If a, b and c are integers, is ab a multiple of 18?

Question

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

TarunKumar1234 · Answer

If a, b and c are integers, is ab a multiple of 18?

Stat1: 2a = 3b
a = 3, 6, 9..... and b = 2,4,6,..... Now, ab can or can't be divisible by 18. Not sufficient

Stat2: 2b = 3c
b= 3, 6, 9.... and c = 2,4,6,..... But, we don't have any idea of a. Not sufficient

Combining both, a = 3, 6, 9..... and, b= 6, 12, 18....Now, ab will be divisible by 18. Sufficient

So, I think C.

twobagels · Answer

From statement 1
b is mult of 2
a is mult of 3

insufficient: a*b could be 6 or 18 or other numbers

From statement 2
c is mult of 2
b is mult of 3

insufficient: we don't know anything about a

Combined: statements 1 and 2 give us that b is a multiple of both 2 and 3 (ie mult of 6) and a is a multiple of 3: a*b is minimum a multiple of 18

Bunuel · Answer

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

Notice that we are told that all variables are integers.

(1) 2a = 3b --> a/b = 3/2. a is a multiple of 3 and b is a multiple of 2. ab must be a multiple of 6 but may or may not be a multiple of 18 ( for example consider (a,b) = (3,2) and (a,b) = 9,6 ). Not sufficient.

(2) 2b = 3c --> b/c = 3/2. b is a multiple of 3. Not sufficient.

(1)+(2) b is a multiple of both 2 and 3, so it must be a multiple of 6 and since a is a multiple of 3, then ab must be a multiple of 6*3 = 18. Sufficient.

Answer: C.

Hope it's clear.

jonNNNNNy · Answer

what about zero? C because b=0 a=0 hence no we can answer ! answer c

Shashi128 · Answer

Taking first statement 2a=3b , a = 3/2 b so b must be multiple of 2 say 2m .a=3m, ab=6m^2 so it depends on value of m to decide if ab is multiple of 18 or not . A is not the answer . Similarly 2b=3c b=3/2c c must be multiple of 2 say c=2p ab = 6p condition can't be determined . B is not the answer .
Now taking both sentence together a= 3/2b let b=2n ,a will be 3n.second statement gives 4n=3c ,c=4/3n as c is integerer so n must be multiple of 3 , n= 3t .a now becomes 6t and b 9t clearly ab is 54t^2 so it is multiple of 18

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siddharthkamat · Answer

I marked E since a, b and c can be 0 as well. Please can the expert reply on this?

Bunuel · Answer

Check the highlighted part:

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself ( \frac{x}{0} = 0 , so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer ( x*0 = 0 , so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 ( x^0 = 1 )

9.  0^0  case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined. You CANNOT take 0 to the negative power.

12.  0! = 1! = 1 .

Hope it helps.

If a, b and c are integers, is ab a multiple of 18?

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