What is the sum of all the numbers less than 496 that divides 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.

We start by factoring 496.

1. Check if 496 is even:
\[
496 \div 2 = 248
\]

2. Check if 248 is even:
\[
248 \div 2 = 124
\]

3. Check if 124 is even:
\[
124 \div 2 = 62
\]

4. Check if 62 is even:
\[
62 \div 2 = 31
\]

5. Now, since 31 is a prime number, we can stop. Therefore, the prime factorization of 496 is:
\[
496 = 2^4 \times 31^1
\]

Next, we will find all the divisors of 496 using this factorization. The number of divisors can be calculated using the formula:

\[
(e_1 + 1)(e_2 + 1) = (4 + 1)(1 + 1) = 5 \times 2 = 10
\]

Thus, 496 has 10 divisors. We can use the prime factorization to find all divisors:

The exponents for the factors are:
- \(2\) can take values from \(0\) to \(4\) (i.e., \(0, 1, 2, 3, 4\))
- \(31\) can take values from \(0\) to \(1\) (i.e., \(0, 1\))

We can list the divisors as follows:

1. \(2^0 \times 31^0 = 1\)
2. \(2^1 \times 31^0 = 2\)
3. \(2^2 \times 31^0 = 4\)
4. \(2^3 \times 31^0 = 8\)
5. \(2^4 \times 31^0 = 16\)
6. \(2^0 \times 31^1 = 31\)
7. \(2^1 \times 31^1 = 62\)
8. \(2^2 \times 31^1 = 124\)
9. \(2^3 \times 31^1 = 248\)
10. \(2^4 \times 31^1 = 496\) (but we will not include this as we're finding numbers less than 496.)

Thus, the divisors of 496 that are less than 496 are:
\[
1, 2, 4, 8, 16, 31, 62, 124, 248
\]

Now, we sum these divisors:

\[
1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248
\]

Calculating this step by step:
- \(1 + 2 = 3\)
- \(3 + 4 = 7\)
- \(7 + 8 = 15\)
- \(15 + 16 = 31\)
- \(31 + 31 = 62\)
- \(62 + 62 = 124\)
- \(124 + 124 = 248\)
- \(248 + 248 = 496\)

Therefore, the total sum is:

\[
1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496
\]

Thus, the sum of all the numbers less than 496 that divide evenly into 496 is:
\[
\boxed{496}
\]