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28 Jan 2026, 22:00
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C
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Difficulty:
65%
(hard)
Question Stats:
52% (01:59) correct
48%
(01:50)
wrong
based on 109
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p, m, and n are positive integers greater than 1 such that p = m x n and m and n do not have any common prime factors. If p has 25 factors, m has r factors, and n has s factors, what is the value of (r + s)?
A. 4
B. 8
C. 10
D. 16
E. 25
A. 4
B. 8
C. 10
D. 16
E. 25
28 Jan 2026, 23:46
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ExpertsGlobal5
p = m x n
Since p has 25 factors, p = a^24 (since 24+1 = 25) or a^4 × b^4 (since (4+1)(4+1) = 5×5 = 25)
Since m and n have no common prime factors and both are greater than 1, we cannot have p = a^24
Thus, p = a^4 × b^4 => m = a^4 and n = b^4 => m has 4+1 = 5 factors i.e. r = 5 and n also has 5 factors i.e. s = 5
=> r + s = 10
Answer C
29 Jan 2026, 10:18
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ExpertsGlobal5
p,m,n are positive integers, which means the values are greater than 0.
But here p,m,n are greater than 1.
p = m * n
p has 25 factors.
m has r factors, and n has s factors.
Let’s take an example for easier understanding: 12 = 2^2 * 3 .
Where m =2 has (2+1)=3 factors = r.
n = 3 has (1+1)= 2 factors = s.
12 has 3*2 = 6 factors.
Let’s jump to the problem: 25 = r*s
The possible values of r*s = (25,1) (5,5).
The case (25,1) is not possible. We cannot have a term with 1 factor.
Hence only valid scenario is (5,5).
Thus, r+s = 5+5 = 10.
Option C.
