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# n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9

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Math Expert
Joined: 02 Sep 2009
Posts: 46280
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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12 Jun 2015, 03:05
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65% (hard)

Question Stats:

60% (01:43) correct 40% (01:43) wrong based on 280 sessions

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n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.
(2) n is odd

Kudos for a correct solution.

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n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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12 Jun 2015, 03:48
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Bunuel wrote:
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.
(2) n is odd

Kudos for a correct solution.

CONCEPT : The ramainder when any number is divided by 5 is the remainder when the unit digit of the Number is divided by 5 therefore such questions require the calculation of the Unit digit of the number (y in this case) only

Question Redefined: y = 9^0 + 9^1 + 9^2 + … + 9^n

Statement 1: n is divisible by 3

@n=3, y = 9^0 + 9^1 + 9^2 + 9^3 = Unit digits are [ 1 + 9 + 1 + 9] = Unit digit of the numbers 0 i.e. Remainder 0 when y is divided by 5
@n=6, y = 9^0 + 9^1 + 9^2 +---+ 9^6 = Unit digits are [ 1 + 9 + 1 + 9 + 1+ 9 + 1] = Unit digit of the numbers 1 i.e. Remainder 1 when y is divided by 5

Hence, NOT SUFFICIENT

Statement 2: n is odd

@n=3, y = 9^0 + 9^1 + 9^2 + 9^3 = Unit digits are [ 1 + 9 + 1 + 9] = Unit digit of the numbers 0 i.e. Remainder 0 when y is divided by 5
@n=5, y = 9^0 + 9^1 + 9^2 + 9^3 + 9^4 + 9^5 = Unit digits are [ 1 + 9 + 1 + 9 + 1 + 9] = Unit digit of the numbers 0 i.e. Remainder 0 when y is divided by 5

with the above working, we can observe that y will always have even number of terms for every odd value of n as the minimum power starts from 0 in each terms.

Also every pair of two consecutive terms in y gives us 0 Unit digit i.e. sum of all pairs of terms in y becomes 0 and the remainder when y is divided by 5 will also become 0

Hence, SUFFICIENT

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Math Expert
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Posts: 46280
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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15 Jun 2015, 05:29
1
Bunuel wrote:
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.
(2) n is odd

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Remember the units digit pattern for 9^x, where x is an integer. The units digit of 9^x is 9 if x is odd, but the units digit is 1 if x is even: a repeating pattern of [9,1].

Now, consider the sums of the powers of 9 up to 9^n:
Attachment:

2015-06-15_1629.png [ 64.33 KiB | Viewed 2406 times ]

The alternating 1's and 9's in the units digits pair to a sum of 10, or a units digit of 0. Thus, the units digit of the sum displays another two-term repeating pattern. The units digit of y is 0 if n is odd, but 1 if n is even.

The remainder when y is divided by 5 depends only on the units digit and will be either 0 or 1 as well. So the rephrased question is “Is n odd or even?”

(1) INSUFFICIENT: If n is a multiple of 3, it may be either odd or even.

(2) SUFFICIENT: If n is odd, the units digit of y is 0, and the remainder is 0 when y is divided by 5.

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n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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03 Dec 2016, 05:19
Bunuel wrote:
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.
(2) n is odd

$$9 \equiv -1 \pmod{5} \implies 9^k \equiv (-1)^k \pmod{5}$$

(1) $$n=3k$$ so

$$\begin{split} y &\equiv (-1)^0 &+(-1)^1 &+(-1)^2 &+...&+(-1)^{3k} &\pmod{5}\\ y &\equiv \quad 1 &+ (-1) &+\quad 1 &+...&+(-1)^k &\pmod{5}\\ \end{split}$$

If $$k$$ is even, for example $$k=2$$, then $$y \equiv 1 \pmod{5}$$
If $$k$$ is odd, for example $$k=1$$, then $$y \equiv 0 \pmod{5}$$

Insufficient.

(2) $$n=2k+1$$ so
$$\begin{split} y &\equiv (-1)^0 &+(-1)^1 &+(-1)^2 &+...&+(-1)^{2k}&+(-1)^{2k+1} &\pmod{5}\\ y &\equiv \quad 1 &+ (-1) &+\quad 1 &+...&+\quad 1 &+(-1) &\pmod{5}\\ y &\equiv \quad 0 &\pmod{5} \end{split}$$

Sufficient

Other solution

$$y \equiv (-1)^0 +(-1)^1 +(-1)^2 +...+(-1)^{n} = \frac{(-1)^{n+1}-1}{(-1)-1}=\frac{-(-1)^{n}-1}{-2}=\frac{(-1)^n+1}{2}\pmod{5}$$

(1) If $$n=3k \implies y \equiv \frac{(-1)^{3k}+1}{2} \pmod{5}$$
If $$k=1 \implies y \equiv \frac{(-1)^3+1}{2}=0 \pmod{5}$$
If $$k=2 \implies y \equiv \frac{(-1)^6+1}{2}=1 \pmod{5}$$

Hence insufficient.

(2) If $$n=2k+1 \implies y \equiv \frac{(-1)^{2k+1}+1}{2}=\frac{(-1)+1}{2}=0 \pmod{5}$$
Hence sufficient.
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n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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02 Mar 2017, 00:38
1
Bunuel wrote:
n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.
(2) n is odd

Kudos for a correct solution.

Hi Bunuel...
I used negative remainders for this one...
9 when divided by 5 gives out a remainder of -1.
Lets see this given series by replacing 9 with -1...

$$(-1)^0 + (-1)^1 + (-1)^2 +(-1)^3..... +(-1)^n$$

So if n is odd, we will have the sum equal to 0.
And if n is even, we will have the sum equal to 1.

Now let's check the statements..
(1) n could be either odd or even. Not sufficient.
(2) N is odd. Sufficient

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Re: n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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17 Mar 2017, 13:40
"Y = 1+ 9+9^2... y = 5k + r. R = 0 if we get 0 or 5 as units digit

S-1) n = 3, 6,9..
If we see the sequence if n = 3
1+ 9 + 81+ 729 is div by 5
n = 6 => consider only units digit 1 + 9 + 1 + 9 + 1+ 9 +1 not div by 5.
NS

S-2) Sufficient if n is always odd n = 3,5 as in S-1
If we see the sequence if n = 3
1+ 9 + 81+ 729 is div by 5
"
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Re: n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9 [#permalink]

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27 Apr 2018, 12:41
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Re: n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9   [#permalink] 27 Apr 2018, 12:41
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