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chetan2u
Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c

Hi chetan2u

I think for clarity, the question must mention that a, b, c are integers and/or different, otherwise it becomes very easy to eliminate both the options.

Statement 1: if we assume a=b=c=2, then a+b+c=6, divisible by 3 but a*b*c=8 not divisible by 3, but if a=b=c=3, then abc will be divisible by 3

again if a=1.1 b=2 and c=2.9, then a+b+c=6, divisible by 3 but a*b*c clearly not divisible by 3

and if a=2, b=5 & c=8, then a+b+c=15, divisible by 3 but a*b*c=80, not divisible by 3

Hence Insufficient

statement 2:implies that a, b & c is an AP series, so if a=2, b=5 & c=8, then abc=80 not divisible by 3

but if a=2, b=4 & c=6, then abc=48, divisible by 3. Insufficient

Combining 1 & 2, again if we have a=b=c=3 or a=2, b=4 & c=6, then abc is divisible by 3

but if a=2, b=5 & c=8, then abc is not divisible by 3.

Insufficient

Option E
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Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c

Hi chetan2u

I think for clarity, the question must mention that a, b, c are integers and/or different

Statement 1: if we assume a=b=c=2, then a+b+c=6, divisible by 3 but a*b*c=8 not divisible by 3, but if a=b=c=3, then abc will be divisible by 3

again if a=1.1 b=2 and c=2.9, then a+b+c=6, divisible by 3 but a*b*c clearly not divisible by 3

and if a=2, b=5 & c=8, then a+b+c=15, divisible by 3 but a*b*c=80, not divisible by 3

Hence Insufficient

statement 2:implies that a, b & c is an AP series, so if a=2, b=5 & c=8, then abc=80 not divisible by 3

but if a=2, b=4 & c=6, then abc=48, divisible by 3. Insufficient

Combining 1 & 2, again if we have a=b=c=3 or a=2, b=4 & c=6, then abc is divisible by 3

but if a=2, b=5 & c=8, then abc is not divisible by 3.

Insufficient

Option E

I agree with niks18. 2) a+c=2b is valid for three consecutive numbers 1,2,3.....3,4,5 as well as APs such as 4,7,10... 5,8,11 etc.
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I feel it's either one. Option 2 is AP. eg: 5,6,7. Or 11,13,15. If you take any 3 numbers in AP one of them should be a multiple of 3.

Option 1: divisible by 3. Take any set of examples. Even this satisfies

Hence each statement alone



Please correct me if wrong

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Can't a,b,c be same ?
In that case, 1+1+1 is divisible by 3 but not its product and 3+3+3 is divisible by 3 and its product too.

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As some others have mentioned here, I guess the question also has a point that a,b and c are distinct numbers

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chetan2u
Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c

Hi chetan2u

I think for clarity, the question must mention that a, b, c are integers and/or different, otherwise it becomes very easy to eliminate both the options.

Statement 1: if we assume a=b=c=2, then a+b+c=6, divisible by 3 but a*b*c=8 not divisible by 3, but if a=b=c=3, then abc will be divisible by 3

again if a=1.1 b=2 and c=2.9, then a+b+c=6, divisible by 3 but a*b*c clearly not divisible by 3

and if a=2, b=5 & c=8, then a+b+c=15, divisible by 3 but a*b*c=80, not divisible by 3

Hence Insufficient

statement 2:implies that a, b & c is an AP series, so if a=2, b=5 & c=8, then abc=80 not divisible by 3

but if a=2, b=4 & c=6, then abc=48, divisible by 3. Insufficient

Combining 1 & 2, again if we have a=b=c=3 or a=2, b=4 & c=6, then abc is divisible by 3

but if a=2, b=5 & c=8, then abc is not divisible by 3.

Insufficient

Option E

Hi..
The Question is 600-700 level, so it doesn't have too deep a secret involved in it.
Very many will go wrong by not differentiating between integers and non integers and few answers on top do point towards that and ofcourse the OA is accordingly written.
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I am considering a , b and c as integers .
is product of abc divisible by 3 ..for this to happen among a,b and c at least one should be 3 or should have a factor 3 .

Now check

statement 1 : a+b+c is divisible by 3 ... a=1 b=1 c=1 a+b+c =3 is divisible by 3 but abc=1 not divisible by 3

or a=4 b =16 c=10 abc not divisible by 3

so statement 1 is insufficient .

Statement 2 : 2b =a+c
i.e terms are in AP ..lets consider terms are a-d , a and a+d ...Sum is 3a which is always divisible by 3 i.e statement 1 always hold correct for sum of 3 terms which are in AP.
now check if a(a-d)(a+d) are divisible by 3 or not
i.e a=10 d=3 abc=10* 7*13 ...means insufficient ..

we already proved statement 2 auto validates statement 1 so both statements together also are not sufficient ..


Hence , answer is E .

Please respond if you feel otherwise
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chetan2u
Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c
Hi @Chetan2u,

I think both the options will be insufficient when a,b,c is a non-integer.

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chetan2u
Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c

I took it this way:

ST1: a + b + c is a multiple of 3, this has many options a=b=c=2, then a+b+c is 6, which is div by 3 but a*b*c is 8 not div by 3. If a=b=c=3, a+b+c is 9 and a*b*c is div by 3.

INSUFFICIENT

ST2: It's kind of giving us the same info that the first statement.

a - b = b - c
a + c = 2b
(We sum b to each side)
a + b + c = 3b (this is telling us already what ST1 said, that the sum of a,b,c is a multiple of 3)

INSUFFICIENT

And since ST1 and ST2 give us the same info, taking them together won't add new info.

HENCE E.
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chetan2u
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Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c

Hi chetan2u

I think for clarity, the question must mention that a, b, c are integers and/or different, otherwise it becomes very easy to eliminate both the options.

Statement 1: if we assume a=b=c=2, then a+b+c=6, divisible by 3 but a*b*c=8 not divisible by 3, but if a=b=c=3, then abc will be divisible by 3

again if a=1.1 b=2 and c=2.9, then a+b+c=6, divisible by 3 but a*b*c clearly not divisible by 3

and if a=2, b=5 & c=8, then a+b+c=15, divisible by 3 but a*b*c=80, not divisible by 3

Hence Insufficient

statement 2:implies that a, b & c is an AP series, so if a=2, b=5 & c=8, then abc=80 not divisible by 3

but if a=2, b=4 & c=6, then abc=48, divisible by 3. Insufficient

Combining 1 & 2, again if we have a=b=c=3 or a=2, b=4 & c=6, then abc is divisible by 3

but if a=2, b=5 & c=8, then abc is not divisible by 3.

Insufficient

Option E

Hi..
The Question is 600-700 level, so it doesn't have too deep a secret involved in it.
Very many will go wrong by not differentiating between integers and non integers and few answers on top do point towards that and ofcourse the OA is accordingly written.
What is correct answer..?

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Hi..
The Question is 600-700 level, so it doesn't have too deep a secret involved in it.
Very many will go wrong by not differentiating between integers and non integers and few answers on top do point towards that and ofcourse the OA is accordingly written.[/quote]
What is correct answer..?

Sent from my Lenovo A6000 using GMAT Club Forum mobile app[/quote]

Hi

Correct answer is E only, as mentioned. Since its not specified that a, b, c are integers - they can be non-integers also, in which case both statements together become insufficient to find the answer.
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chetan2u
Is the product of a, b and c divisible by 3?

(1) a + b + c is divisible by 3.
(2) a - b = b - c


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
\(a - b = b - c\)
\(⇔ a + c = 2b\)
\(a + b + c = ( a + c ) + b = 2b + b = 3b\)

Case 1: \(a = b = c = 1\)
\(abc = 1\) is not divisible by 3.

Case 2: \(a = b = c = 3\)
\(abc = 27\) is divisible by 3.

Both conditions together are not sufficient.

Therefore, the answer is E.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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