📘 Quiz

Solution: To find the maximum value of m for which n!/a^m is an integer, we need to determine how many times the prime factor a appears in the denominator (a^m) and then find the largest value of m such that n! is divisible by a^m. Let's assume a = 12 = 2^2 * 3 So, we are essentially looking for the largest power of 12 that divides n!. To find this, we can use the formula for finding the highest power of a prime p that divides n! (where p is a prime less than or equal to n): Highest power of p that divides n! = floor(n/p) + floor(n/p^2) + floor(n/p^3) + ... In our case, p = 2 and p = 3, and we want to find the highest power of 2 and 3 that divides n!. For p = 2: Highest power of 2 that divides 105! = floor(105/2) + floor(105/2^2) + floor(105/2^3) + ... Now, calculate each term: floor(105/2) = 52 floor(105/2^2) = floor(105/4) = 26 floor(105/2^3) = floor(105/8) = 13 ... Now, add all these values: 52 + 26 + 13 + 6 + 3 + 1 = 101 For p = 3: Highest power of 3 that divides 105! = floor(105/3) + floor(105/3^2) + floor(105/3^3) + ... Now, calculate each term: floor(105/3) = 35 floor(105/3^2) = floor(105/9) = 11 floor(105/3^3) = floor(105/27) = 3 ... Now, add all these values: 35 + 11 + 3 = 49 Since 12 = 2^2 * 3, the power of 12 that divides 105! is limited by the power of 3, which is 49. So, the correct answer is: 50