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Number prop

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vksunder
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Number prop [#permalink] New post   19 Jan 2009, 17:16
Please explain



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Re: Number prop [#permalink] New post   19 Jan 2009, 20:53
vksunder wrote:
Please explain


1. m=13q +2
2. m=17k+2

so 13q=17k, q must be k
q/17=k/13

Test k=4, only 9 is the remainder of q in divison to 17.

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Re: Number prop [#permalink] New post   19 Jan 2009, 21:04
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vksunder wrote:
Please explain


m=13q +2
m=17p+2

13 q = 17 p = integer.

from the above.. q must be mulitple of 17 and p must be multiple of 13

p=k*13


q/17 = p/13 = k*13/13 = k (integer)
remainder zero.

A
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graduateprofessor
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Re: Number prop [#permalink] New post   19 Jan 2009, 21:45
m = 13 q + 2
m = 17 k + 2

which implies 13 q = 17 k

13 q = 17 k = 221 and its multiples

if 13q = 17 k = 221 then q = 17 and 17/17 the remainder is 0
if 13 q = 17 k = 442 then q = 34 and 34/17 the remainder is 0
and so on

answer is A.

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selvae
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Re: Number prop [#permalink] New post   20 Jan 2009, 02:34
m/13 remainder is 2
m/17 remainder is 2
means ==> m should be lcm of {13,17} + 2
so Q should be 13*17
which means q/17 remainder = 0

A
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Re: Number prop [#permalink] New post   20 Jan 2009, 09:59
Since 13q=17p and p, q are integers; therefore, we can say that q is multiple of 17 and p is a multiple of 13. Hence, if the question were to ask for the remainder when p is divided by 13, we will get zero as well.
x2suresh wrote:
vksunder wrote:
Please explain


m=13q +2
m=17p+2

13 q = 17 p = integer.

from the above.. q must be mulitple of 17 and p must be multiple of 13

p=k*13


q/17 = p/13 = k*13/13 = k (integer)
remainder zero.

A
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milind1979
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Re: Number prop [#permalink] New post   20 Jan 2009, 16:34
This might prove a easier way to do it by plugging number.

We need a number 'm' which satisfies our statement. Easiest way to get the number is multiply 13 * 17 and add 2 to it.

which comes out as (13*17)+2 = 223.

Now when you divide 223 with 13 i.e 223/13 we get 17.

Now it easy to find the remainder as 0.

So my answer is also A.

Hope this was easier. :wink:



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Re: Number prop [#permalink]
20 Jan 2009, 16:34
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