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If x and y are positive integers, is y divisible by 3?

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If x and y are positive integers, is y divisible by 3? [#permalink]

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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.
[Reveal] Spoiler: OA

Originally posted by DeeptiM on 15 Aug 2011, 06:43.
Last edited by Bunuel on 05 Apr 2015, 04:56, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 15 Aug 2011, 07:06
DeeptiM wrote:
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.


I think (1) is sufficient

If x = 1, y is not divisible by 3, if x is >1 then y is a fraction and not divisible by 3 obviously.

Would be more interesting is x could be a negative integer.
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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 15 Aug 2011, 09:16
is it A?
A-factorise it=>x(2x-1)(x+5)..insert any value for X,the equation is divisible by 3
B alone is not sufficient as no info about X is available
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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 15 Aug 2011, 20:23
I got to know my mistake..thnx guys..
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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 07 Mar 2015, 11:32
DeeptiM wrote:
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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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 07 Mar 2015, 13:30
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:
[Reveal] Spoiler:
A


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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 04 Apr 2015, 12:45
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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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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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 wrote:
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?

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If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 05 Apr 2015, 10:12
Dear Rich,

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

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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 15 Mar 2016, 18:35
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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If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 11 Jan 2017, 00:03
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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Re: If x and y are positive integers, is y divisible by 3? [#permalink]

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New post 31 Mar 2018, 13:01
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
Re: If x and y are positive integers, is y divisible by 3?   [#permalink] 31 Mar 2018, 13:01
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