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z = x^n - 19

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by sanju09 » Fri Oct 15, 2010 10:33 pm
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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by pradeepkaushal9518 » Sat Oct 16, 2010 12:22 am
z=x^n-19 is Z divisible by 9?

1.x=10

suppose n=1 z=10-19=-9 divisible by 9

n=2 z=100-19=81 divisible by 9

so suff

2.z+981 is mulitple of 9

so z also should be multiple of 9 hence it should be divisible by 9 hence suff

so each suff D
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by neerajkumar1_1 » Sat Oct 16, 2010 1:13 am
pradeepkaushal9518 wrote:z=x^n-19 is Z divisible by 9?

1.x=10

suppose n=1 z=10-19=-9 divisible by 9

n=2 z=100-19=81 divisible by 9

so suff

2.z+981 is mulitple of 9

so z also should be multiple of 9 hence it should be divisible by 9 hence suff

so each suff D
couldnt agree more...
also just to add for statement 1 that the result is constant for any value of n
solving by remainder theorem..
x^n will always leave a remainder of 1
and 19/9 will always leave a remainder of 1
so rem(x^n/9) - rem (19/9) = 1-1 = 0
Hence the expression will leave no remainder when divided by 9
or in other words, will always be divisible by 9...
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by Brian@VeritasPrep » Mon Oct 18, 2010 9:37 am
Hey guys,

Great question, and I just want to add one thing:

Most notably, you should get really comfortable with the premise in statement 1. When you have 10^x - a constant, that number is going to be pretty easy to predict...once you've accounted for the subtraction of "significant digits", the others will all be 9s. In this case, 10^n - 19 will give you:

n = 2: 100-19 = 81
n = 3; 1000 - 19 - 981
n = 4; 10000 - 19 = 9981

Those numbers get awfully repetitive, so for things like "divisible by 3" or "divisible by 9" in which the sum of the digits is critical, you can easily predict what adding more 9s to the sum of the digits will do. Similarly, if they just ask you in a problem solving question to sum the digits, you can multiply the 9s once you know how many digits you'll have, and then account for the other numbers (like 81) at the end.
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