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If a, b and c are integers, is (a·b) a multiple of 18?

Stat 1: 2a = 3b
It means, a = multiple of 3 and b= multiple of 2. Now, (a·b) can be multiple of 6, which means either multiple of 18 or not. Not sufficient

Stat 2: 2b = 3c
It means, b = multiple of 3 and c= multiple of 2. But we don't know anything about a. Not sufficient

Combining both, a = multiple of 3 and b= multiple of 2 & 3. Now, (a·b) can be multiple of 18. Sufficient

So, I think C. :)
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If a, b and c are integers, is (a·b) a multiple of 18?

(1) 2a = 3b
a = 3b/2

Put b = 2, a = 3 and a.b = 6, which is not a multiple of 18

Put b = 6, a = 9 and a.b = 54, multiple of 18
Not sufficient

(2) 2b = 3c
b = 3c/2

Put c = 2, then b = 3 and put a = 1, a.b = 3

Put c = 2, then b = 3 and put a = 6, a.b = 18

Not sufficient

(1) + (2)

2b = 3c
b = 3c/2

2a = 3b
2a = 3(3c/2)
a = 9c/4

Since a, b, and C are integers. Minimum value of c = 4

For c = 4, a = 9, b = 6

a.b = 54

So, For any value of c, a is always a multiple of 9 and b is always a multiple if 6. Thus, a.b is always a multiple of 18

Sufficient

Option C

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