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04 Mar 2012, 23:29
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If a, b, and c are each positive integers greater than 1 is the product abc divisible by 6?
(1) The product ab is even
(2) The product bc is divisible by 3
(1) The product ab is even
(2) The product bc is divisible by 3
05 Mar 2012, 00:45
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If a, b, and c are each positive integers greater than 1 is the product abc divisible by 6?
(1) The product ab is even. No info about c. Not sufficient.
(2) The product bc is divisible by 3. No info about a. Not sufficient.
(1)+(2) abc is divisible by 2 from (1) and by 3 from (2), hence it's divisible by 2*3=6. Sufficient.
Answer: C.
(1) The product ab is even. No info about c. Not sufficient.
(2) The product bc is divisible by 3. No info about a. Not sufficient.
(1)+(2) abc is divisible by 2 from (1) and by 3 from (2), hence it's divisible by 2*3=6. Sufficient.
Answer: C.
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05 Mar 2012, 01:06
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Stmt 1 : The product ab is even, so either a or b must be a multiple of 2. However we can have a , b , c such that none of them is a multiple of 3. For eg 4, 8, 16.
Else we can have one of them a multiple of 3 : 4, 6, 16
The answer can be both Yes & No
Not sufficient.
Stmt 2 : The product bc is divisible by 3, so either c or b must be a multiple of 3. However we can have a , b , c such that none of them is a multiple of 2. For eg 3, 9, 27
Else we can have one of them a multiple of 2 : 3, 6, 18
The answer can be both Yes & No
Not sufficient.
Combining 1 & 2 either a or b must be a multiple of 2 and either b or c is a multiple of 3. So in a , b, c we must have a multiple of 2 and one multiple of 3. Hence sufficient.
Else we can have one of them a multiple of 3 : 4, 6, 16
The answer can be both Yes & No
Not sufficient.
Stmt 2 : The product bc is divisible by 3, so either c or b must be a multiple of 3. However we can have a , b , c such that none of them is a multiple of 2. For eg 3, 9, 27
Else we can have one of them a multiple of 2 : 3, 6, 18
The answer can be both Yes & No
Not sufficient.
Combining 1 & 2 either a or b must be a multiple of 2 and either b or c is a multiple of 3. So in a , b, c we must have a multiple of 2 and one multiple of 3. Hence sufficient.
