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by goyalsau » Tue Dec 07, 2010 3:29 am
Is the six digit number divisible by 13?

(A) The sum of the digits of the number is a multiple of 13.
(B) The number is formed by appending a three digit number to its right such that we get a six digit number.
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Source: — Data Sufficiency |

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by Rahul@gurome » Tue Dec 07, 2010 4:46 am
goyalsau wrote:Is the six digit number divisible by 13?

(A) The sum of the digits of the number is a multiple of 13.
(B) The number is formed by appending a three digit number to its right such that we get a six digit number.
Statement 1: The sum of the digits of the number is a multiple of 13.
There is no such divisibility rule associated with 13.

Not sufficient.

Statement 2: The number is formed by appending a three digit number to its right such that we get a six digit number.

Thus, if the original 3 digit number was abc i.e. (100a + 10b + c), then the 6 digit number will be abcabc i.e. (100000a + 10000b + 1000c + 100a + 10b + c) = (100100a + 10010b + 1001c) = 1001*(100a + 10b + c)

Now, 1001 = 7*11*13 => A multiple of 13
Therefore abcabc is also a multiple of 13.

Sufficient.

The correct answer is B.
Last edited by Rahul@gurome on Tue Dec 07, 2010 4:47 am, edited 1 time in total.
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by beat_gmat_09 » Tue Dec 07, 2010 4:47 am
Can you please Cite the source ?
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by goyalsau » Tue Dec 07, 2010 5:06 am
beat_gmat_09 wrote:Can you please Cite the source ?
Time4education.com
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by shovan85 » Tue Dec 07, 2010 5:07 am
goyalsau wrote:Is the six digit number divisible by 13?

(A) The sum of the digits of the number is a multiple of 13.
(B) The number is formed by appending a three digit number to its right such that we get a six digit number.
(B) Let the number be 670670.
The sum of digits is 26 multiple of 13.
Now, 670670 is divisible by 13. Remainder is 0.

Factorize 670670 = 2 * 5 * 7 * 11 * 13 * 67 = 670 * 2 * 5 * 11 = 670 * 1001
As 1001 is div by 13 670670 is div by 13.

Now Generalize, abc*1001 = abcabc
Always divisible by 13.

Thus sufficient.

(A) Let the number be 100057.
The sum of digits is 13 multiple of 13.
Now, 100057 is not divisible by 13. Remainder is 5.

Let the number be 670670.
The sum of digits is 26 multiple of 13.
Now, 670670 is divisible by 13. Remainder is 0.

Thus Insufficient.

PS: Its better to start with option (B) as we can have multiple options to test for A. If I would not have solved B first I could not have got 670670 for A.

Takeaway is in bold.

IMO B
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by goyalsau » Tue Dec 07, 2010 5:09 am
Rahul@gurome wrote:
goyalsau wrote:Is the six digit number divisible by 13?

(A) The sum of the digits of the number is a multiple of 13.
(B) The number is formed by appending a three digit number to its right such that we get a six digit number.
Statement 1: The sum of the digits of the number is a multiple of 13.
There is no such divisibility rule associated with 13.

Not sufficient.

Statement 2: The number is formed by appending a three digit number to its right such that we get a six digit number.

Thus, if the original 3 digit number was abc i.e. (100a + 10b + c), then the 6 digit number will be abcabc i.e. (100000a + 10000b + 1000c + 100a + 10b + c) = (100100a + 10010b + 1001c) = 1001*(100a + 10b + c)

Now, 1001 = 7*11*13 => A multiple of 13
Therefore abcabc is also a multiple of 13.

Sufficient.

The correct answer is B.
Thanks Rahul I was searching for Divisibility Rule for 13, There is a rule of 7 , 11 and 13,

https://www.youtube.com/watch?v=VB6_pwXD ... re=related

I hope you will like it, It really very interesting....
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