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07 Feb 2024, 02:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
35% (01:40) correct
65%
(01:15)
wrong
based on 83
sessions
History
Date
Time
Result
Not Attempted Yet
If \(p\) is a prime number and \(n\) is a positive integer, how many positive factors does \(3^n*p^2\) have?
(1) \(n = 3\)
(2) \(p\) is an odd number.
(1) \(n = 3\)
(2) \(p\) is an odd number.
Aryaa03
Joined: 12 Jun 2023
Last visit: 19 Feb 2026
Posts: 210
Given Kudos: 159
Location: India
Schools: ISB '27 (A) IIM Calcutta '26 (A) IIMA '26 (D)
GMAT Focus 1: 645 Q86 V81 DI78 
Bunuel
Sol: IMO- Option-E
Given : p is a prime number and n is an integer greater than zero
Asked: Total no. of positive factors of \(3^n*p^2\)
St-1:
\(n = 3\)
For different values of p, prime factorization of \(3^n*p^2\) will vary and accordingly, the number of total postive factors will also vary.
e.g.
for p = 2 , \(3^3*2^2\) will have (3+1)*(2+1) = 12 factors
for p = 3 , \(3^3*3^2\) will have (5+1) = 6 factors
for p = 5 , \(3^3*5^2\) will have (3+1)*(2+1) = 12 factors
Statement is Insufficient
St-2 :
\(p\) is an odd number.
No information about n and multiple cases for p such as 5 and 3 will give different result.
Statement is Insufficient
(1) and (2) together
Same example as in (1) for p= 3 and 5 shows different results. Insufficient
07 Feb 2024, 14:17
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I think E should be the option.
we need to find out the number of positive factors of \(3^{n}*p^{2}\)
Any expression which can be prime factorised in the following way:
\(a^{x}*b^{y}*c^{z}\) has (x+1)*(y+1)*(z+1) number of positive factors
Sattement 1:
n=3
the expression becomes \(3^{3}*p^{2}\)
We shall analyse few cases and see if we get conclusive result.
p is a prime number so let's try for p=2 and p=3
for p=2 it is \(3^{3}*2^{2}\) which has 4*3 = 12 factors
for p=3 it is \(3^{5}\) which has 6 factors
Insufficient.
Sattement 2:
p is an odd number => p is an odd prime
Again if p is not 3 , we have number of factors as (n+1)*3
else if p=3, we have \(3^{n+2}\) so n+3 factors
Not conclusive. Insufficient
Statement 1 and 2 together make our case like this:
\(3^{3}*p^{2}\) which in turn will give us same result as Statement 1 as done.
Neither of the statements is sufficient.
Hence, Option E should be correct.
we need to find out the number of positive factors of \(3^{n}*p^{2}\)
Any expression which can be prime factorised in the following way:
\(a^{x}*b^{y}*c^{z}\) has (x+1)*(y+1)*(z+1) number of positive factors
Sattement 1:
n=3
the expression becomes \(3^{3}*p^{2}\)
We shall analyse few cases and see if we get conclusive result.
p is a prime number so let's try for p=2 and p=3
for p=2 it is \(3^{3}*2^{2}\) which has 4*3 = 12 factors
for p=3 it is \(3^{5}\) which has 6 factors
Insufficient.
Sattement 2:
p is an odd number => p is an odd prime
Again if p is not 3 , we have number of factors as (n+1)*3
else if p=3, we have \(3^{n+2}\) so n+3 factors
Not conclusive. Insufficient
Statement 1 and 2 together make our case like this:
\(3^{3}*p^{2}\) which in turn will give us same result as Statement 1 as done.
Neither of the statements is sufficient.
Hence, Option E should be correct.
