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26 Nov 2020, 07:10
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If a, b, c are three consecutive positive even integers, which of the following must be an integer?
I. \(\frac{(a+b+c)}{2}\)
II. \(\frac{(a+b+c)}{4}\)
III. \(\frac{(a+b+c)}{6}\)
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III
I. \(\frac{(a+b+c)}{2}\)
II. \(\frac{(a+b+c)}{4}\)
III. \(\frac{(a+b+c)}{6}\)
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III
26 Nov 2020, 07:30
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CONCEPT: Numbers/Integers/Even integers
-All even integers can be written as, where k is an integer.
SOLUTION:
Let 2k = The smallest integer; (a)
Let 2k+2 = The next consecutive even integer; (b)
Also 2k+4 = The last consecutive even integer;(c)
Thus a+ b+ c=2k+(2k+2) +(2k+4) =6k+6
Use the options to check:
A. (a+ b+ c)/2=6k+6/2=3k+3(An integer).
B. (a+ b+ c)/4=6k+6/4=1.5k+1.5(Not an integer)
C. (a+ b+ c)/6=6k+6/6=k+1. (An integer).Thus (D)
Hope this helps.
keep studying 
Devmitra Sen
-All even integers can be written as, where k is an integer.
SOLUTION:
Let 2k = The smallest integer; (a)
Let 2k+2 = The next consecutive even integer; (b)
Also 2k+4 = The last consecutive even integer;(c)
Thus a+ b+ c=2k+(2k+2) +(2k+4) =6k+6
Use the options to check:
A. (a+ b+ c)/2=6k+6/2=3k+3(An integer).
B. (a+ b+ c)/4=6k+6/4=1.5k+1.5(Not an integer)
C. (a+ b+ c)/6=6k+6/6=k+1. (An integer).Thus (D)
Hope this helps.
keep studying Devmitra Sen
26 Nov 2020, 07:33
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Bunuel
If a, b, c are three consecutive positive even integers, which of the following must be an integer?
I. \(\frac{(a+b+c)}{2}\)
II. \(\frac{(a+b+c)}{4}\)
III. \(\frac{(a+b+c)}{6}\)
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III
I. \(\frac{(a+b+c)}{2}\)
II. \(\frac{(a+b+c)}{4}\)
III. \(\frac{(a+b+c)}{6}\)
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III
Given: a, b, c are three consecutive positive even integers
Key concept: All EVEN integers can be rewritten as 2k (where k is some integer)
Aside: This also means that all ODD integers can be rewritten as 2k + 1 (where k is some integer)
So......
let \(a = 2k\) (k is an integer)
So, \(b = 2k + 2\) (since a, b, c are consecutive even integers)
And \(c = 2k + 4\)
So, \(a+b+c=2k+(2k+2)+(2k+4) = 6k+6\)
So for each statement we'll replace \(a+b+c\) with \(6k+6\) ....
I. \(\frac{(a+b+c)}{2}=\frac{6k+6}{2}=\frac{2(3k+3)}{2}=3k+3\)
Since k is an integer, we know that 3k+3 must also be an integer.
Statement I is TRUE
II. \(\frac{(a+b+c)}{4}=\frac{6k+6}{4}=\frac{2(3k+3)}{4}=\frac{3k+3}{2}\)
If \(k = 2\), then \(\frac{3k+3}{2}=\frac{3(2)+3}{2} = \frac{9}{2}\), which is NOT an integer.
Statement II is NOT true
III. \(\frac{(a+b+c)}{6}=\frac{6k+6}{6}=\frac{6(k+1)}{6}=k+1\)
Since k is an integer, we know that k+1 must also be an integer.
Statement III is TRUE
Answer: D
