Question

Find the total number of proper factors of the number 35700. Also find the sum of all these factors.

Answer 1

Hint: We first take out 100 as a factor and then find all the prime factors of 100. Then we have to find the prime factors of 375. And taking all that numbers into consideration, we have found the prime factors of 35700.

Complete step-by-step answer:
Let’s start solving this question.
The given number 35700 can be written as $357\times 100$
First we will find the prime factors of 100.
The prime factors of 100 are: $2\times 2\times 5\times 5$
As we have found the value of 100 in terms of prime factors, now we will find it for 357.
The prime factors of 357 are: $3\times 7\times 17$
Now the prime factors of 35700 will be: $2\times 2\times 5\times 5\times 3\times 7\times 17$
Now all the possible factors are: 1, 2, 5, 3, 7, 17, 4, 6, 10, 12, 14, 15, 20, 21, 25, 28, 30, 34, 35, 40, 42, 50, 51, 60, 68, 70, 75, 84, 85, 100, 102, 105, 119, 140, 150, 170, 175, 204, 210, 238, 255, 300, 340, 350, 357, 420, 425, 476, 510, 525, 595, 700, 714, 850, 1020, 1050, 1190, 1275, 1428, 1700, 1785, 2100, 2380, 2550, 2975, 3570, 5100, 5950, 7140, 8925, 11900, 17850, 35700.
Now the total number of factors is 72.
After adding them we get,
The sum of all these factors is 125032.

Note: You can also find out the number of factors of any number N by expressing the N as ${{p}^{a}}{{q}^{b}}{{r}^{c}}$, where p,q,r are nothing but the prime factors of N and a,b,c are the non-negative powers and calculating (a+1)(b+)(c+1). Similarly for the sum of these factors of N we have the direct formula $\dfrac{\left( {{p}^{0}}+{{p}^{1}}+...+{{p}^{a}} \right)\left( {{q}^{0}}+....+{{q}^{b}} \right)\left( {{r}^{0}}+....+{{r}^{c}} \right)}{\left( {{p}^{a}}-1 \right)\left( {{q}^{b}}-1 \right)\left( {{r}^{c}}-1 \right)}$, where p,q,r are prime factors and a,b,c are their respective exponents.