Help me solve this question please

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berks_mum
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Joined: Sat Dec 28, 2013 4:52 pm

Help me solve this question please

Post by berks_mum »

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) ?
aliportico
Posts: 888
Joined: Wed Aug 01, 2007 12:19 pm

Re: Help me solve this question please

Post by aliportico »

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.
moved
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Re: Help me solve this question please

Post by moved »

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
berks_mum
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Joined: Sat Dec 28, 2013 4:52 pm

Re: Help me solve this question please

Post by berks_mum »

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.
moved
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Location: Chelmsford and pleased

Re: Help me solve this question please

Post by moved »

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.
berks_mum
Posts: 939
Joined: Sat Dec 28, 2013 4:52 pm

Re: Help me solve this question please

Post by berks_mum »

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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