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Integer Properties - DS

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by cramya » Sat Dec 13, 2008 12:45 pm
The answer would be B)

Stmt I

8*10^k+j divided by m

k=13

8*10^k would be 8 followed by k zeros with sum of the digits = 8

Depending on the value of j the remainder changes

j=1 remainder 0
j=2 remainder 1

etc...

INSUFF

Stmt II

j=1

Exactly what we need remiander will always be 0

SUFF

Choose B)

Regards,
Cramya
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by cramya » Sat Dec 13, 2008 12:46 pm
Also if possible please post DS questions in the DataSufficiency section of the forum
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by Brent@GMATPrepNow » Sat Dec 13, 2008 5:32 pm
The OA is B and I'll post these questions in the DS section from now on.
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by logitech » Sat Dec 13, 2008 5:48 pm
Brent Hanneson wrote:The OA is B and I'll post these questions in the DS section from now on.
Cheers,
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I warned you about Cramya!!! Watch out!
LGTCH
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by sohosweets » Sun Dec 14, 2008 10:59 am
8*10^k would be 8 followed by k zeros with sum of the digits = 8

Depending on the value of j the remainder changes

j=1 remainder 0
j=2 remainder 1


hmm .. i don't get it .. u mentioned "8*10^k would be 8 followed by k zeros" ... that i get ... but why do i care about the "sum of the digits = 8?"
and what about the nine mentioned in the question?
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by Brent@GMATPrepNow » Sun Dec 14, 2008 3:10 pm
The earlier poster is referring to a divisibility rule that says "If a number x is such that the sum of its digits is divisible by 9, then the original number, x, is divisible by 9"

Take 1000000320048 for example. It would take a while to determine whether this number is divisible by 9. However, we can use the rule. The sum of the digits is 18, and 18 is divisible by 9. So, we can conclude that 1000000320048 is divisible by 9.

Back to the original question. We know that 8 x 10^k will be an 8 followed by several 0's

If j=1 [from statement (2)], then 8 * 10^k + j will be 8 followed by several 0's and then 1 (e.g., 80000001)
We can see that the sum of the digits will always equal 9, so this number will be divisible by 9.
Last edited by Brent@GMATPrepNow on Mon May 30, 2011 6:52 am, edited 1 time in total.
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easier way

by magros80 » Sat Dec 20, 2008 5:12 pm
You can also see that 8 x 10^k will always have a remainder of 8 when divided by 9.
That is, 8 x 10^1 = 80, which has a remainder of 8 when divided by 9.
8 x 10^2 = 800 which also has a remainder of 8 when divided by 9.
. . . etc.
Therefore, the only variable we need to concern ourselves with is what j is.
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by paritosh_b » Fri Nov 05, 2010 6:00 am
Rule of divisibility by 9:
If the sum of the digits of the number is divisible by 9, so is the number itself.

Stat 1:
8*10^13+j.Depending upon the value of J,sum of the digits can change.
So,Insufficient.

Stat 2:
8*10^K+1.Sum of digits is 9.
9/9.
So,the number is divisible by 9.
So,Sufficient.

Ans:B
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by thebigkats » Tue Nov 09, 2010 12:58 pm
for a given value of k, the remainder would be different for different 'j'. So it is not sufficient to know the value of k
however if j = 1 then you always get a number whose digit-sum = 9, hence predictable / constant remainder.
Hence (B)
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by Abhishek009 » Wed Nov 10, 2010 2:55 am
Brent Hanneson wrote:Source: Beat The GMAT Practice Questions

If j and k are positive integers, what is the remainder when 8 * 10^k + j is divided by 9?
(1) k = 13
(2) j = 1
Indeed a cool question ,

We know any number is divisible by 9 , if and only if the sum of the digits of the number is divisible by 9.

Now it is given :

8 * 10^k + j

k can take any value from 0 to infinity but the numeric value will remain the same which is 8 in this case.

Numberic value/Digital Sum = Sum of all the digits of a number .

Say

If k = 0 we get 8 (Digital Sum is 8)

If k = 2 we get 800 (Digital Sum 8)

If k = 4 we get 80000 (Digital Sum 8)


Thus this part has the same digital sum which is 8.


Hence we can't get anything from this statement.

From statement 2

We get j = 1

Now the digital sum of 8 * 10^k + j will come as 9 ; irrespective of any value of k ( As we have seen).

Thus statement 2 helps us to find the answer.
Abhishek
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by ishansottani » Tue Nov 16, 2010 3:50 am
Brent Hanneson wrote:Source: Beat The GMAT Practice Questions

If j and k are positive integers, what is the remainder when 8 * 10^k + j is divided by 9?
(1) k = 13
(2) j = 1
in the beginning i thought u meant to say 8*10^(k+j) but once i got that cleared the problem is pretty simple as mentioned in the explanations mentioned above IMO B
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by jd792104 » Mon Nov 22, 2010 10:48 am
Brent

What level DS question would this be - I am thinking that this will be in the medium / medium easy range (600 - 650? or lower?)

What do you think?

thanks
Jd
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by anirudhbhalotia » Thu Dec 02, 2010 10:36 am
Fantastic questions..totally squeezed my tiny brains by trying to apply the concepts! But....

...why the options are not mentioned....A, B, C, D, E?
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by Shubhu@MBA » Sat Jan 08, 2011 9:28 pm
The answer will be B.

This can be solved by basic division rule.

8*10^k , if we add the digits of the number formed by this than we will get 8(because the number will be in the form of
8 for k=0
80 for k=1
800 for k=2
8000 for k=3

Now if we add 1 to this we will get the sum of digits of that number as 9. The number thus formed will be like
9 for k=0 & j=1
81 for k=1 & j=1
801 for k=2 & j=1
8001 for k=3 & j=1

Any number is divisible by 9 if the sum of it's digit is directly divisible by 9.

So we see the value of k doesn't matter while deciding the divisibility of this number.

Hence only B is required.

Hope this explains the question well.

A fairly simple one though(compared to the previous question I use to get in my emails)
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