SATHEE: Permutations And Combinations Question 202

Permutations And Combinations Question 202

Question: The sum of all positive divisors of 960 is [Karnataka CET 2000]

Options:

A) 3048

B) 3087

C) 3047

D) 2180

Show Answer

Answer:

Correct Answer: A

Solution:

Given number is 960, we know that $ 960=2^{6}\times 3^{1}\times 5^{1}. $ Therefore bases are $ p_1=2,p_2=3 $ and $ p_3=5. $ and powers $ a_1=6,a_2=1 $ and $ a_3=1. $ Thus sum of all the positive divisors of 960 $ =( \frac{p_1^{a_1+1}-1}{p_1-1} )( \frac{p_2^{a_2+1}-1}{p_2-1} )( \frac{p_3^{a_3+1}-1}{p_3-1} ) $ $ =( \frac{{2^{6+1}}-1}{2-1} )( \frac{{3^{1+1}}-1}{3-1} )( \frac{{5^{1+1}}-1}{5-1} )=(127)(4)(6)=3048 $ .

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Permutations And Combinations Question 202

Question: The sum of all positive divisors of 960 is [Karnataka CET 2000]

Options:

Answer:

Solution:

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