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Updated on: 06 Nov 2022, 06:46
Originally posted by gmatophobia on 06 Nov 2022, 06:36.
Last edited by gmatophobia on 06 Nov 2022, 06:46, edited 2 times in total.
Last edited by gmatophobia on 06 Nov 2022, 06:46, edited 2 times in total.
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gmatophobia
DS Question for the day
The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?
(1) N has exactly four factors, and D is a factor of N
(2) D = (3^P) + 2
Source: Magoosh | Difficulty: Hard
The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?
(1) N has exactly four factors, and D is a factor of N
(2) D = (3^P) + 2
Source: Magoosh | Difficulty: Hard
C
Braintree
Good question! Answer: C. From statement (1), D can be a prime or it can also be 1. From statement (2) - Prime numbers don’t follow a definite algorithm. Combining the statements, we can eliminate the possibility of 1, and thus the answer is a definite yes.
Very well explained 
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Here’s a Warm-up question for the quiz - Problem Solving
If |x^2-x+1|=2x-1, then what is the sum of all possible values of x ?
A. 0
B. 1
C. 2
D. 3
E. 4
Source: GMATClub Tests | Difficulty : Hard
Updated on: 06 Nov 2022, 07:09
Originally posted by gmatophobia on 06 Nov 2022, 07:03.
Last edited by gmatophobia on 06 Nov 2022, 07:09, edited 1 time in total.
Last edited by gmatophobia on 06 Nov 2022, 07:09, edited 1 time in total.
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kapil1995
c.2
The trick in this question is 2x - 1 >= 0
as the LHS is modulus
Another one - Problem Solving
If |x | + |y | = -x – y and xy does not equal 0, which of the following must be true?
A. x + y > 0
B. x + y 0
D. x – y 0
Source: Manhattan GMAT | Difficulty : Medium
06 Nov 2022, 07:17
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If |x^2-x+1|=2x-1
Case 1:
=> x^2 - x + 1 = 2x - 1
x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x = 1/2
Case 2:
=> x^2 - x + 1 = -2x + 1
x^2 + x = 0
x(x+1) = 0
x = 0/-1
But for both these values RHS is < 0 but LHS is modulus which cannot be negative so we reject these 2 values
Sum = 1+2 = 3
Case 1:
=> x^2 - x + 1 = 2x - 1
x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x = 1/2
Case 2:
=> x^2 - x + 1 = -2x + 1
x^2 + x = 0
x(x+1) = 0
x = 0/-1
But for both these values RHS is < 0 but LHS is modulus which cannot be negative so we reject these 2 values
Sum = 1+2 = 3
)
