A sequence, a1=64, a2=66, a3=67, an=8+an-3, which of the following is in the sequence?
105
786
966
1025
Given:
\(a_1 = 64\)
\(a_2 = 66\)
\(a_3 = 67\)
...
\(a_n = 8 + a_{n - 3}\)
So \(a_4 = 8 + a_1\) = 8 + 64
\(a_5 = 8 + a_2\) = 8 + 66
\(a_6 = 8 + a_3\) = 8 + 67
\(a_7 = 8 + a_4\) = 8 + 8 + 64
\(a_8 = 8 + a_5\) = 8 + 8 + 66
and so on...
So any number that belongs to this sequence will be sum of one of 64/66/67 and some number of 8s.
105 - 64 = 41 which is not a multiple of 8. 41 is 1 more than a multiple of 8 so when you subtract 2/3 out of 41 (in effect subtracting 66/67 out of 105), you will still not get a multiple of 8. Hence 105 is not in this sequence.
786 - 64 = 720 which is divisible by 8 hence it will be in the sequence. This is your answer and ideally you should stop here. But if you want to check the remaining two options:
966 - 64 = 902 which is not divisible by 8. Neither are 900 and 899. Or say that 902 is 6 more than a multiple of 8 so when you subtract 2/3 out of it, you will still not get a multiple of 8.
1025 - 64 = 961 which is not divisible by 8 and is 1 more than a multiple of 8 so when you subtract 2/3 out of it, it will still not give a multiple of 8.