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? (CAT 2005)
I. 1000≤n≤1200
II. Every digit in n is odd
Then how many elements of S are divisible by 3? (CAT 2005)
Answer: Option A
:
A
Ans: (a)
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 another 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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:
A
Ans: (a)
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 another 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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