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25 Apr 2013, 01:55
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D
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Difficulty:
35%
(medium)
Question Stats:
70% (01:45) correct
30%
(01:54)
wrong
based on 420
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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
(1) x = 10; n is a positive integer
(2) z + 981 is a multiple of 9
25 Apr 2013, 03:04
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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.
Answer: D.
(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.
Answer: D.
General Discussion
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
\(z=10^1-19=-9\) 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
D
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
D
