If the product represented by 128! is divisible by 18^n, what is the largest possible value of n?
is this from spirit of math
Yes
To determine the largest possible value of n such that the product represented by 128! is divisible by 18^n, we need to analyze the prime factorization of 18 and the prime factorization of 128!.
The prime factorization of 18 is 2 * 3^2.
To find the prime factorization of 128! (128 factorial), we need to consider all the numbers from 1 to 128 and decompose them into prime factors.
First, we consider the multiples of 2. The number 2 appears as a factor in half of the numbers from 1 to 128, which is 128/2 = 64 times. However, some of the numbers from 1 to 128 have more than one factor of 2, so we need to account for those multiples as well.
Next, we consider the multiples of 3. The number 3 appears as a factor in approximately 128/3 = 42.67 numbers. However, some of the numbers have more than one factor of 3.
Finally, we consider the multiples of any prime number greater than 3. In this case, 18 is already the greatest prime factor in the expression 18^n.
To determine the largest possible value of n, we need to find the highest power of each prime factor in the prime factorization of 128!.
For the prime factor 2, we need to count the number of 2's in the prime factorization of 128!. There are 64 numbers from 1 to 128 that are divisible by 2, producing a total of 64 factors of 2. However, some numbers have more than one factor of 2. To find the number of numbers with more than one factor of 2, we divide 128 by 2^2 (since 2^2 gives the next power of 2), which gives us 32. Consequently, there are 32 numbers that have more than one factor of 2. Continuing this process, we divide 128 by 2^3 to find the number of numbers with at least three factors of 2, which is 16. We continue dividing until the power of 2 exceeds 128.
Therefore, the highest power of 2 in the prime factorization of 128! is 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127.
For the prime factor 3, we apply the same logic. The number of numbers divisible by 3 is 128/3 = 42.67, but some numbers have more than one factor of 3. Continuing the calculation process, we find that the highest power of 3 in the prime factorization of 128! is 42 + 14 + 4 + 1 = 61.
Since 18 = 2 * 3^2, the maximum power of 18 that divides 128! is determined by the lowest power of its prime factors, which are 2 and 3. In other words, we need to find the minimum of the exponents of 2 and 3 in the prime factorization of 128!.
Hence, the largest possible value of n is the exponent of 18 in the prime factorization of 128!, which is min(127, 61) = 61. Therefore, the largest possible value of n is 61.
hi divergent
135
18 = 2*3^2
so look at how many times 9 divides 128!