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Divisibility and Exponents combined. Please help!

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Divisibility and Exponents combined. Please help!

by adt29 » Wed Dec 21, 2011 4:49 am
x is a positive number. If 9^(x) + 9^(x+1)+ 9^(x+2)+ 9^(x+3)+ 9^(x+4)+ 9^(x+5)= y, is y divisible by 5?

(1) 5 is a factor of x. (2) x is an integer.


D

I did the problem this way.
Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.

Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'.

Now if x=1/2 which is not an integer, 9^(1/2) = 3. And 3 times a number that ends in 0, will give you a number that is divisible by 5.

So doesn't statement 2 get refuted? What am I missing here?
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by Anurag@Gurome » Wed Dec 21, 2011 5:10 am
adt29 wrote:x is a positive number. If 9^(x) + 9^(x+1)+ 9^(x+2)+ 9^(x+3)+ 9^(x+4)+ 9^(x+5)= y, is y divisible by 5?

(1) 5 is a factor of x. (2) x is an integer.


D

I did the problem this way.
Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.

Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'.

Now if x=1/2 which is not an integer, 9^(1/2) = 3. And 3 times a number that ends in 0, will give you a number that is divisible by 5.

So doesn't statement 2 get refuted? What am I missing here?

9^(x) + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4) + 9^(x+5)
= 9^x (1 + 9 + 9² + 9^3 + 9^4 + 9^5)
= 9^x[(1 + 9) + 9²(1 + 9) + 9^4(1 + 9)]
= 9^x(1 + 9)(1 + 9² + 9^4)
= 9^x * 10 * (1 + 9² + 9^4) = y, which implies that 10 is a factor of y if x is a positive integer. So, we can say that 5 is also a factor of y, if x is a positive integer.

(1) 5 is a factor of x which implies x is a positive integer; SUFFICIENT.

(2) x is an integer and given that x is a positive number together implies that x is a positive integer; SUFFICIENT.

The correct answer is D.
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by GmatMathPro » Wed Dec 21, 2011 7:21 am
adt29 wrote:x is a positive number. If 9^(x) + 9^(x+1)+ 9^(x+2)+ 9^(x+3)+ 9^(x+4)+ 9^(x+5)= y, is y divisible by 5?

(1) 5 is a factor of x. (2) x is an integer.


D

I did the problem this way.
Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.

Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'.

Now if x=1/2 which is not an integer, 9^(1/2) = 3. And 3 times a number that ends in 0, will give you a number that is divisible by 5.

So doesn't statement 2 get refuted? What am I missing here?
You're right that x=1/2 would make the expression divisible by 5. However, that does not refute any of the statements, because x=1/2 is not the only value that works. As Anurag demonstrates, the expression will also be divisible by 5 if x is any positive integer.
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by adt29 » Wed Dec 21, 2011 7:38 am
Thanks, guys.

I guess then my question becomes one of debating the philosophies of data sufficency questions.

The fact of the matter is that x=1/2 IS an option that works in the statement. So in statement 1, since 1/2 is not a factor of 5, the statement is sufficent, as it omits the possibility of any integers/non-integers that are not divisbile by 5.

In statement 2, however, x is an integer is the given statement. So does it mean that x=1/2 is "not allowed" to be one of the values, and thus given that the statement says x is a positive integer the fact that the number will end in zero no matter what means this statement is sufficient as well?

Anurag and GMATMathPro, I think this is slightly different from what you say.

Also, does my way of solving the problem,
"Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.

Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'. "
make sense?

Thank you for your time.
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by GmatMathPro » Wed Dec 21, 2011 12:01 pm
adt29 wrote:Thanks, guys.

I guess then my question becomes one of debating the philosophies of data sufficency questions.

The fact of the matter is that x=1/2 IS an option that works in the statement. So in statement 1, since 1/2 is not a factor of 5, the statement is sufficent, as it omits the possibility of any integers/non-integers that are not divisbile by 5.

In statement 2, however, x is an integer is the given statement. So does it mean that x=1/2 is "not allowed" to be one of the values, and thus given that the statement says x is a positive integer the fact that the number will end in zero no matter what means this statement is sufficient as well?
I guess I'm not really sure what your point is here. Statement 1's sufficiency has nothing to do with whether it rules out x=1/2. It has to do with the fact that it guarantees x is a positive integer, and this guarantees that the answer to the question "is y divisible by 5?" is yes.

In general, I would discourage you from zeroing in on one particular solution like this on data sufficiency questions. Instead, try to think about the larger classes of numbers that will prove sufficiency. Before we look at the statements we can see that if x is restricted to the positive integers, that will be sufficient to answer the question definitively. However, it would also be sufficient if we knew that x=(2n+1)/2, where n is a positive integer. Thus, a similar question with the same outcome could have been written with statements that centered around these non-integer solutions. But this isn't the path the question writer happened to take.
Also, does my way of solving the problem,
"Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.

Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'. "
make sense?
Yes, this is a fine way of looking at it. I assume the other part of your argument is that even powers of 9 end in 1.
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by adt29 » Wed Dec 21, 2011 5:53 pm
Thanks, GMATMathPro.

I understand what you're saying. One last question. If one of the statements was that x=(2n+1)/2, the the statement would also be sufficient, but the answer would be that y is NOT divisible by 5.

Correct?

Thanks!
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by GmatMathPro » Thu Dec 22, 2011 7:47 am
adt29 wrote:Thanks, GMATMathPro.

I understand what you're saying. One last question. If one of the statements was that x=(2n+1)/2, the the statement would also be sufficient, but the answer would be that y is NOT divisible by 5.

Correct?

Thanks!
No, it would be sufficient, but y would still be divisible by 5 for the same reason that x=1/2 works. Numbers of the form (2n+1)/2 are just odd numbers over 2, so we're be dealing with values like x=3/2, 5/2, 7/2...

Looking at the factored expression, 9^x * 10 * (1 + 9² + 9^4) = y, y will be divisible by 5 any time that 9^x is an integer. if we plug in something like 3/2 for x, we'd get 9^(3/2) which is the same as [9^(1/2)]^3 which is the same as 3^3 which is 27, an integer. We could do this process for any number of the form (2n+1)/2, so 9^x would always be an integer, and y would always be divisible by 5.
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by mankey » Thu Dec 22, 2011 8:19 am
Dear Anurag, GMATPro: What is the problem if "x is not a positive integer"? Please explain.

Thanks
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by GmatMathPro » Thu Dec 22, 2011 1:52 pm
mankey wrote:Dear Anurag, GMATPro: What is the problem if "x is not a positive integer"? Please explain.

Thanks
Mankey.
Are you asking how the problem would change if one of the statements was "x is not a positive integer"?
If so, that statement would be insufficient to answer the question "is y divisible by 5?" because sometimes y will be divisible by 5(like when x=1/2) and sometimes y will not be divisible by 5(like when x=0.1).
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by ArunangsuSahu » Fri Dec 30, 2011 9:49 am
Power of '9' has cycle of 9 and 1 in the unit place

as total 6 terms always the sum will be in the multiples of 10 combining the units and tens place.

So divisible by 5 and both the statements are SUFFICIENT
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