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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?

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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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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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?

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: 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.

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

How do we know that when m is divided by 17 the quotient is different?

You'd get the same answer if you set them equal: 13q + 2 = 17q + 2 --> q = 0 --> remainder when dividing 0 by 17 is 0.

So, the quotients might or might not be the same but in any case you should consider more general case and denote them with different variables.
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