The sum of all the factors of a positive number 'k' greater than 1 is 'p'. Which of these could be the ratio of ‘p’ to ‘k’?
Answers
Given :- The sum of all the factors of a positive number 'k' greater than 1 is 'p'. Which of these could be the ratio of ‘p’ to ‘k’ ?
A) 2 : 5
B) 6 : 5
C) 3 : 7
D) 1 : 1
Answer :-
we know that, if N is a positive integer , such that,
- N = p^a * q^b * r^c where p , q and r are prime factors of N . then,
- sum of factors of N = [(p⁰ + p¹ + p² + ____ p^a) * (q⁰ + q¹ + q² + ____ q^b) * (r⁰ + r¹ + r² + ____ r^c)] / [(p^a - 1)(q^b - 1)(r^c - 1)] .
- Also, if N > 1 , sum of factors of N > N or we can say that, (sum of factors of N / N ) > 1 .
we have, given that, k is a positive integer and ,
- k > 1 .
- sum of factors of k = p .
then,
→ (sum of factors / given integer) must be > 1 .
→ (p / k) > 1 .
checking options now, we get,
A) 2 : 5
→ p/k > 1
but
→ 2/5 < 1 .
therefore, option (A) is not possible.
B) 6 : 5
→ p/k > 1
and
→ 6/5 > 1 .
therefore, option (B) is possible.
C) 3 : 7
→ p/k > 1
but
→ 3/7 < 1 .
therefore, option (C) is not possible.
D) 1 : 1
→ p/k > 1
but
→ 1/1 < 1 .
therefore, option (D) is not possible.
Hence , we can conclude that, the ratio of ‘p’ to ‘k’ can be 6 : 5 (B) .
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Given : The sum of all the factors of a positive number 'k' greater than 1 is 'p'.
To Find : Which of these could be the ratio of ‘p’ to ‘k’ ?
A) 2 : 5
B) 6 : 5
C) 3 : 7
D) 1 : 1
Solution :
k > 1
Hence k has at least 2 factors 1 and k
Sum of all factors ≥ k + 1
The sum of all the factors = p
=> p ≥ k + 1
=> p/k ≥ 1 + 1/k
2 : 5 => 2/5 = 1 - 3/5 does not satisfy 1 + 1/k
6 : 5 => 6/5 = 1 + 1/5 satisfy 1 + 1/k
3 : 7 => 3/7 = 1 - 4/7 does not satisfy 1 + 1/k
1 : 1 => 1 /1 = 1 does not satisfy 1 + 1/k
Hence ratio of ‘p’ to ‘k’ can be 6 : 5 in the given options
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