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

Question

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.

Bunuel · Accepted Answer

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 .

TehMoUsE · Answer

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 :-D

hunterdonald · Answer

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)

thefibonacci · Answer

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.

dabral · Answer

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

taleesh · Answer

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.

dabral · Answer

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

MorningRunner · Answer

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....

AdlaT · Answer

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

lipsi18 · Answer

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

Hence answer is D

Thanks,

stonecold · Answer

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

chetan2u · Answer

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..

stonecold · Answer

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

kumarparitosh123 · Answer

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.

Sent from my Lenovo TAB S8-50LC using  GMAT Club Forum mobile app

Shishou · Answer

There is no need to calculate.Think carefully. When you factor out 9^x what you have are a sum of integers.You can always check if that sum is divisible by 5 by adding but we don't need to.
It will depend on the value of x in 9^x.

Statement 1: if x is a multiple of 5,then x is an integer.no value of 9^x will be divisible by 5.And since we can add the values in the bracket then we can know that y is not divisible by 5

Statement 2 :Same result as Statement 1.Since x is an integer 9^x will never be divisible by 5.we can add everything else in the bracket.Sufficient

Answer is D

kpop1234567890 · Answer

Why do we even need the statements if we can anyway figure out from qestion stem that y is divisible by 5

chetan2u · Answer

x is not given as integer but as positive number, so it could be 0.1 or 0.6. In these cases, y would not be an integer.

kpop1234567890 · Answer

I am sorry can you explain more? Even if x i snot an integer when y is divided by 5 it can cross out the 5 multiple in numerator

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

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