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If z = x^n – 19, is z divisible by 9?
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25 Apr 2013, 01:55
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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;
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25 Apr 2013, 02:04
If \(z = x^n  19\), is z divisible by 9? 1) x = 10; n is a positive integer\(z=10^119=9\) divisible by 9 \(z=10^219=81\) divisible by 9 \(z=10^319=991\) 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 9this means that \(\frac{z + 981}{9}=\frac{z+9*9*11}{9}=integer\) so one of the factors of z must be 9. Sufficient D
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Re: If z = x^n – 19, is z divisible by 9? 1) x = 10;
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25 Apr 2013, 02:58
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 \(z= 10^n 19 = 10^n109 = 10(10^{n1}1)9\) . Thus, z/9 = \(\frac{10(10^{n1}1)9}{(101)}\) \(a^x1\) is always divisible by (a1) 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. D.
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Re: If z = x^n – 19, is z divisible by 9?
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25 Apr 2013, 03:04
If z = x^n – 19, is z divisible by 9?(1) x = 10; n is a positive integer > \(z = x^n  19=10^n19=(10^n1)18\). Now, 10^n1 is always a multiple of 9 (for positive integer n, 10^n1 = 9, 99, 999, ...) and 18 is also a multiple of 9, thus \(x=(10^n1)18=(a \ multiple \ of \ 9)(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)\). 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. Answer: D.
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If z = x^n  19, is z divisible by 9?
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14 Jul 2015, 08:27
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  19Question : 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 9Statement 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, SUFFICIENTStatement 2: z + 981 is a multiple of 9981 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(c1) i.e. a multiple of 9 Hence, z must be a multiple of 9 as well SUFFICIENTAnswer: Option D
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Re: If z = x^n – 19, is z divisible by 9?
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05 May 2017, 12:26
Bunuel wrote: If z = x^n – 19, is z divisible by 9?
(1) x = 10; n is a positive integer > \(z = x^n  19=10^n19=(10^n1)18\). Now, 10^n1 is always a multiple of 9 (for positive integer n, 10^n1 = 9, 99, 999, ...) and 18 is also a multiple of 9, thus \(x=(10^n1)18=(a \ multiple \ of \ 9)(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)\). 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.
Answer: D. 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?
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05 May 2017, 12:30
llamsivel wrote: Bunuel wrote: If z = x^n – 19, is z divisible by 9?
(1) x = 10; n is a positive integer > \(z = x^n  19=10^n19=(10^n1)18\). Now, 10^n1 is always a multiple of 9 (for positive integer n, 10^n1 = 9, 99, 999, ...) and 18 is also a multiple of 9, thus \(x=(10^n1)18=(a \ multiple \ of \ 9)(a \ multiple \ of \ 9)=(a \ multiple \ of \ 9)\). 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.
Answer: D. 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/(nonzero integer) = 0 = integer.
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Re: If z = x^n – 19, is z divisible by 9?
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25 Dec 2018, 08:28
I was wondering whether we could also say 19 has a remainder of 1 and 10 has a remainder of 1, hence 10^n19 has a remainder of 0?
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Re: If z = x^n – 19, is z divisible by 9?
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25 Dec 2018, 08:28






