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z = x^n - 19

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z = x^n - 19

by sanju09 » Sat Apr 04, 2009 6:14 am
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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Source: — Data Sufficiency |
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by Sumit69 » Sun Apr 05, 2009 6:57 am
IMO Ans is D
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by sacx » Sun Apr 05, 2009 10:33 am
Stmt I

z = x^n - 19

it will always be a multiple of 9

when n = 1, z = -0009
when n = 2, z = 0081
when n = 3, z = 0981
when n = 4, z = 9981

if you observe the pattern here, the units digit is always 1 and the tens digit is 8 and the other digits will either be 0 or 9 (except when n =1)

the sum of digits of any such number will always be divisible by 9

Sufficient

Stmt II

z + 981 is a multiple of 9

z/9 + 981/9 = Integer
z/9 + 109 = Integer

therefore z/9 has to be an integer to satisfy the equation. Hence z has to be evenly divisible of 9

Sufficient

D
Last edited by sacx on Mon Apr 06, 2009 1:35 am, edited 1 time in total.
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by cubicle_bound_misfit » Sun Apr 05, 2009 5:30 pm
just one correction to sacx

for n=1 z = -9 which does not follow the pattern.

From n>2 it will always follow the pattern of 9* 10^n-1 +81
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by mjjking » Sun Apr 05, 2009 10:34 pm
Cubicle, you're right. Yet, is -9 a multiple of 9?
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by sacx » Mon Apr 06, 2009 1:36 am
thanks cubicle_bound_misfit

I have edited my previous post
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by sanju09 » Mon Apr 06, 2009 2:48 am
The question reads "is z divisible by 9" this should not be taken as "is z a multiple of 9", for the reason that when we look for the multiples of a positive integer, we look for the positive integers only, whereas while checking divisibility, we do not mind sign, we only make sure that when the absolute value of the integer (z here) is divided by 9, the remainder is zero. So when we put x = 10 and n = 1, we get 10 ^ 1 - 19 = -9, which is divisible by 9, and hence it is not violating the so called pattern! :)
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by vittalgmat » Tue Apr 07, 2009 12:15 am
IMO D as well.
Followed same procedure/logic as that of sacx.

Thanks for the excellent explanations everyone.
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