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

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(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

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$16 	imes 13 	imes 156$

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 . . . . . . .

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