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PostPosted: Wed Jun 18, 2014 1:21 pm 
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Joined: Sat Dec 28, 2013 3:52 pm
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x divided by 2 leaves a remainder of 1.
x divided by 3 leaves a remainder of 2.
x divided by 4 leaves a remainder of 3.
x divided by 5 leaves a remainder of 4.
x divided by 6 leaves a remainder of 5.

Find the smallest value of x.

Is there a method to solve such problems or just trial and error, using (LCM -1) ?


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PostPosted: Wed Jun 18, 2014 2:02 pm 
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Yup, LCM of 2, 3, 4, 5 and 6, take away 1. Hopefully not too much trial and error involved there? 2 and 3 go into 6, so they're covered. LCM of 4 and 6 is 12. LCM of 12 and 5 is 60.


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PostPosted: Thu Jun 19, 2014 10:08 am 
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Writing out the values for 6 doesn't take long.
11, 17, 23, 29, 35, 41, 47, 53, 59

Divide by 5 and look at remainders.

R1, r2, r3, r4, r0, repeat

Divide by 4

R3, r1,r3, r1

Solution found very quickly


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PostPosted: Thu Jun 19, 2014 10:31 am 
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Thanks 'Aliportico' and 'moved'.

I was wondering if there is a generic formula, (LCM-n) where n is the remainder or some relation between divisor and remainder.


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PostPosted: Thu Jun 19, 2014 12:23 pm 
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Developing systematic methods and reasoning in maths is essential. Solving problems such as this require a logical approach and no more really. If we teach a bunch of formulae then we are just giving children more and more to remember. Maths is fundamentally an approach to problem solving.


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PostPosted: Sat Jun 21, 2014 7:27 pm 
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Agreed, that is what/how maths should be. However, in a time constrained exam where passing the exam is the only goal, temptation to find a short cut/ formula takes over the fundamentals.


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