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

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!

m-13q=2
m-17s=2

13q=17s

r=0

A

Thanks. Just to add, if q=0, it is divisible by 17.. and hence would not leave any remainder.

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.

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?

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.

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

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.

Thank You Engr2012 for the rapid response. Appreciated

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 \(\frac{0}{17} = 0\)

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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