The sum of the divisors of 9000 is . . . . . | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) First,find the prime factorization of $9000$: $9000 = 9 	imes 1000 = 3^2 	imes 10^3 = 3^2 	imes (2 	imes 5)^3 = 2^3 	imes 3^2 	imes 5^3$. Th

Vedclass · Accepted Answer

$16 	imes 13 	imes 156$

Vedclass · Answer

(A) First,find the prime factorization of $9000$: $9000 = 9 	imes 1000 = 3^2 	imes 10^3 = 3^2 	imes (2 	imes 5)^3 = 2^3 	imes 3^2 	imes 5^3$. The sum of divisors of a number $N = p_1^{a} 	imes p_2^{b} 	imes p_3^{c}$ is given by the formula: $S = \frac{p_1^{a+1}-1}{p_1-1} 	imes \frac{p_2^{b+1}-1}{p_2-1} 	imes \frac{p_3^{c+1}-1}{p_3-1}$. For $9000 = 2^3 	imes 3^2 	imes 5^3$: Sum $= (2^0 + 2^1 + 2^2 + 2^3) 	imes (3^0 + 3^1 + 3^2) 	imes (5^0 + 5^1 + 5^2 + 5^3)$. Sum $= (1 + 2 + 4 + 8) 	imes (1 + 3 + 9) 	imes (1 + 5 + 25 + 125)$. Sum $= (15) 	imes (13) 	imes (156)$. Note: The provided option $A$ is $16 	imes 13 	imes 156$. Re-evaluating the sum of powers of $2$: $(2^0 + 2^1 + 2^2 + 2^3) = 1+2+4+8 = 15$. Given the options,there is a slight discrepancy in the first term ($16$ vs $15$). However,$15 	imes 13 	imes 156 = 30420$. The sum of divisors is $30420$.

The sum of the divisors of $9000$ is . . . . . . .

Similar Questions

If $\alpha$ and $\beta$ are the greatest common divisors of $n(n^2-1)$ and $2n(n^2+2)$ respectively for all $n \in N$, then $\alpha \beta=$

What is the total number of divisors of $9600$,including $1$ and $9600$?

$A$ natural number $n$ such that $n!$ ends in exactly $1000$ zeros is

If the constant term in the expansion of $\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}$ is $2^{k} \cdot l$,where $l$ is an odd integer,then the value of $k$ is equal to

If the integer represented by $100!$ has $K$ consecutive zeroes at the end,then $K=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module