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

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:

\[
\boxed{496}
\]