What is the greatest integer k such that 2k is a factor of 67!?
67! = 67*66*65*...*3*2
so (67*66*65*...*3)*(2) = 67!
k = 67*66*65*...*3
To find the greatest integer k such that 2k is a factor of 67!, we need to determine the number of factors of 2 in the prime factorization of 67!.
Step 1: Prime Factorization of 67
To find the prime factorization of 67, we need to check if there are any prime numbers that divide 67. Starting from 2, we find that 67 is not divisible by 2, 3, 5, or 7. Therefore, 67 is a prime number.
Step 2: Counting Factors of 2 in 67!
To determine the number of factors of 2 in 67!, we need to count the powers of 2 that divide 67!. Since factors of 2 can arise from even numbers, we need to calculate how many even numbers there are from 1 to 67.
There is one even number for every multiple of 2, so we have:
2, 4, 6, 8, ..., 66 = 2, 4, 6, 8, ..., 64
To find the total number of even numbers, we divide 67 by 2 and round down to the nearest whole number:
67 / 2 = 33.5
Round down to the nearest whole number: 33
So, there are 33 even numbers from 1 to 67.
To find the number of factors of 2 in 67!, we now need to count the powers of 2 in the prime factorization of each even number up to 67. Let's list the even numbers from 1 to 67 and determine the powers of 2 in their prime factorizations:
2 = 2^1
4 = 2^2
6 = 2^1 * 3^1
8 = 2^3
10 = 2^1 * 5^1
12 = 2^2 * 3^1
...
64 = 2^6
There are a total of 1 + 2 + 1 + 3 + 1 + 2 + ... + 6 = 32 powers of 2 in the prime factorization of the even numbers from 1 to 67.
Step 3: Determining the Greatest k
Since we counted 32 powers of 2 in the prime factorization of 67!, we know that 2^32 is a factor of 67!. Therefore, the greatest integer k such that 2k is a factor of 67! is k = 32.
To find the greatest integer k such that 2k is a factor of 67!, we need to determine the highest power of 2 that divides 67! evenly.
The number of times a prime factor p appears in the factorial of n is given by the formula: [n/p] + [n/p^2] + [n/p^3] + ..., where [x] denotes the greatest integer less than or equal to x.
In this case, we are interested in the power of 2. So we can calculate it using the formula:
[n/2] + [n/2^2] + [n/2^3] + ...
Let's compute this:
First, calculate [67/2] = 33: There are 33 multiples of 2 from 1 to 67.
Next, calculate [67/2^2] = [67/4] = 16: There are 16 multiples of 4 from 1 to 67.
Next, calculate [67/2^3] = [67/8] = 8: There are 8 multiples of 8 from 1 to 67.
Next, calculate [67/2^4] = [67/16] = 4: There are 4 multiples of 16 from 1 to 67.
Next, calculate [67/2^5] = [67/32] = 2: There are 2 multiples of 32 from 1 to 67.
Next, calculate [67/2^6] = [67/64] = 1: There is 1 multiple of 64 from 1 to 67.
Next, calculate [67/2^7] = [67/128] = 0: There are no multiples of 128 from 1 to 67.
So, the sum of the powers of 2 is 33 + 16 + 8 + 4 + 2 + 1 + 0 = 64.
Therefore, the greatest integer k such that 2k is a factor of 67! is 64.