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Divisibility test

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Divisibility test

by krishna239455 » Tue Mar 27, 2012 7:37 am
Problem is stated below:
x is a positive number
(9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is Y divisible by 5?
A)5 is a factor of x
B)x is a integer

I have doubt whether this can be a data suffeciency problem because,
I can simplify the above expression as: 9^x(1+9^1+9^2+9^3+9^4+9^5)=y
and i know that even power of 9 gives digit with 1 in units place and odd poer gives digit with 9 in units place. With this logic if i add all the terms in the bracket i should get digit with 0 in units place. This digit will always be divisible by 5. No need of statements given below. Kindly correct me if i am wrong!!!!
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by Mike@Magoosh » Tue Mar 27, 2012 2:23 pm
Hi, there. I'm happy to give my two cents with this. :)

Prompt:
x is a positive number
(9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is y divisible by 5?

I totally agree with your approach. . . .

y = (9^x)*(1+9^1+9^2+9^3+9^4+9^5)

The parenthesis indeed is a multiple of 10, so must be divisible by 5. If x = 0 or a positive integer, than y will definitely be divisible by 5.

IF x is a positive rational number, y will not be a whole number. If x is a positive irrational number, for example a logarithm, then y could be either a non-integer or even an integer not divisible by 5 --- (for example, if x = log(17/5)/log(9) --- but that's miles and miles beyond what you need to know for the GMAT.)

I am deeply suspicious of this question, because the "catch" in it seems to depend on an understanding of logarithms, which again is way way beyond the realm of GMAT math. I think someone was trying too hard to write a challenging DS question, and lost all track of what math is actually tested on the GMAT.

So, in short, I agree with you that this is a flawed DS question.

Here's a much more GMAT-like DS question involving laws of exponents.
https://gmat.magoosh.com/questions/1003
When you submit your answer to that question, the following page will have a complete video explanation. At Magoosh, each of our over 800+ questions has its own video explanation. If you are looking for high quality GMAT prep material at a surprisingly affordable price, now is particular good time to check out Magoosh, because we are having a sale that ends later this week.

Does everything I've said make sense? Let me know if you have any further questions.

Mike :)
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by krishna239455 » Tue Mar 27, 2012 9:44 pm
Thanks Mike
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by rahulvsd » Thu Mar 29, 2012 7:44 am
Hi Mike,

Could you please explain how Statement 2 in the Magoosh example (from the link you have provided) is insufficient ?
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by Mike@Magoosh » Thu Mar 29, 2012 10:45 am
rahulvsd wrote:Hi Mike,

Could you please explain how Statement 2 in the Magoosh example (from the link you have provided) is insufficient ?
Prompt: If x and y are integers and xy ≠ 0, what is the value of [x^(-2y)]/[y^(2x)]?

Well, that expression simplifies a little to

[x^(-2y)]/[y^(2x)] = 1/[x^(2y)*y^(2x)] = 1/[x^y*y^x]^2

Statement #2: xy = y/x

Well, with this expression, first multiply by x to clear the fraction, and we get:

(x^2)*y = y

Since we know y does not equal zero, we can divide by y, which results in:
x^2 = 1
x = +/-1
(Always remember you need to include the +/- when you take the squareroot of a variable.)

Thus, we know x = +/-1 and we have absolutely no idea what y is --- y could be absolutely an integer other than zero. That's why statement #2 is insufficient.

Does this make sense? Let me know if you have any further questions.

Mike :)
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by [email protected] » Fri Mar 30, 2012 12:47 am
Well there are 2 cases in here...

Case (i): when x is a positive integer divisible by 5, in that case the value of x can only be 0, or multiples of 5.

In that case see the equation and overall there are 6 terms i.e even number of terms.

we know that number 9 has a cyclicity of 9 and 1. only 2 numbers

if x = 0, then the terms last digit would be - 1,9,1,9,1,9. the total comes out to be 10 in short the last digit has a 0 which is divisible by 5.

same for 5 and 10, the total comes out to be 10 and the last digit has a 0 in the end.

In case i, the answer comes out to be A, as we do not want the statement 2.


Case (ii): The value of x can be a negative multiple of 5, in that case the overall Y will be in decimal terms...

IN case of decimal terms it is very difficult Eg: 9^-1 which is (1/9)

and also 9^-2, the answer is (1/81) so when you will add the 6 numbers you will still get that the overall numerator has 0 in the last place.

the numerator has 0 in its last place, the denominator is 9 and when you multiply the overall fraction by 5 the answer is a decimal fraction left by the denominator 9 but the numerator is divisible 5.

In this case the answer comes out to be E, then as the resulting number is having 9 as the numerator...


Hope my explanation is helpful to you guyzzzzz


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by sanju09 » Fri Mar 30, 2012 3:15 am
krishna239455 wrote:Problem is stated below:
x is a positive number
(9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is Y divisible by 5?
A)5 is a factor of x
B)x is a integer

I have doubt whether this can be a data suffeciency problem because,
I can simplify the above expression as: 9^x(1+9^1+9^2+9^3+9^4+9^5)=y
and i know that even power of 9 gives digit with 1 in units place and odd poer gives digit with 9 in units place. With this logic if i add all the terms in the bracket i should get digit with 0 in units place. This digit will always be divisible by 5. No need of statements given below. Kindly correct me if i am wrong!!!!
Received a PM from [email protected] to comment:


We can take

y = 9^x (1 + 9 + 9^2 + 9^3 + 9^4 + 9^5) = 9^x {(9^6) - 1}/8.

The part {(9^6) - 1}/8 is a positive integer divisible by 5 because

(9^6) - 1 = {(9^3) - 1} {(9^3) + 1} = {8 (81 + 9 + 1)} {10 (81 - 9 + 1)}.

Now, only one thing can hamper y to be divisible by 5, and that is if x is other than a whole number. Let's browse the statements:

I. If 5 is a factor of x, then y is divisible by 5 if x is a whole number and not divisible by 5 if x is a negative integer. Insufficient

II. If x is an integer, then y is divisible by 5 if x is non negative and not divisible by 5 if x is negative. Insufficient

Combining the two statements, we know that x is an integer divisible by 5; and [spoiler]still, y is divisible by 5 if x is non negative and not divisible by 5 if x is negative. Insufficient

Take E[/spoiler]
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by [email protected] » Fri Mar 30, 2012 9:15 am
Thank You Sanju 09!!! this means that my calculations were right...

I took some time in the trial and error, but finally got it...

Thanks once again...
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