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:

\[ \boxed{496} \]