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.
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Check if 496 is even: \[ 496 \div 2 = 248 \]
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Check if 248 is even: \[ 248 \div 2 = 124 \]
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Check if 124 is even: \[ 124 \div 2 = 62 \]
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Check if 62 is even: \[ 62 \div 2 = 31 \]
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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:
- \(2^0 \times 31^0 = 1\)
- \(2^1 \times 31^0 = 2\)
- \(2^2 \times 31^0 = 4\)
- \(2^3 \times 31^0 = 8\)
- \(2^4 \times 31^0 = 16\)
- \(2^0 \times 31^1 = 31\)
- \(2^1 \times 31^1 = 62\)
- \(2^2 \times 31^1 = 124\)
- \(2^3 \times 31^1 = 248\)
- \(2^4 \times 31^1 = 496\) (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: \[ \boxed{496} \]