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19 Dec 2018, 01:45
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D
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FRESH GMAT CLUB CHALLENGE PROBLEM
If x is a prime number, what is the value of x + c?
(1) \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\)
(2) \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\)
M35-67
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09 Nov 2021, 04:57
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Bunuel
Official Solution:
If \(x\) is a prime number, what is the value of \(x + c\)?
(1) \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\)
Re-write as: \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} + |c|=0\). We have that the sum of three non-negative values (the absolute value of some number, the square root of some number and the absolute value of some number) is 0. This to be true, each of the three must be 0.
So:
(i) \(a + b + c - 2x=0\), which gives \(a + b + c=2x\);
(ii) \(x^2 - a - b - c=0\), which gives \(a + b + c=x^2\);
(iii) \(|c|=0\), which gives \(c=0\).
Equate (i) and (ii): \(2x=x^2\). This gives that \(x=0\) or \(x=2\). It's given that \(x\) is a prime number, so \(x=2\)
\(x + c=2+0=2\). Sufficient.
(2) \( \sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\)
Here we also have that the sum of three non-negative values is 0. This to be true, each of the three must be 0.
So:
(i) \(a + b + c - 1 + x=0\), which gives \(a + b + c = 1 - x\)
(ii) \(\frac{1}{1-x}- a - b - c=0\), which gives \(a + b + c =\frac{1}{1-x}\)
(iii) \(c^2=0\), which gives \(c=0\).
Equate (i) and (ii): \(1 - x=\frac{1}{1-x}\). This gives that \((1-x)^2=1\).
Takes the square root: \(|1-x|=1\);
\(x=0\) or \(x=2\). It's given that \(x\) is a prime number, so \(x=2\)
\(x + c=2+0=2\). Sufficient.
Answer: D
General Discussion
19 Dec 2018, 09:27
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If x is a prime number, what is the value of x + c?
(1) \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\)
The RHS - |c| can only be possible as 0, because the LHS has SUM of a modulus and a square root, both of which are positive. so it will be positive and -|c| cannot be positive
Now, the LHS, \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} =0\) , but since both terms are positive, they should be 0 themselves too..
so a + b + c - 2x=0......a+b+c=2x...(i)
Also \(\sqrt{(x^2 - a - b - c)^2} =0......=>x^2 - a - b - c=0.........x^2=a+b+c\)...(ii)
From (i) and (ii), we get \(a+b+c=2x=x^2........x^2-x=0....x(x-2)=0\), so x is 0 or 2
But x is a prime number, so x+c=2+0=2
sufficient
(2) \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{x-1}- a - b - c| + \sqrt{c^2}=0\)
All terms on LHS are positive, so their SUM can only be 0 when all terms are 0..
\(\sqrt{c^2}=0, so \ \ c=0.\)
\(|\frac{1}{x-1}- a - b - c|=0..............\frac{1}{x-1}= a + b\)...(i)
\(\sqrt{(a + b + c - x+1)^2}=0.........a + b + c - x+1 =0..........a+b=x-1.\)...(ii)
from (i) and (ii).. \(\frac{1}{x-1}= a + b=x-1........x-1=\frac{1}{x-1}......(x-1)^2=1\), so x can be 0 or 2
But x is prime number so 2
hence x+c=2+0=2
suff
D
Note:- Bunuel, there seems to be a typo in statement II.
Either it is 1/(1-x) and not 1/(x-1) or it is a+b+c-x+1 and not a+b+c-1+x...
Pl relook
(1) \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\)
The RHS - |c| can only be possible as 0, because the LHS has SUM of a modulus and a square root, both of which are positive. so it will be positive and -|c| cannot be positive
Now, the LHS, \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} =0\) , but since both terms are positive, they should be 0 themselves too..
so a + b + c - 2x=0......a+b+c=2x...(i)
Also \(\sqrt{(x^2 - a - b - c)^2} =0......=>x^2 - a - b - c=0.........x^2=a+b+c\)...(ii)
From (i) and (ii), we get \(a+b+c=2x=x^2........x^2-x=0....x(x-2)=0\), so x is 0 or 2
But x is a prime number, so x+c=2+0=2
sufficient
(2) \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{x-1}- a - b - c| + \sqrt{c^2}=0\)
All terms on LHS are positive, so their SUM can only be 0 when all terms are 0..
\(\sqrt{c^2}=0, so \ \ c=0.\)
\(|\frac{1}{x-1}- a - b - c|=0..............\frac{1}{x-1}= a + b\)...(i)
\(\sqrt{(a + b + c - x+1)^2}=0.........a + b + c - x+1 =0..........a+b=x-1.\)...(ii)
from (i) and (ii).. \(\frac{1}{x-1}= a + b=x-1........x-1=\frac{1}{x-1}......(x-1)^2=1\), so x can be 0 or 2
But x is prime number so 2
hence x+c=2+0=2
suff
D
Note:- Bunuel, there seems to be a typo in statement II.
Either it is 1/(1-x) and not 1/(x-1) or it is a+b+c-x+1 and not a+b+c-1+x...
Pl relook
