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9 ^ (x) + ... +9 ^ (x + 5) = y

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9 ^ (x) + ... +9 ^ (x + 5) = y

by sanju09 » Mon Mar 30, 2009 3:52 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.

OA D
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by bluementor » Mon Mar 30, 2009 6:56 am
y = (9^x)*(1 + 9^1 + 9^2 + 9^3 + 9^4 + 9^5)

Now, notice the following the pattern:
The unit digit of 9 = 9
The unit digit of 9^2 = 1
The unit digit of 9^3 = 9
The unit digit of 9^4 = 1
The unit digit of 9^5 = 9
… and so on…

Therefore:
(1 + 9^1) will yield a multple of 10, since unit digit of 0 is resulted when the unit digits of 1 and 9^1 are summed. The same thing can be deduced for (9^2 + 9^3) and (9^4 + 9^5)

So, 1 + 9^1 + 9^2 + 9^3 + 9^4 + 9^5 = some multiple of 10.

Therefore y = (9^x)*(some multiple of 10) is divisible by 5, provided that 9^x is an integer. And this will be the case if x can be proved to be an integer.

Both statements individually satisfy that x = integer. Hence D.

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Re: 9 ^ (x) + ... +9 ^ (x + 5) = y

by Ian Stewart » Mon Mar 30, 2009 6:56 am
sanju09 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.

OA D
You never see real GMAT questions about divisibility that don't specify in advance that the quantities involved are integers. Statement 1, for example, which is a nebulous way of saying 'x is an integer, and is divisible by 5', is not the kind of statement you'd ever see on a GMAT unless you were told in the question that x was an integer.

In any case, it's a factorization question:

9^(x) + 9^(x+1) + 9^(x+2) + 9^(x+3) + 9^(x+4) + 9^(x+5) = 9^x (1 + 9 + 9^2 + 9^3 + 9^4 + 9^5)
= 9^x ((1 + 9) +9^2(1+9) + 9^4(1+9))
= 9^x (1+9)(1 + 9^2 + 9^4)
= 9^x * 10 * (1 + 9^2 + 9^4)

So 10 (and therefore 5) is a factor of y if x is a positive integer. Since we can conclude from each statement that x is a positive integer, each is sufficient.
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