lazyashell wrote:
If a, b, c are three consecutive positive even integers, which of the following must be an integer?
Indicate all that's possible.
A. \(\frac{(a+b+c)}{2}\)
B. \(\frac{(a+b+c)}{4}\)
C. \(\frac{(a+b+c)}{6}\)
D. \(\frac{(a+b+c)}{12}\)
E. \(\frac{(a+b+c)}{15}\)
All even integers can be written as \(2k\), where \(k\) is an integer.
So, let \(2k\) = the smallest integer (\(a\))
So, \(2k + 2\) = the next consecutive even integer (\(b\))
And \(2k + 4\) = the last consecutive even integer (\(c\))
So, \(a + b + c = 2k + (2k+2)+ (2k+4) = 6k + 6\)
We get:
A. \(\frac{(a+b+c)}{2}=\frac{6k+6}{2}=3k+3\).
Definitely an integer. B. \(\frac{(a+b+c)}{4}=\frac{6k+6}{4}=1.5k+1.5\).
This need not be an integer.C. \(\frac{(a+b+c)}{6}=\frac{6k+6}{6}=k+1\).
Definitely an integer.D. \(\frac{(a+b+c)}{12}=\frac{6k+6}{12}=0.5k+0.5\).
This need not be an integer.E. \(\frac{(a+b+c)}{15}=\frac{6k+6}{15}=0.4k+0.4\).
This need not be an integer.Answer: A, C