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03 Nov 2010, 09:52
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Hi everybody!
I am stuck with a solution method for two similar questions from the area of number properties.
Here is the first:
The sum of three consecutive even integers is always divisible by which of the following?
I. 2
II. 3
III. 4
IV. 8
A) I and II only
B) I and IV only
C) II and IV only
D) I, II and III only
E) I, II and IV only
OA: D
The solution method is the following:
Sum of three integers = n+n+2+n+4= 3n+6
3n+6 is divisible by 2, 3, 4, but not by 8 for all n=(2,4,6,8...)
(the Source: Winners' Guide to GMAT Math - Part II)
Here is the second question:
Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is
a) 12
b) 6
c) 8
d) 10
e) 14
OA: e
The solution method is the following:
Let the three even numbers be (x-2), x, (x+2)
Then 3(x-2)-2(x+2)=2
3x-6-2x-4=2, x=2
The third number= 12+2=14 Hence: e
(the Source: Math Question Bank for GMAT Winners)
But there is there is one point I cannot understand:
The starting point of both questions is the same because they both deal with three consecutive even integers. This is why both solution methods can presumably be applied alternatively to each other.
In other words, both questions must be soluable regardless whether we take three consecitive even integers as n, n+2, n+3 or as (x-2), x, (x+2).
However, it does not work if we take (x-2), x, (x+2) as a starting point for the first question or if we take n, n+2, n+3 as a starting point for the second question.
Pls., can someone disprove my remark and give better explanation?
I am stuck with a solution method for two similar questions from the area of number properties.
Here is the first:
The sum of three consecutive even integers is always divisible by which of the following?
I. 2
II. 3
III. 4
IV. 8
A) I and II only
B) I and IV only
C) II and IV only
D) I, II and III only
E) I, II and IV only
OA: D
The solution method is the following:
Sum of three integers = n+n+2+n+4= 3n+6
3n+6 is divisible by 2, 3, 4, but not by 8 for all n=(2,4,6,8...)
(the Source: Winners' Guide to GMAT Math - Part II)
Here is the second question:
Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is
a) 12
b) 6
c) 8
d) 10
e) 14
OA: e
The solution method is the following:
Let the three even numbers be (x-2), x, (x+2)
Then 3(x-2)-2(x+2)=2
3x-6-2x-4=2, x=2
The third number= 12+2=14 Hence: e
(the Source: Math Question Bank for GMAT Winners)
But there is there is one point I cannot understand:
The starting point of both questions is the same because they both deal with three consecutive even integers. This is why both solution methods can presumably be applied alternatively to each other.
In other words, both questions must be soluable regardless whether we take three consecitive even integers as n, n+2, n+3 or as (x-2), x, (x+2).
However, it does not work if we take (x-2), x, (x+2) as a starting point for the first question or if we take n, n+2, n+3 as a starting point for the second question.
Pls., can someone disprove my remark and give better explanation?
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03 Nov 2010, 10:12
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feruz77
You should get the same answer whether you choose \(x\) or \(x-2\) for the first even number.
1. The sum of three consecutive even integers is always divisible by which of the following?
I. 2
II. 3
III. 4
IV. 8
A. I and II only
B. I and IV only
C. II and IV only
D. I, II and III only
E. I, II and IV only
Let the first even # be \(x\). Note that as \(x\) is even then \(x=2k\), for some integer \(k\):
\(x+(x+2)+(x+4)=3x+6=3*2k+6=6(k+1)\) --> so the sum of 3 consecutive even integers is divisible by 2, 3 and 6.
Now, if take the first integer to be \(x-2\) then:
\((x-2)+x+(x+2)=3x=3*2k=6k\) --> so the sum of 3 consecutive even integers is divisible by 2, 3 and 6.
Answer: A.
2. Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is
A. 12
B. 6
C. 8
D. 10
E. 14
Let the first even # be \(x\) (so the 3 consecutive integers are \(x\), \(x+2\), and \(x+4\)), then: \(3x=2(x+4)+2\) --> \(x=10\) --> third # is \(x+4=10+4=14\).
Now, if we take the first integer to be \(x-2\) (so the 3 consecutive integers are \(x-2\), \(x\), and \(x+2\)) then: \(3(x-2)=2(x+2)+2\) --> \(x=12\) --> third # is \(x+2=12+2=14\)
Answer: E.
Hope it's clear.
03 Nov 2010, 10:42
Kudos
Bookmarks
feruz77
I'm not sure if there's a typo in the question above, or an error in the original source, but the sum of three consecutive even integers is certainly not always divisible by 4. Take 0+2+4, or 4+6+8 for example. Bunuel's solution above is perfect except that he must have thought they were asking about divisibility by 6, not by 4; as the question is written, the answer is A, not D. [edit - I guess Bunuel fixed that now]
The second question you've asked is not worded in a way at all similar to what you'll see on the test (you won't see the word 'thrice' on the GMAT, and the question contains a comma splice).
