shraddhaFebBorn
GeorgeKo111
Before we solve this problem we should note the following properties:
(Odd Number) * (Odd Number) = (Odd Number)
(Even Number) * (Even Number) = (Even Number)
(Odd Number) * (Even Number) = (Even Number)
(Even Number) / (Even Number) = (Even Number)
as per the given: a is an even integer and b is an odd integer:
a^3= (Even Number) * (Even Number) * (Even Number)= (Even Number)
and b^2= (Odd Number) * (Odd Number) = (Odd Number)
(a^3)(b^2)= (Odd Number) * (Even Number) = (Even Number) and 8 is an even number so (a^3)(b^2)/8= Always even(here the question stem did not specify if integer or not so we can assume it can be both)
the correct answer is A.
Can somebody explain why E is the answer?
If a is an even integer and b is an odd integer, then \(\frac{a^3b^2}{8}\)?A. Always even
B. Always odd
C. Always a fraction
D. Could be a fraction
E. Always an integer
a is an integer, so a = 2k, for some integer k.
\(\frac{a^3b^2}{8}=\frac{(2k)^3b^2}{8}=\frac{8k^3b^2}{8}=k^3b^2=integer\).
Answer: E
A is not always true. Consider a = 2. In this case, \(\frac{a^3b^2}{8}=b^2=odd\)
B is not always true. Consider a = 4. In this case, \(\frac{a^3b^2}{8}=8b^2=even\)