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# If z = x^n – 19, is z divisible by 9?

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If z = x^n – 19, is z divisible by 9?  [#permalink]

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25 Apr 2013, 00:55
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If z = x^n – 19, is z divisible by 9?

(1) x = 10; n is a positive integer
(2) z + 981 is a multiple of 9
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Re: If z = x^n – 19, is z divisible by 9? 1) x = 10;  [#permalink]

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25 Apr 2013, 01:04
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If $$z = x^n - 19$$, is z divisible by 9?

1) x = 10; n is a positive integer
$$z=10^1-19=-9$$ divisible by 9
$$z=10^2-19=81$$ divisible by 9
$$z=10^3-19=991$$ divisible by 9
We can see a pattern here. Whan a number is divisible by nine? When the sum of its digit is a multiple of nine, here is always the case as you can see. Going on with the values of $$n$$ you'll see that the resulting number will have the form 9XXX( 9 x times)81 and the sum of those digits will always be a multiple of nine. Sufficient

2) z + 981 is a multiple of 9
this means that $$\frac{z + 981}{9}=\frac{z+9*9*11}{9}=integer$$ so one of the factors of z must be 9. Sufficient
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Re: If z = x^n – 19, is z divisible by 9? 1) x = 10;  [#permalink]

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25 Apr 2013, 01:58
rajatr wrote:
If z = x^n – 19, is z divisible by 9?

1) x = 10; n is a positive integer
2) z + 981 is a multiple of 9

From F.S 1, we know that $$z= 10^n -19 = 10^n-10-9 = 10(10^{n-1}-1)-9$$ . Thus, z/9 = $$\frac{10(10^{n-1}-1)-9}{(10-1)}$$

$$a^x-1$$ is always divisible by (a-1) for positive integer x, thus the above expression is always divisible by 9. Sufficient.

From F.S 2, we know that (z+981) = 9k, where k is an integer. Thus, k = $$\frac{z}{9}+\frac{981}{9}$$=$$\frac{z}{9}$$+ 109. Thus, as k is an integer, thus z/9 has to be an integer,--> z is divisible by 9.Sufficient.

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Re: If z = x^n – 19, is z divisible by 9?  [#permalink]

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25 Apr 2013, 02:04
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If z = x^n – 19, is z divisible by 9?

(1) x = 10; n is a positive integer --> $$z = x^n - 19=10^n-19=(10^n-1)-18$$. Now, 10^n-1 is always a multiple of 9 (for positive integer n, 10^n-1 = 9, 99, 999, ...) and -18 is also a multiple of 9, thus $$x=(10^n-1)-18=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

(2) z + 981 is a multiple of 9. Since 981 is a multiple of 9, then we have that $$z+(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$ --> $$z=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

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If z = x^n - 19, is z divisible by 9?  [#permalink]

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14 Jul 2015, 07:27
Priyank38939 wrote:
If z = x^n - 19, is z divisible by 9?
(1) x = 10; n is a positive integer
(2) z + 981 is a multiple of 9

Given : z = x^n - 19

Question : is z divisible by 9?

CONCEPT: The Number will be divisible by 9 if the Sum of the digits of the number is a multiple of 9

Statement 1: x = 10; n is a positive integer

@n=1, z = 10^1 - 19 = -9 Divisible by 9
@n=2, z = 10^2 - 19 = 82 Divisible by 9
@n=3, z = 10^3 - 19 = 981 Divisible by 9
i.e. The result is always Divisible by 9. Hence,
SUFFICIENT

Statement 2: z + 981 is a multiple of 9

981 is a multiple of 9 and also (z + 981) is a multiple of 9 as well which is possible only when z also is a multiple of 9 because if 9a + b = 9c then b = 9(c-1) i.e. a multiple of 9
Hence, z must be a multiple of 9 as well
SUFFICIENT

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Re: If z = x^n – 19, is z divisible by 9?  [#permalink]

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05 May 2017, 11:26
Bunuel wrote:
If z = x^n – 19, is z divisible by 9?

(1) x = 10; n is a positive integer --> $$z = x^n - 19=10^n-19=(10^n-1)-18$$. Now, 10^n-1 is always a multiple of 9 (for positive integer n, 10^n-1 = 9, 99, 999, ...) and -18 is also a multiple of 9, thus $$x=(10^n-1)-18=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

(2) z + 981 is a multiple of 9. Since 981 is a multiple of 9, then we have that $$z+(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$ --> $$z=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

For statement 2, why can't Z=0 as well as a multiple of 9?
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Re: If z = x^n – 19, is z divisible by 9?  [#permalink]

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05 May 2017, 11:30
1
llamsivel wrote:
Bunuel wrote:
If z = x^n – 19, is z divisible by 9?

(1) x = 10; n is a positive integer --> $$z = x^n - 19=10^n-19=(10^n-1)-18$$. Now, 10^n-1 is always a multiple of 9 (for positive integer n, 10^n-1 = 9, 99, 999, ...) and -18 is also a multiple of 9, thus $$x=(10^n-1)-18=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

(2) z + 981 is a multiple of 9. Since 981 is a multiple of 9, then we have that $$z+(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$ --> $$z=(a \ multiple \ of \ 9)-(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)$$. Sufficient.

For statement 2, why can't Z=0 as well as a multiple of 9?

z can be 0 but this won't change the answer because 0 is a multiple of every integer, 0/(non-zero integer) = 0 = integer.
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Re: If z = x^n – 19, is z divisible by 9?  [#permalink]

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25 Dec 2018, 07:28
I was wondering whether we could also say 19 has a remainder of 1 and 10 has a remainder of 1, hence 10^n-19 has a remainder of 0?
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Re: If z = x^n – 19, is z divisible by 9? &nbs [#permalink] 25 Dec 2018, 07:28
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