Given:
x, y and z are integers.
To prove:
If 
is even, then at least one of x, y or z is even.
Solution:
We know that,
Product of two odd integers is always odd. ...(i)
Difference of two odd integers is always even. ...(ii)
Sum of an even integer and an odd integer is odd. ...(iii)
Let as assume x, y and z all are odd, then is even.
is always odd. [Using (i)]
is always odd. [Using (i)]
is always even. [Using (ii)]
is always odd. [Using (iii)]
is always odd.
So, out assumption is incorrect.
Thus, at least one of x, y or z is even.
Hence proved.
Answer: the last pos term is 3
Details:
213
207
201
195
189
183
177
171
165
159
153
147
141
135
129
123
117
111
105
99
93
87
81
75
69
63
57
51
45
39
33
27
21
15
9
3
-3
Answer:
Greater
Step-by-step explanation:
just because
Answer:
-68
Step-by-step explanation:
-28 + - 40 =
2 + 4 = 6
8 + 0 = 8 making
-68
hope this helped
