digits (16) plus the sum of the asterisks (let's call it S). So, 16 + S must be divisible by 3.Therefore, S ≡ (-16) mod 3. Let's compute -16 mod 3.16 divided by 3 is 5 with a remainder of 1, so 16 ≡ 1 mod 3. Therefore, -16 ≡ -1 ≡ 2 mod 3.So, S ≡ 2 mod 3. So, the sum of the five asterisks must be congruent to 2 modulo 3.So, now, we have five variables (digits) each taking values in {0,2,4,5,7,9}, and we need the sum of these five digits to be ≡2 mod 3.We need to count the number of 5-tuples (d1,d2,d3,d4,d5) where each di ∈ {0,2,4,5,7,9} and d1 + d2 + d3 + d4 + d5 ≡2 mod 3.This seems like a problem that can be approached using generating functions or using modular arithmetic properties.Alternatively, since each digit contributes a certain value mod 3, we can model this as a problem of counting the number of sequences where the sum mod 3 is 2.First, let's note the residues of each digit in {0,2,4,5,7,9} modulo 3.Compute each digit mod 3:0 mod 3 = 02 mod 3 = 24 mod 3 = 15 mod 3 = 27 mod 3 = 19 mod 3 = 0So, the digits and their residues:0:02:24:15:27:19:0So, for each digit, we can categorize them by their residue:Residue 0: 0,9Residue 1:4,7Residue 2:2,5So, for each asterisk, the possible residues are 0,1,2, with the following counts:Residue 0: 2 choices (0,9)Residue 1: 2 choices (4,7)Residue 2: 2 choices (2,5)So, each digit contributes a residue of 0,1, or 2, each with 2 choices.Now, we have five digits, each contributing a residue, and we need the total sum of residues to be ≡2 mod 3.This is a classic problem in combinatorics, often solved using generating functions or dynamic programming.Let me think about how to model this.Each digit can contribute 0,1, or 2, each with multiplicity 2. So, for each digit, the generating function is 2x^0 + 2x^1 + 2x^2.Since we have five digits, the generating function for the total sum is (2 + 2x + 2x^2)^5.We need the coefficient of x^{3k + 2} in this expansion, summed over all k such that 3k + 2 ≤ 5*2=10.But actually, since we're working modulo 3, we can consider the exponents modulo 3.So, the generating function modulo 3 is (2 + 2x + 2x^2)^5 mod (x^3 - 1).But perhaps a better approach is to use the multinomial theorem or to compute the number of ways using the concept of convolution.Alternatively, since each digit is independent, we can model this as a state machine where each state represents the current sum modulo 3, and we transition through each digit's contribution.Let me define dp[i][j] as the number of ways to choose the first i digits such that their sum ≡j mod 3.We need dp[5][2].Initialize dp[0][0] = 1, dp[0][1] = 0, dp[0][2] = 0.Then, for each digit from 1 to 5, we update the dp table based on the possible residues.For each step, the number of ways to reach state j is the sum over all possible residues r of (number of ways to reach state (j - r) mod 3 in the previous step) multiplied by the number of choices for residue r.But since each digit has 2 choices for each residue, we can factor that in.Wait, actually, for each digit, the number of ways to contribute residue r is 2 for each r in {0,1,2}.So, for each digit, the transition is:dp[i][j] = sum_{r=0}^{2} dp[i-1][(j - r) mod 3] * 2But since each residue r has 2 choices, it's 2 * dp[i-1][(j - r) mod 3].Wait, no, actually, for each digit, the number of ways to contribute residue r is 2, so for each transition, it's 2 * dp[i-1][(j - r) mod 3].But since we have three residues, each contributing 2 ways, the total number of ways for each transition is 2 * (sum over r of dp[i-1][(j - r) mod 3]).But actually, no, because for each residue r, the number of ways is 2, so for each j, dp[i][j] = sum_{r=0}^{2} (number of ways to get residue r) * dp[i-1][(j - r) mod 3].Since number of ways to get residue r is 2 for each r, it's:dp[i][j] = 2 * (dp[i-1][(j - 0) mod 3] + dp[i-1][(j - 1) mod 3] + dp[i-1][(j - 2) mod 3])But wait, no, that would be if each residue had 1 way, but since each residue has 2 ways, it's actually:dp[i][j] = 2 * (dp[i-1][(j - 0) mod 3] + dp[i-1][(j - 1) mod 3] + dp[i-1][(j - 2) mod 3])But that seems incorrect because it would imply that each transition is multiplied by 2, but actually, each residue has 2 choices, so for each residue r, the number of ways is 2, so the total number of ways is 2 * (sum over r of dp[i-1][(j - r) mod 3]).Wait, no, actually, for each digit, the number of ways to contribute residue r is 2, so for each j, dp[i][j] = sum_{r=0}^{2} (2 * dp[i-1][(j - r) mod 3]).But that would mean dp[i][j] = 2 * (dp[i-1][j] + dp[i-1][(j - 1) mod 3] + dp[i-1][(j - 2) mod 3]).But that can't be right because it would lead to exponential growth, which isn't the case.Wait, perhaps I need to think differently. Each digit contributes a residue r with multiplicity 2. So, for each digit, the number of ways to contribute residue r is 2. Therefore, for each digit, the generating function is 2 + 2x + 2x^2.So, for five digits, the generating function is (2 + 2x + 2x^2)^5.We need the coefficient of x^{3k + 2} in this expansion.But since we're working modulo 3, we can consider the exponents modulo 3.Let me compute (2 + 2x + 2x^2)^5 mod (x^3 - 1).But perhaps a better approach is to use the fact that (2 + 2x + 2x^2) = 2(1 + x + x^2).And 1 + x + x^2 is the cyclotomic polynomial for the 3rd roots of unity, so (1 + x + x^2)^n can be evaluated using roots of unity.But maybe that's overcomplicating.Alternatively, note that (2 + 2x + 2x^2) = 2(1 + x + x^2).So, (2(1 + x + x^2))^5 = 2^5 (1 + x + x^2)^5.We need the coefficient of x^{3k + 2} in this.But (1 + x + x^2)^5 can be expanded, and the coefficients correspond to the number of ways to get each residue.But perhaps it's easier to compute the total number of sequences and then divide by 3, but that might not be exact.Wait, the total number of sequences is 6^5 = 7776.Since each digit has 6 choices, and 5 digits.But we need the number of sequences where the sum ≡2 mod 3.In general, for problems like this, if the number of choices for each digit is uniform across residues, the number of sequences for each residue class is roughly equal.But in this case, each digit has 2 choices for each residue 0,1,2.So, for each digit, the number of ways to contribute residue 0,1,2 is equal (2 each). Therefore, the total number of sequences is 6^5, and the number of sequences for each residue class should be equal.Therefore, the number of sequences where the sum ≡0 mod 3, ≡1 mod 3, or ≡2 mod 3 should each be 6^5 / 3 = 7776 / 3 = 2592.Therefore, the number of ways is 2592.Wait, but let me verify this because sometimes when the number of choices isn't a multiple of the modulus, the counts aren't exactly equal.But in this case, each digit has exactly 2 choices for each residue, so the generating function is symmetric, and the number of sequences for each residue class should indeed be equal.Therefore, the number of ways is 6^5 / 3 = 2592.So, putting it all together:- We fixed the last digit as 5 to satisfy divisibility by 25.- Then, we had 5 remaining asterisks, each with 6 choices, but with the condition that their sum plus the fixed digits' sum (16) is divisible by 3.- Since the sum of the fixed digits is 16 ≡1 mod 3, we needed the sum of the asterisks to be ≡2 mod 3.- Given the symmetry in the choices for each digit, the number of valid sequences is 6^5 / 3 = 2592.Therefore, the total number of ways is 2592.

💡First, I need to determine the length of the train. The train is moving at a speed of 36 kilometers per hour (km/h) and passes a telegraph post in approximately 4 seconds.To calculate the distance the train travels in this time, I'll convert the speed from km/h to meters per second (m/s). Since 1 kilometer equals 1000 meters and 1 hour equals 3600 seconds, the conversion is as follows:36 km/h = (36 × 1000) / 3600 = 10 m/sNow that the speed is in meters per second, I can calculate the distance by multiplying the speed by the time:Distance = Speed × Time = 10 m/s × 4 s = 40 metersTherefore, the length of the train is 40 meters.

💡First, I need to find the value of the expression ((x + 5)(y + 5)) given that (x + y = -3) and (xy = 1).I'll start by expanding the expression:[(x + 5)(y + 5) = xy + 5x + 5y + 25]Next, I'll group the terms involving (x) and (y):[xy + 5(x + y) + 25]Now, I'll substitute the given values (x + y = -3) and (xy = 1) into the expression:[1 + 5(-3) + 25]Calculating the multiplication:[1 - 15 + 25]Finally, I'll perform the addition and subtraction to find the result:[1 - 15 + 25 = 11]

💡I need to determine the mode of the list: 3, 4, 5, 5, 5, 5, 7, 11, 21.First, I'll count how many times each number appears in the list.- The number 3 appears once.- The number 4 appears once.- The number 5 appears four times.- The number 7 appears once.- The number 11 appears once.- The number 21 appears once.Comparing the counts, the number 5 has the highest frequency with four occurrences.Therefore, the mode of the list is 5.

💡Okay, so I have this equation to solve: [(x - 2)^4 + (x - 6)^4 = 32.] Hmm, quartic equations can be tricky, but maybe there's a way to simplify this. Let me think about how to approach it.First, I notice that both terms are raised to the fourth power, which is even, so they will always be non-negative. The sum of two non-negative numbers equals 32, which is a positive number. That makes sense. Now, looking at the terms ((x - 2)^4) and ((x - 6)^4), I see that they are symmetric around some point. Maybe I can make a substitution to simplify the equation.Let me try to center the equation around the midpoint between 2 and 6. The midpoint is ((2 + 6)/2 = 4). So, if I let ( z = x - 4 ), then ( x = z + 4 ). That should center the equation around zero, which might make things easier.Substituting ( x = z + 4 ) into the equation:[( (z + 4) - 2 )^4 + ( (z + 4) - 6 )^4 = 32.]Simplifying inside the parentheses:[(z + 2)^4 + (z - 2)^4 = 32.]Okay, that looks better. Now, I can expand both terms. Let me recall the binomial expansion for ((a + b)^4) and ((a - b)^4). The expansion of ((z + 2)^4) is:[z^4 + 4 cdot z^3 cdot 2 + 6 cdot z^2 cdot 2^2 + 4 cdot z cdot 2^3 + 2^4]Which simplifies to:[z^4 + 8z^3 + 24z^2 + 32z + 16.]Similarly, the expansion of ((z - 2)^4) is:[z^4 - 4 cdot z^3 cdot 2 + 6 cdot z^2 cdot 2^2 - 4 cdot z cdot 2^3 + 2^4]Which simplifies to:[z^4 - 8z^3 + 24z^2 - 32z + 16.]Now, adding these two expanded forms together:[(z^4 + 8z^3 + 24z^2 + 32z + 16) + (z^4 - 8z^3 + 24z^2 - 32z + 16).]Let me combine like terms:- (z^4 + z^4 = 2z^4)- (8z^3 - 8z^3 = 0)- (24z^2 + 24z^2 = 48z^2)- (32z - 32z = 0)- (16 + 16 = 32)So, the equation simplifies to:[2z^4 + 48z^2 + 32 = 32.]Subtracting 32 from both sides:[2z^4 + 48z^2 = 0.]I can factor out a common term of (2z^2):[2z^2(z^2 + 24) = 0.]Dividing both sides by 2:[z^2(z^2 + 24) = 0.]Setting each factor equal to zero:1. (z^2 = 0) leads to (z = 0).2. (z^2 + 24 = 0) leads to (z^2 = -24), so (z = pm sqrt{-24} = pm 2isqrt{6}).Now, translating back to (x):Since (z = x - 4), we have:- For (z = 0): (x - 4 = 0) ⇒ (x = 4).- For (z = 2isqrt{6}): (x - 4 = 2isqrt{6}) ⇒ (x = 4 + 2isqrt{6}).- For (z = -2isqrt{6}): (x - 4 = -2isqrt{6}) ⇒ (x = 4 - 2isqrt{6}).So, the solutions are (x = 4), (x = 4 + 2isqrt{6}), and (x = 4 - 2isqrt{6}).Let me double-check my steps to make sure I didn't make any mistakes. I made a substitution to center the equation, expanded both terms correctly, combined like terms, factored, and solved for (z). Then, I translated back to (x). Everything seems to add up. I think these are the correct solutions.