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When integer m is divided by 13, the quotient is q and the r [#permalink]
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06 Apr 2009, 13:08
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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
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Last edited by Bunuel on 29 Jan 2013, 04:16, edited 1 time in total.
Renamed the topic and added OA.



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Re: Remainder [#permalink]
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06 Apr 2009, 13:29
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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. pradeep
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Re: Remainder [#permalink]
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06 Apr 2009, 16:40
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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]
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06 Apr 2009, 23:59
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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]
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07 Apr 2009, 05:20
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]
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05 May 2009, 00:22
m13q=2 m17s=2
13q=17s
r=0
A



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Re: Remainder [#permalink]
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06 May 2009, 00:45
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]
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10 May 2010, 23:18
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 (m2) from eqn 2. Therefore, q = (m2)/3
Now substituting the values in eqn 1: => m = 13(m2)/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]
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01 Jun 2011, 08:32
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]
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13 Jun 2011, 01:17
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13q = 17p lcm = 13 * 17. thus q = 17. hence remainder = 0
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Re: Remainder [#permalink]
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13 Jun 2011, 10:26
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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]
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29 Jan 2013, 03:08
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



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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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11 Aug 2015, 12:18
Hi stewart,
Can this be the alternative approach?
m=13q+2 2=13(0)+2 15=13(1)+2 28=13(2)+2
and
m=17p+2 2=17(0)+2 19=17(1)+2 36=17(2)+2
Thus we know p=q and that's 0, so 0/17 = 0.



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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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11 Aug 2015, 13:22
jaspreets wrote: Hi stewart,
Can this be the alternative approach?
m=13q+2 2=13(0)+2 15=13(1)+2 28=13(2)+2
and
m=17p+2 2=17(0)+2 19=17(1)+2 36=17(2)+2
Thus we know p=q and that's 0, so 0/17 = 0. Yes, you are correct in your approach. This question can be solved either by algebra as shown by Ian above or by plugging a few values as you have done. The trick here is to realise that you are finding a number that gives a remainder of 2 with both 13 and 17.
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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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11 Aug 2015, 14:49
Thank You Engr2012 for the rapid response. Appreciated



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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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29 Jan 2017, 01:25
seofah 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 Here's my take on this We are given \(\frac{m}{13}\) = q (rem 2) and \(\frac{m}{17}\) = _ r 2 One possible value for m that satisfies both the conditions is \(m = 2\). When \(m = 2\), \(q\) will be \(0 (\frac{2}{13}\) gives quotient \(q= 0\) and rem r= 2), which follows \(\frac{0}{17} = 0\)
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Re: When integer m is divided by 13, the quotient is q and the r [#permalink]
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29 Jan 2017, 09:46
seofah 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 m=2+(13*17)=223 q=223/13=17 17/17 gives remainder of 0 A




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