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Good DS.

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Source: — Data Sufficiency |

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by [email protected] » Fri Oct 18, 2013 5:58 pm
Hi rakeshd347,

This DS question has a great little Number Property behind it, although most people don't realize it. Sometimes the best way to figure out that a pattern exists is to PROVE it.

Here, we're told that N is a NON-NEGATIVE integer (so it can be either 0 or a positive integer). We're asked if (10^N) + 8 is evenly divisible by 18? This is a Yes/No question.

I'm going to show you the pattern right up front:
If N = 0, then 10^0 + 8 = 9 and the answer is NO
If N = 1, then 10^1 + 8 = 18 and the answer is YES
If N = 2, then 10^2 + 8 = 108 and the answer is YES (pattern: 108 = 18 + 90)
If N = 3, then 10^3 + 8 = 1008 and the answer is YES (pattern: 1008 = 108 + 900)
If N = 4, then 10^4 + 8 = 10008 and the answer is YES (pattern: 10008 = 1008 + 9000)

18 divides evenly into 90, so it divides into 900, 9000, etc.
As N gets bigger, each total = a multiple of 90 + the prior total, so 18 will ALWAYS divide evenly into the total. The "exception" is when N = 0.

Fact 1: N is prime.
We have proof (in the above list) that the answer is ALWAYS YES.
Fact 1 is SUFFICIENT

Fact 2: N is even
If N = 0, then the answer is NO
If N = 2, then the answer is YES
Fact 2 is INSUFFICIENT

Final Answer: A

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Rich
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by theCodeToGMAT » Fri Oct 18, 2013 11:44 pm
(10^n + 8)/18

= (10^n + 8 + 10 - 10)/18

= (10^n - 10 + 18)/18

= 10(10^(n-1))/18 + 18/18

To find: (10^(n-1) - 1)/9

For every value greater than 1 the equation is true as all digits will result in "9"

Statement 1:
N > 1
SUFFICIENT

Statement 2:
N can be 0
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]
R A H U L
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by Brent@GMATPrepNow » Sat Oct 19, 2013 6:47 am
rakeshd347 wrote:If n is an non-negative integer, is (10^n)+8 divisible by 18?

(1) n is a prime number.

(2) n is even.
Target question: Is (10^n)+8 divisible by 18?

This is a great candidate for rephrasing the target question.

IMPORTANT: In order for (10^n)+8 to be divisible by 18, it must be divisible by 9 AND by 2.
The good thing is that (10^n)+8 is already divisible by 9 for ALL non-negative integer values of n.
We know this because all integers divisible by 9 are such that the sum of their digits is divisible by 9. For example, 504, 11709 and 8838 are divisible by 9 because the sums of their integers are 9, 18 and 27 (respectively), and 9, 18 and 27 are all divisible by 9.

Let's examine some values of (10^n)+8
n = 0: (10^n)+8 = 1 + 8 = 9
n = 1: (10^n)+8 = 10 + 8 = 18
n = 2: (10^n)+8 = 100 + 8 = 108
n = 3: (10^n)+8 = 1000 + 8 = 1008
n = 4: (10^n)+8 = 10000 + 8 = 10008
As you can see, the sum of the digits of (10^n)+8 will ALWAYS be 9. So, (10^n)+8 is ALWAYS divisible by 9.

So, as you can see, we've already taken care of the "divisible by 9" part.
So, in order to determine whether (10^n)+8 is divisible by 18, we need only determine whether (10^n)+8 is divisible by 2.
This means we can rephrase the target question as follows:

REPHRASED target question: Is (10^n)+8 even? (i.e., divisible by 2)

Since 8 is ALWAYS even, we can see that (10^n)+8 will be even whenever (10^n) is even. Furthermore, we can see that (10^n) will be even as long as n ≠ 0. In other words, (10^n) will be even AS LONG AS n > 0. So, we can rephrase the target question one last time.

REPHRASED target question: Is n > 0 ?

Now that we've taken a moment to rephrase the target question to such a great extent, the statements should take no time at all to analyze.

Statement 1: n is a prime number.
In n is prime, then n is definitely greater than zero
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: n is even
case a: n = 2, in which case n is greater than zero
case b: n = 0, in which case n is not greater than zero
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent

NOTE: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
Brent Hanneson - Creator of GMATPrepNow.com
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by Kamakshi Chawla » Sat Dec 24, 2016 10:10 am
But what we study is 0 is either odd not even.
So in that case even statement 2 is sufficient and answer is D.
Am i right?
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by Kamakshi Chawla » Sat Dec 24, 2016 10:12 am
Brent@GMATPrepNow wrote:
rakeshd347 wrote:If n is an non-negative integer, is (10^n)+8 divisible by 18?

(1) n is a prime number.

(2) n is even.
Target question: Is (10^n)+8 divisible by 18?

This is a great candidate for rephrasing the target question.

IMPORTANT: In order for (10^n)+8 to be divisible by 18, it must be divisible by 9 AND by 2.
The good thing is that (10^n)+8 is already divisible by 9 for ALL non-negative integer values of n.
We know this because all integers divisible by 9 are such that the sum of their digits is divisible by 9. For example, 504, 11709 and 8838 are divisible by 9 because the sums of their integers are 9, 18 and 27 (respectively), and 9, 18 and 27 are all divisible by 9.

Let's examine some values of (10^n)+8
n = 0: (10^n)+8 = 1 + 8 = 9
n = 1: (10^n)+8 = 10 + 8 = 18
n = 2: (10^n)+8 = 100 + 8 = 108
n = 3: (10^n)+8 = 1000 + 8 = 1008
n = 4: (10^n)+8 = 10000 + 8 = 10008
As you can see, the sum of the digits of (10^n)+8 will ALWAYS be 9. So, (10^n)+8 is ALWAYS divisible by 9.

So, as you can see, we've already taken care of the "divisible by 9" part.
So, in order to determine whether (10^n)+8 is divisible by 18, we need only determine whether (10^n)+8 is divisible by 2.
This means we can rephrase the target question as follows:

REPHRASED target question: Is (10^n)+8 even? (i.e., divisible by 2)

Since 8 is ALWAYS even, we can see that (10^n)+8 will be even whenever (10^n) is even. Furthermore, we can see that (10^n) will be even as long as n ≠ 0. In other words, (10^n) will be even AS LONG AS n > 0. So, we can rephrase the target question one last time.

REPHRASED target question: Is n > 0 ?

Now that we've taken a moment to rephrase the target question to such a great extent, the statements should take no time at all to analyze.

Statement 1: n is a prime number.
In n is prime, then n is definitely greater than zero
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: n is even
case a: n = 2, in which case n is greater than zero
case b: n = 0, in which case n is not greater than zero
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent

NOTE: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100



But what we study is 0 is either odd not even.
So in that case even statement 2 is sufficient and answer is D.
Am i right?
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by [email protected] » Sat Dec 24, 2016 2:05 pm
Hi Kamakshi Chawla,

On a Number Line, integers follow the same pattern in either direction: odd-even-odd-even-odd-even, etc.

-3 = odd
-2 = even
-1 = odd
0 = even
1 = odd
2 = even
etc.

So 0 IS even. It is NEITHER positive nor negative though - it's what is referred to as a 'null value.'

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by Brent@GMATPrepNow » Sat Dec 24, 2016 5:13 pm
If integer y can be written as the product 2k, for some integer k, then y is EVEN.
For example, since we can write 6 as (2)(3), we can conclude that 6 is even.
Likewise, since we can write -10 as (2)(-5), we can conclude that -10 is even.

Similarly, since we can write 0 as (2)(0), we can conclude that 0 is even.

Cheers,
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