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28 Nov 2023, 06:50
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
18% (02:37) correct
82%
(02:16)
wrong
based on 28
sessions
History
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Time
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Not Attempted Yet
If the number of factors of a positive integer, P, that are in the form 5k is 30, and P=56^a* P1^b*P2^c, where 5, P1, P2 are distinct prime factors and a,b & c are positive integers, then what is the value of a+b+c?
1) The total number of factors of P is even
2) O is divisible by 125, but not divisible by 625
a) Statement 1) ALONE is sufficient but statement
2) ALONE isn't sufficient b) Statement 2) ALONE is sufficient but statement 1) ALONE isn't sufficient
c) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE isn't sufficient
d) EACH statement ALONE is sufficient
e) Statement 1) and 2) TOGETHER aren't sufficient
1) The total number of factors of P is even
2) O is divisible by 125, but not divisible by 625
a) Statement 1) ALONE is sufficient but statement
2) ALONE isn't sufficient b) Statement 2) ALONE is sufficient but statement 1) ALONE isn't sufficient
c) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE isn't sufficient
d) EACH statement ALONE is sufficient
e) Statement 1) and 2) TOGETHER aren't sufficient
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
04 Dec 2023, 02:01
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can someone elloborate the solution?
Spectre26
Joined: 20 May 2023
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30 Jul 2024, 17:19
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I think there are typos in the question.
Typo 1. P=56^a* P1^b*P2^c, where 5, P1, P2 are distinct prime factor
Since the underlined portion is given in the question,
P is I guess, 5^a*P1^b*P2^c (assuming this for solving the question)
From this total no of factors of P in the form 5k is given,
a cant be 0
So, (a)*(b+1)*(c+1) = 30
Statement (1), No of factors of P is even,
It can be understood, amonghts a, b or c, one or all are odd numbers.
Not sufficient to get a+b+c
Statement (2)
Typo 2. O is divisible by 125. From where O is coming, shouldn't it be P?
If O is P and it is divisible by 125 but not 625, then, P must have 5^3
i.e. a is 3
if a is 3,
(b+1)*(c+1) = 10
b+1 and c+1 can be 5 or 2
Note: It cant be 10 or 1 as it is given a,b,c are positive integers
and b+1 or c+1 = 1 will give either b or c as 0
Hence a+b+c = 5+2+3 = 10
Correct Option: B
Typo 1. P=56^a* P1^b*P2^c, where 5, P1, P2 are distinct prime factor
Since the underlined portion is given in the question,
P is I guess, 5^a*P1^b*P2^c (assuming this for solving the question)
From this total no of factors of P in the form 5k is given,
a cant be 0
So, (a)*(b+1)*(c+1) = 30
Statement (1), No of factors of P is even,
It can be understood, amonghts a, b or c, one or all are odd numbers.
Not sufficient to get a+b+c
Statement (2)
Typo 2. O is divisible by 125. From where O is coming, shouldn't it be P?
If O is P and it is divisible by 125 but not 625, then, P must have 5^3
i.e. a is 3
if a is 3,
(b+1)*(c+1) = 10
b+1 and c+1 can be 5 or 2
Note: It cant be 10 or 1 as it is given a,b,c are positive integers
and b+1 or c+1 = 1 will give either b or c as 0
Hence a+b+c = 5+2+3 = 10
Correct Option: B
