Let S be a set of positive integers such that every element n of S satisfies the conditions

(a)1000
≤n<=1200

(b) every digit in n is odd

Then how many elements of S are divisible by 3?

a. 9

b. 10

c. 11

d. 12The correct option is A 9

The 100th and 1000th position value will be only 1.

Divisibility rule for 3 is sum of digits.

We already have a sum of 2, we need to make it to 6, 9 or any other multiple of 3.

Units digit = a

Tens digit = b

Possible combinations of a+b=4,7,10,13,16; where both "a and b" are odd.

Thus, the possibility of unit and tens digits are (1,3),(1,9),(3,1),(3,7),(5,5),(7,3),(7,9),(9,1),(9,7).

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