The question asks for the sum of all the divisors of the number 256.
Divisors Calculation for 256
First, find the prime factorization of 256.
256 is a power of 2:
\(256 = 2^8\)
Sum of Divisors Formula
For a number \(N\) expressed as a product of prime factors \(N = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}\), the sum of its divisors, denoted by \(\sigma(N)\), is calculated using the formula:
\(\sigma(N) = \left(\frac{p_1^{a_1+1}-1}{p_1-1}\right) \times \left(\frac{p_2^{a_2+1}-1}{p_2-1}\right) \times ... \times \left(\frac{p_k^{a_k+1}-1}{p_k-1}\right)\)
In the case of 256, we have only one prime factor, \(p_1 = 2\), and its exponent is \(a_1 = 8\). So, \(N = 2^8\).
Applying the Formula
Substitute the values into the formula:
\(\sigma(256) = \sigma(2^8) = \frac{2^{8+1}-1}{2-1}\)
Simplify the expression:
\(\sigma(2^8) = \frac{2^9-1}{1}\)
\(\sigma(2^8) = 2^9 - 1\)
Final Calculation
Calculate the value of \(2^9\):
\(2^9 = 512\)
Now, subtract 1:
\(\sigma(256) = 512 - 1 = 511\)
The sum of all the divisors of 256 is 511.
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