Question
A is set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?
(CAT 1998)
(CAT 1998)
Answer: Option B
:
B
Ans: (b)
This can be solved by Chinese remainder theorem, but as the common difference is constant, it is a special case of Chinese remainder concept.
Required number of the set is calculated by the LCM of (2, 3, 4, 5, 6) - (common difference)
In this case, common difference = (2-1) = (3-2) = (4-3) = (5- 4) = (6 - 5) = 1.
All integers of the set will be given by (60n - 1) ; If n = 1, (60 - 1) = 59; If n = 2,((60*2) - 1) = 119; Since range of the set A is between 0 and 100, hence there will exist only one number i.e., 59.
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:
B
Ans: (b)
This can be solved by Chinese remainder theorem, but as the common difference is constant, it is a special case of Chinese remainder concept.
Required number of the set is calculated by the LCM of (2, 3, 4, 5, 6) - (common difference)
In this case, common difference = (2-1) = (3-2) = (4-3) = (5- 4) = (6 - 5) = 1.
All integers of the set will be given by (60n - 1) ; If n = 1, (60 - 1) = 59; If n = 2,((60*2) - 1) = 119; Since range of the set A is between 0 and 100, hence there will exist only one number i.e., 59.
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