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I got to know my mistake..thnx guys..
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DeeptiM
If x and y are positive integers, is y divisible by 3?
(1) Y=2x^3+9x^2-5x.
(2) x is an odd number.


Bunuel could you help to solve this question.

Regards,
Ammu
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Hi All,

GMAT questions are often based on patterns (in math, in logic, in grammar, etc.), so if you don't immediately see a pattern, there might still be one there...you just might have to do a little bit of work to find it.

This DS question is perfect for TESTing VALUES. While there is one "big" calculation to consider (in Fact 1), the individual math "steps" involved are NOT difficult. You just have to be ready to do a little work.

We're told that X and Y are POSITIVE INTEGERS. We're asked if Y is divisible by 3. This is a YES/NO question.

Fact 1: Y=2(X^3)+9(X^2)-5X

IF....
X = 1
Y = 2(1) + 9(1) - 5(1) = 6 and the answer to the question is YES.

IF....
X = 2
Y = 2(8) + 9(4) - 5(2) = 42 and the answer to the question is YES.

IF...
X = 3
Y = 2(27) + 9(9) - 5(3) = 120 and the answer to the question is YES.

It certainly looks like there's a pattern here: Y will ALWAYS be divisible by 3.
Fact 1 is SUFFICIENT

Fact 2: X is an odd number

This tells us nothing about Y.
Fact 2 is INSUFFICIENT

Final Answer:
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If x and y are positive integers, is y divisible by 3?
(1) Y=2x^3+9x^2-5x.
(2) x is an odd number.

Statement (1) is insufficient because if X=1 than Y is divisible by 3 i.e. Y=2X^3+9X^2-5X => Y=2+9-5=6 which is divisible by Y;
if X=2 then Y is not divisible by Y i.e. Y=2X^3+9X^2-5X => Y= 2*8+9*4-5*2=16+32-10=38 which is not divisible by 3.
Statement (2) is clearly insufficient.

(1)+(2) sufficient, because if we put odd numbers as indicated in statement (2) into Y=2x^3+9x^2-5x we will have a number divisible by 3 e.g.:

X=1 has been shown above let's try X=3, X=5 and etc:

X=3
Y=2*3^3+9*3^2-5*3=2*27+9*9-15=3(18+27+5) which is divisible by 3;

X=5
Y=2*5^3+9*5^2-5*5=2*125+9*25-25=250-25+225=225+225=450 which is divisible by 3;

Let's try X=17 then:

Y=2*17^3+9*17^2-5*17=17(2*289+9*17-5)=17(578+153-5)=17*726 and 726 is divisible by 3.

Hence, the answer is C.
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Hi ziyavutdinov,

Your approach to TEST VALUES is a good one, but you made a mistake in one of your calculations. As a result, you've chosen the wrong answer.

Take a look at the calculation when X = 2.....

ziyavutdinov
If x and y are positive integers, is y divisible by 3?
(1) Y=2x^3+9x^2-5x.
(2) x is an odd number.

Statement (1) is insufficient because if X=1 than Y is divisible by 3 i.e. Y=2X^3+9X^2-5X => Y=2+9-5=6 which is divisible by Y;
if X=2 then Y is not divisible by Y i.e. Y=2X^3+9X^2-5X => Y= 2*8+9*4-5*2=16+32-10=38 which is not divisible by 3.

Hence, the answer is C.

Once you've dealt with that issue, what would you do next? And what answer would you choose?

GMAT assassins aren't born, they're made,
Rich
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Dear Rich,

I acknowledge I've made a calculation mistake. Thanks for correcting me. My answer is A.

ziyavutdinov
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Answer is A and here is a quick proof by contradiction.

If 2x^3 + 9x^2 - 5x is not divisible by 3, then 2x^3 - 2x + (9x^2 - 3x) is not divisible by 3.

Since there always exists a k1 where 3k1 = (9x^2 - 3x), then there must not exist a k2 where 3k2 = 2x^3 -2x (otherwise 3(k1 + k2) = y, y divisible by 3)

Thus, 2x^3 -2x is not divisible by 3

However, 2x^3 -2x = 2 * x *(x-1)*(x+1) is clearly divisible by 3

Therefore, the contradiction is false and the statement is true.
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This is such an amazing question.
It is all about patterns.
Here is what i did in this Question=>

We need to see if y/3 is an integer or not.
We are given that a and y are positive integers.
Statement 1=>
\(y = 2x^3 + 9x^2 - 5x\)
Taking out x as a factor we get -> \(x[2x^2+9x-5]=> x(2x-1)(x+5)\)
Now putting in the values of x=> y is always a multiple of 3.
Hence sufficient.

Statement 2=>
No clue of y=> Not sufficient.

Hence A.
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Statement 1
\(y=2x^3+9x^2-5x = x(2x^2+9x-5)=x(2x-1)(x+5)\)

If one of {\(x, 2x-1,x+5\)} is divisible by 3, then y is divisible by 3.

Let's say we have 3 consecutive integers {\(x,x+1,x+2\)}. One of these will be divisible by 3. Let's check each one.

Case 1 (\(x\) is div by 3)
Clearly, y is div by 3 since x is factor of y.

Case 2 (\(x+1\) is div by 3)
\(x+1=3a\) for some integer \(a\)
\(x=3a-1\)
Substituting into \(2x-1\) we get \(2(3a-1)-1=6a-3=3(2a-1)\), which is div by 3.

Case 3 (\(x+2\) is div by 3)
If x+2 is div by 3, then so is x+5, which is a factor of y.

Thus y is div by 3.
Sufficient

Statement 2
Not sufficient

Answer: A
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