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Question
Let S be a set of positive integers such that every element n of S satisfies the conditions:
I. 1000≤n≤1200
II. Every digit in n is odd
Then how many elements of S are divisible by 3?
Let S be a set of positive integers such that every element n of S satisfies the conditions:
I. 1000≤n≤1200
II. Every digit in n is odd
Then how many elements of S are divisible by 3?
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Solution
The 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).
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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