Question

The sum of the natural numbers which are divisors of 100 is,

Answer 1

Answer:

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Answer 2

Answer:

The sum of the natural numbers which are the divisors of 100 is 217.

Step-by-step explanation:

If a natural number N can be written in terms of the product of its prime factors,

$N=p^{x} q^{y}$

• then the sum of the divisors of that number is given by

       $S=\big(p^{0}+p^{1}+...+p^{x}\big) \times \big(q^{0}+q^{1}+...+q^{y}\big)$

• and the total number of divisors is given by

        $n=\big(x+1\big) \times \big(y+1\big)$

Given the number 100.

By prime factorization, 100 can be written as

$100=2\times2\times5\times5=2^{2} \times 5^{2}$

Using the formula for the number of divisors, we get

$n=\big(2+1\big) \times \big(2+1\big)=3 \times 3=9$

Therefore, the number of divisors of 100 is 9.

And they are 1, 2, 4, 5, 10, 20, 25, 50, 100.

Using the formula for sum of divisors, we get

$S=\big(2^{0}+2^{1}+2^{2}\big) \times \big(5^{0}+5^{1}+5^{2}\big)$

  $=\big(1+2+4\big) \times \big(1+5+25\big)$

  $=7 \times 31=217$

Therefore, the sum of the natural numbers which are the divisors of 100 is 217.

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