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In the following finite sequence of integers, how many terms are divisible by their immediately preceding terms? 8, 3, 4, 9, 3, 5, 9, 5, 9, 9, 4,
Question

In the following finite sequence of integers, how many terms are divisible by their immediately preceding terms?
8, 3, 4, 9, 3, 5, 9, 5, 9, 9, 4, 5, 6, 3, 3, 5, 7, 2, 3, 9, 9

A.

3

B.

4

C.

5

D.

6

Correct option is C


Concept:
let a,b be two integers and we said a is divisible by b iff a=bc where c is also an integer.
Solution:
The sequence is : 
8, 3, 4, 9, 3, 5, 9, 5, 9, 9, 9, 4, 5, 6, 3, 3, 5, 7, 2, 3, 9, 9 
Here, immediately preceding means immediately before as in the given sequence 8 is immediately preceding term of 3
now, we have to check  how many terms are divisible by their immediately preceding terms.
Now, 3 is not divisible by 8 , 4 is not divisible by 3 continue like that 
we get there are only 5 terms which are  divisible by their immediately preceding terms.
i.e. 9 is divisible by 9
9 is divisible by 9
3 is divisible by 3
9 is divisible by 3
9 is divisible by 9
Hence, Option (c) is true

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