Question

What is the sum of all the numbers less than 496 that divide evenly into 496

Answer 1

To find the sum of all the numbers less than 496 that divide evenly into 496, we first need to determine the divisors of 496.

1. **Find the divisors of 496**:
- Start by finding the prime factorization of 496.
- 496 is an even number, so we begin dividing by 2:
\[
496 \div 2 = 248
\]
\[
248 \div 2 = 124
\]
\[
124 \div 2 = 62
\]
\[
62 \div 2 = 31
\]
- 31 is a prime number, so the complete prime factorization of 496 is:
\[
496 = 2^4 \times 31^1
\]

2. **List the divisors**:
- The divisors can be found by taking all combinations of the prime factors:
- The exponents range from 0 to 4 for \(2\) and from 0 to 1 for \(31\):
- \(2^0, 2^1, 2^2, 2^3, 2^4\) gives us the values \(1, 2, 4, 8, 16\).
- \(31^0, 31^1\) gives us the values \(1, 31\).
- We combine these:
- For \(31^0\):
- \(1, 2, 4, 8, 16\)
- For \(31^1\):
- \(31, 62, 124, 248, 496\) (multiplying each of the previous divisors by 31).
- Therefore, the complete list of divisors is:
- \(1, 2, 4, 8, 16, 31, 62, 124, 248, 496\)

3. **Sum the divisors less than 496**:
- Now we sum all the divisors except for 496. The divisors less than 496 are:
\[
1, 2, 4, 8, 16, 31, 62, 124, 248
\]
- We compute the sum:
\[
1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248
\]

Adding these in pairs for easier calculations:
\[
(1 + 248) + (2 + 124) + (4 + 62) + (8 + 31) + 16
\]
Which equals:
\[
249 + 126 + 66 + 39 + 16 = 496
\]

Thus, the sum of all the numbers less than 496 that divide evenly into 496 is:

\[
\boxed{496}
\]