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When integer m is divided by 13, the quotient is q and the r

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When integer m is divided by 13, the quotient is q and the r [#permalink] New post 06 Apr 2009, 12:08
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When integer m is divided by 13, the quotient is q and the remainder is 2. When m is divided by 17, the remainder is also 2. What is the remainder when q is divided by 17?

A. 0
B. 2
C. 4
D. 9
E. 13
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Jan 2013, 03:16, edited 1 time in total.
Renamed the topic and added OA.
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Re: Remainder [#permalink] New post 06 Apr 2009, 12:29
So q is 17 because if m = 13q + 2 and m is also 17(z)+2

So when 17 is divided by 17 the remainder is 0.

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Re: Remainder [#permalink] New post 06 Apr 2009, 15:40
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botirvoy wrote:
When integer m is divided by 13, the quotient is q and the remainder is 2. When m is divided by 17, the remainder is also 2. What is the remainder when q is divided by 17?

A. 0
B. 2
C. 4
D. 9
E. 13

Detailed explanations please.


From the definition of quotients and remainders, we have:

m = 13q + 2
m = 17a + 2

(note that the quotient is different in the second case). So we have

13q + 2 = 17a + 2
13q = 17a

and since this equation involves only integers, the primes that divide the right side must divide the left, and vice versa. That is, q must be divisible by 17, and a must be divisible by 13. If q is divisible by 17, the remainder is zero when you divide q by 17.

Of course, if you can see that q = 17 is one possible value for q here, you can use that to get the answer of zero quickly as well.
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Re: Remainder [#permalink] New post 06 Apr 2009, 22:59
Got 0 as well but am I right in thinking that 0 is another possible value of q?
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Re: Remainder [#permalink] New post 07 Apr 2009, 04:20
shkusira wrote:
Got 0 as well but am I right in thinking that 0 is another possible value of q?


Yes, perfectly correct - and that makes the question quite easy!
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Re: Remainder [#permalink] New post 04 May 2009, 23:22
m-13q=2
m-17s=2

13q=17s

r=0

A
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Re: Remainder [#permalink] New post 05 May 2009, 23:45
IanStewart wrote:
shkusira wrote:
Got 0 as well but am I right in thinking that 0 is another possible value of q?


Yes, perfectly correct - and that makes the question quite easy!

Thanks. Just to add, if q=0, it is divisible by 17.. and hence would not leave any remainder.
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Re: Remainder [#permalink] New post 10 May 2010, 22:18
We can solve this using a simple equation and deriving the values:-
m = 13q + 2 - 1
m = 17p + 2 - 2

=> 13q + = 17p + 2
=> 13q = 17p
=> q = 17p/3
Here, 17p is equal to (m-2) from eqn 2.
Therefore, q = (m-2)/3

Now substituting the values in eqn 1:-
=> m = 13(m-2)/3 + 2
From here, m = 2.

Now using M, the value of Q can be derived from eqn 1.
=> 2 = 13q + 2
=> q = 0

Now if we divide Q by any number henceforth, the remainder would always be 0.
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Re: Remainder [#permalink] New post 01 Jun 2011, 07:32
So, I understand how to get to here

m=(13q)+2
m=(17x)+2

13q=17x

But, once I reduce the equation to 13q=17x, I am unable to make any deductions...can someone provide a clear explanation on how to use algebra to derive the values when we still have variables?
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Re: Remainder [#permalink] New post 13 Jun 2011, 00:17
13q = 17p
lcm = 13 * 17.

thus q = 17.

hence remainder = 0
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Re: Remainder [#permalink] New post 13 Jun 2011, 09:26
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amit2k9 wrote:
13q = 17p
lcm = 13 * 17.

thus q = 17.

hence remainder = 0


Careful here; if 13q = 17p, all you can say is that q is a multiple of 17, and that p is a multiple of 13. There is no way to find the actual value of q or p, and you certainly cannot be sure that q=17. It could be that q=34 and p=26, for example.

In general, if you see an equation like 13q = 17p, and if q and p are integers, then 13q and 17p are *the same number*. So they must have the same divisors. Since 17 is a divisor of 17p, it must be a divisor of 13q, so q must be divisible by 17.

Alternatively you can rewrite the equation as p = 13q/17, and since p is an integer, 13q/17 must be an integer, from which again we have that 13q is divisible by 17, so q is divisible by 17.
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Re: Remainder [#permalink] New post 29 Jan 2013, 02:08
Can't I just say q=0, so
(1) m=13q+2 => m=13*0+2 <=> m=2
(2) m=17k+2 => 2=17k+2 <=> k=0
--> 0 divided by 17 will obviously result in a reminder of 0
Re: Remainder   [#permalink] 29 Jan 2013, 02:08
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