When a, b and c are consecutive positive even integers such that abc

Question

When  a, b  and  c  are consecutive positive even integers such that  a>b>c , which of the following must be an odd integer?

A. \frac{(a-c)}{2}  
 B. \frac{(c-a)}{2}  
 C. \frac{(a+c)}{2}  
 D. \frac{(a+c)}{4}  
 E. \frac{(a-c)}{4}

chetan2u · Answer

Two operations..
1) c-a..
Since they are consecutive positive even integers, and a>b>c, c-a or a-c will be |4|
So it will always leave an EVEN integer when divided by 2..
So eliminate A and B
But what about div by 4, (a-c)/4=4/4=1, so ODD..C is the answer

But let's see the addition
2) a+c
It is nothing but 2b..
2b will be even when div by 2 or 4 when B is multiple of 4, so eliminate C and D
Example.. a>b>c => 10>8>6, so 2*8/4=4, even number

Ans C

GMATGuruNY · Answer

Test a=6, b=4 and c=2 in the five answer choices.
Eliminate any answer choice that does not yield an odd integer.

A:  \frac{a-c}{2} = \frac{6-2}{2} = 2  --> Eliminate A.
B:  \frac{a-c}{2} = \frac{2-6}{2} =- 2  --> Eliminate B.
C:  \frac{a+c}{2} = \frac{6+2}{2} = 4  --> Eliminate C.
D:  \frac{a+c}{4} = \frac{6+2}{4} = 2  --> Eliminate D.

E

BrentGMATPrepNow · Answer

If a, b and c are consecutive EVEN integers, and a>b>c, then we we know that b is 2 greater than c , and a is 2 greater than b So, we can write: b = c + 2 a = c + 4 Now let's check the answer choices from E to A -----ASIDE---------------- This is one of those questions that require us to check/test each answer choice. In these situations, always check the answer choices from E to A , because the correct answer is typically closer to the bottom than to the top. For more on this strategy, see my article: https://www.gmatprepnow.com/articles/han ... -questions ------------------------------------------------ E) (a - c)/4 = c + 4 ) - (c)]/4 = 4/4 = 1 (ODD!) Answer: E Cheers, Brent

EgmatQuantExpert · Answer

Solution

Given:  
• a,b and c are consecutive positive even integers and a>b>c.

To find:  
• Among the given options which one is an odd integer.

Approach and Working:   
• Since, a,b and c are consecutive positive integer and a>b>c, a = c+4, b=c+2.

Now, let us check every option.

A.  \frac{(a−c)}{2}= \frac{(c+4 - c)}{2}= 2 
B.  \frac{(c−a)}{2}= \frac{-(a-c)}{2}= -2 
C.  \frac{(a+c)}{2}= \frac{2c+4}{2}= c+2= b 
D.  \frac{(a+c)}{4}= \frac{2c+4}{4}= \frac{c}{2+1}

• If c= 2 then  \frac{c}{2}+1= 2  and if c= any other number than 2 then  \frac{c}{2}+1 = odd
• Hence, we can say if  \frac{c}{2}+1  is odd or not.

E.  \frac{(a-c)}{4}= \frac{4}{4}= 1 
Hence, the correct answer is option E.

Answer: E

ScottTargetTestPrep · Answer

We can let c = x, so b = x + 2 and a = x + 4.

We see that (a - c)/4 =  /4 = 4/4 = 1, which is an odd integer.

Answer: E

MathRevolution · Answer

=>

Write  a = 2n + 2, b = 2n  and  c = 2n – 2 . We check each of the alternatives.

A. \frac{( a – c )}{2} = \frac{( 2n + 2 – ( 2n – 2 ) )}{2} = \frac{4}{2} = 2. 
 B. \frac{( c – a )}{2} = \frac{( 2n – 2 – ( 2n + 2 ) )}{2} = \frac{-4}{2} = -2. 
 C. \frac{( a + c )}{2} = \frac{( 2n + 2 + 2n – 2 )}{2} = \frac{4n}{2} = 2n. 
 D. \frac{( a + c )}{4} = \frac{( 2n + 2 + 2n – 2 )}{4} = \frac{4n}{4} = n. 
 E. \frac{( a – c )}{4} = \frac{( 2n + 2 – ( 2n – 2 ) )}{4}= \frac{4}{4} = 1

Only option E is guaranteed to be odd. Therefore, the answer is E.
Answer : E

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When a, b and c are consecutive positive even integers such that a>b>c

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