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# x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)

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Math Expert
Joined: 02 Sep 2009
Posts: 51197
x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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24 Oct 2014, 08:26
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85% (hard)

Question Stats:

53% (01:45) correct 47% (01:41) wrong based on 243 sessions

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Tough and Tricky questions: Factors.

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.

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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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24 Oct 2014, 08:48
The problem can be reduced by factoring out 9^x => 9^x(1+9+9^2+9^3+9^4+9^5)=y
Now, the last 2 digits of number in the parenthesis would be 30. So, we are down to 9^x(....30)=y
This would be divisible by 5 as long as 9^x is not a fraction.

From 1) 5 is a factor of x => X cannot be a fraction; X has to be integer and X is +ve from the question stem = Sufficient
From 2) X is an integer and X is positive => sufficient.

Answer should be : d (each alone is sufficient)
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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02 Nov 2014, 00:57
Bunuel wrote:

Tough and Tricky questions: Factors.

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.

9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4) + 9^(x+5) = y, is y divisible by 5 means:
is {9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4) + 9^(x+5)} mod 5 = 0 ?
or {(-1)^x + (-1)^(x+1) + (-1)^(x+2) + (-1)^(x+3) + (-1)^(x+4) + (-1)^(x+5)} mod 5 = 0
now we just need to make sure whether x is an integer or not (it is already mentioned that x is a +ve)
as it wouldn't matter if x is ever or odd coz for even x, remainder would be 0 and for odd x the remainder would again be 0.

1) 5 is a factor of x.
So x is an integer (odd or even)
sufficient.

2) x is an integer
sufficient.

D.
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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03 Nov 2014, 03:46
1
Bunuel wrote:

Tough and Tricky questions: Factors.

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.

Y is only divisible by 5 if its digit number is either 0 or 5

If x is an integer $$9^x$$ can only have the digit numbers 1 and 9

$$y = 9 + 1 + 9 +1 +9 +1 = 10$$
or
$$y = 1 + 9 +1 +9 +1 +9= 10$$

(keep in mind that $$x>0$$ so we can't have fracions)

$$y$$ will be divisible by 5 if x is an integer!

(1) 5 is a factor of x $$--->$$ x is an integer, SUFFICIENT

(1) x is an integer $$--->$$ x is an integer, SUFFICIENT

Hence, OA i D

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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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03 Nov 2014, 05:45
Bunuel

I have a couple of comments here on why the GMAT would not test a question in this format. First leaving x as a positive real number introduces ambiguity about the concept of divisibility of the expression 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4) + 9^(x+5)=66430(9^x). I know you eliminate this outcome with the statements, but in my opinion GMAT will not phrase a main stem in this format. Technically it does not have to be that way, but GMAT seems to have some unwritten rules and this one seems to be one of those. Instead they are more likely to announce in the main stem that x is an integer, which restricts the 9^x + .... expression to integer values.

I would be curious to know of an official GMAT question that is phrased in this fashion.

Cheers,
Dabral

Cheers,
Dabral
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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11 Nov 2014, 16:04
I have a question here - if x is an integer = it can be any integer 0 or 1,2,3,4,5. so how can it is correct.
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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11 Nov 2014, 22:31
taleesh

The expression $$9^x + 9^{x+1} + 9^{x+2} + 9^{x+3} + 9^{x+4} + 9^{x+5} = y$$ can be rewritten by factoring $$9^x$$, which yields $$9^x(1 + 9^1 + 9^2 + 9^3 + 9^4+ 9^5) = 9^x(66430)=y$$. Here x is constrained to be a positive integer and $$9^x$$ would be some integer, however it is being multiplied by a number that ends in 0, which means the entire number is a multiple of 10, which is always divisible by 5. Note here that the number 66430 need not be computed, instead one can look at the units digits of powers of 9 and conclude that it has to end in 0.

Cheers,
Dabral
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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24 Mar 2015, 05:05
Bunuel wrote:

Tough and Tricky questions: Factors.

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.

Hi Bunuel,
I have a basic question~
Should factors be always integers?

Can decimals be called factors?

For example, I write 1.5 = 0.3*5 -- Here, can I say 0.3 and 5 are factors of 1.5?

I know it is a very basic question.. I just want to clear my doubt....
Math Expert
Joined: 02 Sep 2009
Posts: 51197
Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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24 Mar 2015, 05:29
2
2
navneetb wrote:
Bunuel wrote:

Tough and Tricky questions: Factors.

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.

Hi Bunuel,
I have a basic question~
Should factors be always integers?

Can decimals be called factors?

For example, I write 1.5 = 0.3*5 -- Here, can I say 0.3 and 5 are factors of 1.5?

I know it is a very basic question.. I just want to clear my doubt....

On the GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
(i) $$a$$ is an integer;
(ii) $$b$$ is an integer;
(iii) $$\frac{a}{b}=integer$$.
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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Updated on: 25 Mar 2015, 11:33
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?

whether X is a Odd integer or Even integer the above expression will have 0 as its unit digit.

$$9^1$$=9 as unit digit

$$9^2$$=1 as unit digit

$$9^3$$=9 as unit digit

if X is odd integer-->$$9^{(odd)}$$+$$9^{(even)}$$+$$9^{(odd)}$$+$$9^{(even)}$$+$$9^{(odd)}$$+$$9^{(even)}$$

xxxx9+xxxx1+xxxx9+xxxx1+xxxx9+xxxx1=xxxxxxx0, which is divisible by 5.

if x is even integer--->$$9^{(even)}$$+$$9^{(odd)}$$+$$9^{(even)}$$+$$9^{(odd)}$$+$$9^{(even)}$$+$$9^{(odd)}$$

xxxxx1+xxxxx9+xxxxx1+xxxxx9+xxxx1+xxxx9=xxxxxxx0, which is divisible by 5

Statement (1): 5 is a factor of x.

X is an even integer or odd integer. Value of y will have 0 in its unit place. so SUFFICIENT.

Statement (2): x is an integer.

X is an even integer or odd integer. Value of y will have 0 in its unit place. so SUFFICIENT.

Ans. D

Originally posted by AdlaT on 24 Mar 2015, 12:01.
Last edited by AdlaT on 25 Mar 2015, 11:33, edited 4 times in total.
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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24 Mar 2015, 23:34
Last digit of 9*9=1, 9*9*9=9, 9*9*9=1 so on....
hence whatever may be the value of X, the sum of six terms in this order will be 1+9+1+9+1+9=last digit =0; hence divisible by 5

1. x is multiple of 5; hence value would be, Y= 9+1+9+1+9+1=last digit=0 ; hence divisible by 5; sufficient
2. X is an integer; x can be 1,2,3,4.....; but Y= 1+9+1+9+1+9=last digit =0; hence divisible by 5; sufficient

Thanks,
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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Updated on: 15 Mar 2016, 10:08
Hey chetan2u thank you for responding to our queries ..
Here in this question i choose B ..
I saw someone write rule that If GMAT says factors consider only positive terms
is it necessary?
thanks
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Originally posted by stonecold on 15 Mar 2016, 09:01.
Last edited by stonecold on 15 Mar 2016, 10:08, edited 1 time in total.
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Posts: 7106
Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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15 Mar 2016, 09:13
Chiragjordan wrote:
Hey chetan2u thank you for responding to our queries ..
Here in this question i choose B ..
I had saw someone write rule that If GMAT says factors consider only positive terms
is it necessary?
thanks

Hi Chiragjordan,

Yes when ever we are talking of factors or remainders or square roots, they are always positive..
But why did you choose B here, Clearly A too is sufficient..
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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15 Mar 2016, 10:13
chetan2u wrote:
Chiragjordan wrote:
Hey chetan2u thank you for responding to our queries ..
Here in this question i choose B ..
I had saw someone write rule that If GMAT says factors consider only positive terms
is it necessary?
thanks

Hi Chiragjordan,

Yes when ever we are talking of factors or remainders or square roots, they are always positive..
But why did you choose B here, Clearly A too is sufficient..

I got confused actually ...
With that Tic tic clock ticking away i choose B
coz in A i thought what if x is -25
so the denominator will come into play...
But as you said positive values only...
I forgot the GOLDEN rule..

Thanks
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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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27 Jul 2017, 01:17
If we solve the exponents we get as y=(91*730)* 9^x
And irrespective of the value of X, this set will always be divisible by 5 since 730 is divisible b 5.
Thus each statement alone is sufficient.
Thus D.

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Re: x is a positive number. If 9^x + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4)  [#permalink]

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