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Mismatch of answers - Manhanttan Numbers Chapter
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27 Mar 2016, 03:20
27 Mar 2016, 03:20
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Hello,
I have been on the numbers chapter recently ans I was solving the hard problem set, after completion of the test I started verifying my answers.
Following were three questions for which the answer given at the back of the book do not satisfy my understanding of math.
1:If \(x^2\)/4 is an integer greater than 50, then what is the smallest possible value of \(x^2\)?
OA : 256
However what I deduce through my calculation was 204. As 204 is divisible by 4. The answer provided in manhanttan is on the assumption that x is an integer (which is not mentioned anywhere in the question or rules I have referred).
2:x,y,z are consecutive integers, where x<y<z. Which of the following must be divisible by 3?
A: xyz
B: (x+1)yz
C: (x+2)yz
D: (x+3)yz
E: (x+1)(y+1)(z+1)
F: (x+1)(y+2)(z+3)
OA: A,D,E,F
3: b,c and d are consecutive even integers such that 2<b<c<d. What is the largest positive integer that must be a divisor of bcd?
OA: 48
I need help understanding (both 2 and 3), is there a proper way of solving this types of question of it is just completely based on understanding.
I have been on the numbers chapter recently ans I was solving the hard problem set, after completion of the test I started verifying my answers.
Following were three questions for which the answer given at the back of the book do not satisfy my understanding of math.
1:If \(x^2\)/4 is an integer greater than 50, then what is the smallest possible value of \(x^2\)?
OA : 256
However what I deduce through my calculation was 204. As 204 is divisible by 4. The answer provided in manhanttan is on the assumption that x is an integer (which is not mentioned anywhere in the question or rules I have referred).
2:x,y,z are consecutive integers, where x<y<z. Which of the following must be divisible by 3?
A: xyz
B: (x+1)yz
C: (x+2)yz
D: (x+3)yz
E: (x+1)(y+1)(z+1)
F: (x+1)(y+2)(z+3)
OA: A,D,E,F
3: b,c and d are consecutive even integers such that 2<b<c<d. What is the largest positive integer that must be a divisor of bcd?
OA: 48
I need help understanding (both 2 and 3), is there a proper way of solving this types of question of it is just completely based on understanding.
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GRE 1: Q165 V161
Re: Mismatch of answers - Manhanttan Numbers Chapter
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27 Mar 2016, 05:54
27 Mar 2016, 05:54
Expert Reply
Explanation
1. Given that \(x^2/4>50\) is an integer, \(x^2\) can be 51. So you are right in saying since nothing is given about x it is not okay to assume its an integer.
2. Since x, y , z are consecutive integers you can write:
x Now it must be understood that one among x, y and z must be divisible by 3 since they are consecutive three integers. For example, lets take [5,6,7] or [10,11,12] or [51,52,53]
Now use the above expressions in the options and check:
- xyz = One among x,y,z must be divisible by 3. So xyz must be divisible by 3.
- (x+1)yz = \(y^2z\) Its not possible to say with certainty that either y or z is divisible by 3. Not enough information.
- (x+2)yz = \(yz^2\) Its not possible to say with certainty that either y or z is divisible by 3. Not enough information.
- (x+3)yz = Now it must be divisible by 3 since one of x,y,z is a multiple of 3. Similarly one of (x+3),y,z is a multiple of 3.
- (x+1)(y+1)(z+1) = Now it must be divisible by 3 since one of x,y,z is a multiple of 3. Similarly one of (x+1),y+1,z+1 is a multiple of 3.
- (x+1)(y+2)(z+3) = Consider three case. Case 1: x is divisible by 3. y and z is not. Then in the expression, y = x+1 = 3n +1 , then (y+2) = 3n + 3. So the expression becomes divisible by 3. Case 2: y is divisible by 3. Then the term (x+1) becomes divisible by 3 since (x+1) = (y-1+1) = y. Case 3: z is divisible by 3. (z+3) becomes divisible by 3 then.Now it must be divisible by 3
3. b,c and d are consecutive even integers such that 2<b<c<d.
Let's assume b = 2n, then c = 2n+2 , d = 2n+4 and n>1 since 2<b<c<d.
Then the expression bcd = 2n(2n+2)(2n+4)= 8n(n+1)(n+2). Now n(n+1)(n+2) is the product of 3 consecutive integers which is always divisible by 6. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48.
gmatclubot
Re: Mismatch of answers - Manhanttan Numbers Chapter [#permalink]
27 Mar 2016, 05:54
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