Stiv wrote:
If x is even integer, which of the following must be an odd integer?
A. \(\frac{3x}{2}\)
B. \(\frac{3x}{2} + 1\)
C. \(3x^2\)
D. \(\frac{3x^2}{2}\)
E. \(\frac{3x^2}{2} + 1\)
One can spot right away that if \(x\) is any even number then \(x^2\) is a multiple of 4, which makes \(\frac{x^2}{2}\) an even number and therefore \(\frac{3x^2}{2}+1=3*even+1=even+1=odd\).
Answer: E.
If you don't notice this, then one also do in another way. Let \(x=2k\), for some integer k, then:
A. \(\frac{3x}{2}=\frac{3*2k}{2}=3k\) --> if \(k=odd\) then \(3k=odd\) but if \(k=even\) then \(3k=even\). Discard;
B. \(\frac{3x}{2}+1=\frac{3*2k}{2}+1=3k+1\) --> if \(k=odd\) then \(3k+1=odd+1=even\) but if \(k=even\) then \(3k+1=even+1=odd\). Discard;
C. \(3x^2\) --> easiest one as \(x=even\) then \(3x^2=even\), so this option is never odd. Discard;
D. \(\frac{3x^2}{2}=\frac{3*4k^2}{2}=6k^2=even\), so this option is never odd. Discard;
E. \(\frac{3x^2}{2}+1=\frac{3*4k^2}{2}=6k^2+1=even+1=odd\), thus this option is always odd.
Answer: E.
Similar question to practice:
if-a-and-b-are-positive-integers-such-that-a-b-and-a-b-are-88108.htmlHope it helps.