Question

How many 8's are present in the following sequence of numbers which
are exactly divisible by both its preceding and following numbers? 38 68876834826 6 2 8 2 4 8 6 3 7 4 8 4 ​

Answer 1

Answer:

8

Step-by-step explanation:

Answer 2

Answer:

Three 8's are present.

Step-by-step explanation:

Consider the sequence of numbers as follows:

$3\ 8\ 6\ 8\ 8\ 7\ 6\ 8\ 3\ 4\ 8\ 2\ 6\ 6\ 2\ 8\ 2\ 4 \ 8\ 6\ 3\ 7\ 4\ 8\ 4$

Write all the set of those $3$ numbers such that $8$ is present in between them.

(a) 3  8  6

Preceding number = 3

Following number = 6

Clearly, the number $8$ is not divisible by both $3$ and 6.

(b) 6  8  8

Preceding number = 6

Following number = 8

Thus, the number $8$ is not divisible by 6 but is divisible by 8.

(c) 8  8  7

Preceding number = 8

Following number = 7

Thus, the number $8$ is divisible by 8 but not divisible by 7.

(d) 6  8  3

Preceding number = 6

Following number = 3

Thus, the number $8$ is not divisible by both 6 and 3.

(e) 4  8  2

Preceding number = 4

Following number = 2

Thus, the number $8$ is divisible by both $4$ and 2.

(f) 2  8  2

Preceding number = 2

Following number = 2

Thus, the number $8$ is divisible by both $2$ and 2.

(g) 4  8  6

Preceding number = 4

Following number = 6

Thus, the number $8$ is divisible by 4 but not divisible by 6.

(h) 4  8  4

Preceding number = 4

Following number = 4

Thus, the number $8$ is divisible by both $4$ and 4.

Therefore, three 8's are present.

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