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An Euler Question

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mc3003
 Post subject: An Euler Question
 Post Posted: Fri Aug 19, 2011 5:03 am 
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I have a question here:

Find the lowest possible integer whose remainder is 10 when it is divided by 11, 12 when it is divided by 13, and 16 when it is divided by 17.


Last edited by mc3003 on Sat Aug 20, 2011 7:02 pm, edited 1 time in total.

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laiusergiu
 Post subject: Re: An Euler Question
 Post Posted: Fri Aug 19, 2011 3:31 pm 
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Let the lowest possible integer, which satisfies these conditions, be denoted by n:

Let's translate the statements into mathematical relations (where a,b,c are integers)
n = 11a + 10 = 11a+11-1 = 11(a+1) - 1
n = 13b + 12 = 13b+13-1 = 13(b+1) - 1
n = 17c + 16 = 17c+17-1 = 17(c+1) - 1

Since n = 11(a+1) - 1 --> n+1 = 11(a+1) i.e (n+1) is a multiple of 11
Since n = 13(b+1) - 1 --> n+1 = 13(b+1) i.e (n+1) is a multiple of 13
Since n = 17(c+1) - 1 --> n+1 = 17(a+1) i.e (n+1) is a multiple of 17

The smallest positive number that is a multiple of 11, 13, and 17 is 11*13*17=2431 --> n+1=2431 -->n=2430


Last edited by laiusergiu on Mon Aug 22, 2011 9:00 am, edited 1 time in total.

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mc3003
 Post subject: Re: An Euler Question
 Post Posted: Sat Aug 20, 2011 7:01 pm 
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Thanks honey ^^
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ranjithsmiles
 Post subject: Re: An Euler Question
 Post Posted: Sat Aug 20, 2011 9:59 pm 
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This is one of the LCM Model.
Any number ,which when divided by p,q or r leaving respective remainders of s,t and u
where (P-s)=(q-t)=(r-u)=V(say),
it will be of the form
k(LCM of p,q and r)-v,
the smallest such no will be obtained by substituting k=1

In this case ,
v=1
l.c.m( 11,13,17) = 2431
k=1
Ans: 2431-1 = 2430
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mithunsam
 Post subject: Re: An Euler Question
 Post Posted: Sun Aug 21, 2011 11:35 am 
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Posts: 76
If you look at the divisors and reminders...
11-10 = 1
13-12 = 1
17-16 = 1

That means, we need to subtract 1 from the resulting number.

Now the smallest possible multiple of 11, 13, & 17 (let us say N) is nothing but the multiplications of these 3 numbers (because, all these numbers are prime).

Therefore N = 11*13*17 = 2431

Now, look at the sentence, which I highlighted above.

To get our final answer, subtract 1 from 2431 => 2430.
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jnelson0612
 Post subject: Re: An Euler Question
 Post Posted: Sun Sep 11, 2011 11:20 pm 
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Great minds at work in this thread!

Of course, I need to ask what is the original source of this problem?

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Jamie Nelson
ManhattanGMAT Instructor
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