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# Prove that the numbers of divisors of 8400 is 58 only (excluding the number itself and one).Also find the sum of divisors.

Last updated date: 24th Mar 2023
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Hint- Write the factors of 8400.

We have to prove that the number of divisors of 8400 is 58 only and we have to exclude the number itself including 1and also the sum of the divisors.
Now the factors of $8400 = 3 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7$…………………….. (1)
So we can also write this as
$8400 = {3^a}{2^b}{5^c}{7^d} = {3^1}{2^4}{5^2}{7^1}$
Using equation 1
Now it is clear that a varies from 0 to 1, similarly b varies from 0 to 4 , c varies from 0 to 2 and d varies from 0 to 1.
So the total number of divisors is $2 \times 5 \times 3 \times 2 = 60$ using the concept that numbers between 0 and 1 are 2 in total and extending it to all b, c, and d as well.
Now it’s being told that we have to exclude the divisor 1 and the divisor 8400.
Hence the total number of divisors are $60 - 2 = 58$
Now we have to find the sum of divisors
Now for any divisor of the form ${3^a}{2^b}{5^c}{7^d}$ the sum is given as
$Sum = \sum\nolimits_0^1 {{3^a}} \sum\nolimits_0^4 {{2^b}} \sum\nolimits_0^2 {{5^c}} \sum\nolimits_0^1 {{7^d}}$
This gives us
$\therefore (1 + 3)(1 + 2 + {2^2} + {2^3} + {2^4})(1 + 5 + {5^2})(1 + 7)$
On solving this we get
$4 \times 31 \times 31 \times 8 = 30752$

Note- Solving such problems requires the basics of finding the factors of the given number in the problem statement. Then we can simply express them in forms of $factor{s^{repetition}}$to obtain the final answer.