Question

How many no(n) r there b/w 1 & 200 such that n/2,n/3 & 2n +1/5 r all composite natural no. ?
a]0
b]1
c]2
d]3

Answer 1

The number of numbers between 1 and 200 such that n/2, n/3 and (2n +1)/5 are all composite natural numbers are 7.

• It is to be noted that only whole numbers are divisile by 2 and 3.
• For a whole number, denoted by n, to be divisible by 2 and 3 it must be a multiple of 6.
• The multiples of 6 between 1 to 10 that leave a reminder 0, when multiplied with 2, added 1 and the sum is divided by 5 are 12 and 42, that is, the second and seventh multiples.
• The pattern is followed in the twelth, seventeenth, twenty-second and twenty-seventh and so on.
• Since we are to limit our answer only the natural numbers between 1 and 200, we get the suitable numbers as 12, 42, 72, 102, 132, 162, 192.

Answer 2

Answer:

Option C -> 2

Step-by-step explanation:

Here n should be divisbile by 2 & 3, hence it is a multiple of 6.

Let n = 6 k

As (2n+1) should be divisble by 5, k = 2, 7, 12, 17, 22, 27 and 32.

Additionally it is required by the question that  n/2,n/3 & (2n +1)/5 should be all composite natural numbers. This is valid only when k = 27 & 32.

Hence, n = 162 & 192. So there are only 2 numbers that satisfy this condition.