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Number Properties: Divisibility/Remainders

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by II » Thu Aug 07, 2008 1:53 pm
Hi,

I got the right answer on this one ... but am interested in finding out the various approaches to solving this.

One question I have is regarding statement (2).

I am ok with choosing any 7 numbers greater than 7 ... for example:
8, 9, 10, 11, 12, 13, 14
For example ...
If you divide 8 by 7 or 8/7 you get 1 remainder 1.
If you divide 9 by 7 or 9/7 you get 1 remainder 2. etc etc

If you choose the 7 consecutive numbers as:
1, 2, 3, 4, 5, 6, and 7.
how can you define 1/7 ... so 1 divided by 7 gives you what remainder ??? since you get a fraction.

Hope this makes sense. Thanks in advance for your help.
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Source: — Data Sufficiency |
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by parallel_chase » Thu Aug 07, 2008 6:40 pm
Statement I

The range of the remainders is 6. But we dont know anything about the numbers.

The numbers could be 7,14,21,28,35,42,48
The remainders will be 0,0,0,0,0,0,6, the range will be 6

Therefore, Statement is insufficient.

Statement II

The numbers are consecutive. Pick any set of 7 integers.

numbers 18,19,20,21,22,23,24
remainders 4,5,6,0,1,2,3

Just to confirm pick another set of numbers

numbers 1,2,3,4,5,6,7
remainders 1,2,3,4,5,6,0


Hence Sufficient

Therefore B is the answer.

Kindly note the below post for finding the remainders.

Hope this helps. Let me know if you still have any doubts
Last edited by parallel_chase on Thu Aug 07, 2008 8:15 pm, edited 1 time in total.
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by Ian Stewart » Thu Aug 07, 2008 8:07 pm
parallel_chase wrote: Here is technique for finding remainders when dividend is less than the divisor.

1 divided by 7 remainder = 3
10 divided by 7 remainder = 3

2 divided by 7 remainder = 6
20 divided by 7 remainder = 6

Similarly you can calculate remainder for any two integers.
I'm not sure what you're doing there, p_chase! When you divide 1 by 7, the remainder is 1, and the quotient is zero:

1 = 7*0 + 1

When you divide 2 by 7, the remainder is 2 and the quotient is zero:

2 = 7*0 + 2

Remember, when we divide n by d, we can always write:

n = qd + r

where 0 <= r < d. When we do this, r is the remainder, and q is the quotient. Whenever you divide positive integer n by positive integer d and n is less than d, the quotient will be zero, and the remainder will be n.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by parallel_chase » Thu Aug 07, 2008 8:19 pm
Ian just to be clear about remainders.

1 will always be the remainder when 1 is divided by 2,3,4,5,6,7,8,....n
2 will always be the remainder when 2 is divided by 3,4,5,6,7,8,....n
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by Ian Stewart » Fri Aug 08, 2008 4:04 am
parallel_chase wrote:Ian just to be clear about remainders.

1 will always be the remainder when 1 is divided by 2,3,4,5,6,7,8,....n
2 will always be the remainder when 2 is divided by 3,4,5,6,7,8,....n
Yes, that's exactly right.
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by pepeprepa » Fri Aug 08, 2008 4:27 am
For n,k positive, if n<k so the remainder of n/k is n

for negative fractions that is another story
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