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 Post subject: CLSB Group 2 - Help
PostPosted: Sat Nov 02, 2013 5:22 pm 
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Joined: Sun Mar 11, 2012 12:04 pm
Posts: 25
Question 21

A positive whole number less than 100 has remainder 2 when it is divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5. What is the number?

The answer is 59. Please show ALL working out. Thank you.


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Sat Nov 02, 2013 5:35 pm 
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Question 21

A positive whole number less than 100 has remainder 2 when it is divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5. What is the number?

The last bit is the most restrictive so start with that.

remainder 4 when divided by 5: possibilities are 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 63, 69, 74, 79, 84, 89, 93, 99 (all one less than the multiples of 5)

remainder 3 when divided by 4: test the above or list 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99 (all one less than multiples of 4)

so how many in common so far ....19, 39, 59, 79, 99 (one less than multiple of 20)

remainder 2 when divided by 3 not 19, 39, 79 or 99 so 59 is the answer ... or you can think of it as looking for one less than multiples of 60.


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Sat Nov 02, 2013 5:50 pm 
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Guest55

Thank you for the rapid response. NLM


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Sat Nov 02, 2013 6:52 pm 
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Joined: Sat Dec 03, 2011 3:14 pm
Posts: 625
A really simple and quick way to do this is as follows -

work out the LCM of all four numbers. The LCM of 2, 3, 4, 5 is 60. As the remainder for 5 is 4 which is 1 less than its multiple, simply take away 1 from 60. All other numbers will give you the required remainders as well. The next numbers in this series will be 119, 179, etc (multiples of 60 - 1)

You can try that question with any series of numbers. For eg. - substitute the numbers as follows - 6 r 5, 7 r 6, 8 r 7, 9 r 8.
Their LCM is 504.
504 - 1 gives you 503.
503 / 6 = 83 r. 5;
503 / 7 = 71 r. 6;
503 / 8 = 62 r. 7;
503 / 9 = 55 r. 8

So your answer for this question will be 503


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Thu Nov 07, 2013 11:19 pm 
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Leanmeamum Thanks so much for this sorry to be dim but why do you also need to see what the LCM of 2 is ? Also how on earth are dc supposed to know these tricks of the trade. Everytime I think we are getting somewhere with maths another of these things come along. GS stage 2 Exam is 8 days time now!


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Fri Nov 08, 2013 8:35 am 
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Location: Warwickshire
The original question was "A positive whole number less than 100 has remainder 2 when it is divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5" So one less than 3 x 4 x 5 - you can ignore 2 which wasn't mentioned, and in any case is a factor of 4 so anything which has 4 as a factor will have 2 as well.


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Fri Nov 08, 2013 8:40 am 
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iamfrazzled wrote:
Leanmeamum Thanks so much for this sorry to be dim but why do you also need to see what the LCM of 2 is ? Also how on earth are dc supposed to know these tricks of the trade. Everytime I think we are getting somewhere with maths another of these things come along. GS stage 2 Exam is 8 days time now!



Sorry my mistake - I added 2 to the question.

Maths problem solving is quite tricky but the best way to overcome this is to practise as much as possible.

LCM or the Lowest Common Multiple is very easy to work out. The LCM means the smallest number which can be divided by all the numbers in a given list of numbers without leaving a remainder.

LCM can be calculated in 2 ways -

1) Foe eg. we need to find the LCM of 2, 6, 8 The child can list all the multiples of all the numbers

2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26 .....................
6 = 6, 12, 18, 24, 30, 36, .............
8 = 8, 16, 24, 32, .............

We can see that 24 is the first number which is common to all 3. There the LCM for 2, 6 and 8 is 24

2) The second method is as follows - This method is difficult to show without symbols but I'll try to explain it

we write all 3 numbers next to each other and start dividing them together. All numbers are divided by the same number and if they can't be divided they are divided by their own factors. We need to use prime numbers to divide them as bigger numbers can give a wrong answer

2, 6, 8 divided by 2 = 1, 3, 4 (I'm taking these numbers and dividing them again as you can see in the next line)
1, 3, 4 divided by 3 = 1, 1, 4
1, 1, 4 divided by 2 = 1, 1, 2
1, 1, 2 divided by 2 = 1, 1, 1

We can't divide them any further as all numbers have been reduced to 1

Then we take the numbers we used to divide the group and multiply them. So we take 2 X 3 X 2 X 2 = 24 to give us the LCM for 2, 6 and 8


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Fri Nov 08, 2013 12:50 pm 
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I suppose the question is, how did you know to start to solve the problem by first finding the LCM of 3, 4 and 5?


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Fri Nov 08, 2013 1:31 pm 
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mystery wrote:
I suppose the question is, how did you know to start to solve the problem by first finding the LCM of 3, 4 and 5?


The way I would explain it to a child would be this:

Because the number has remainder 2 when it is divided by 3 then it must be 1 less than an exact multiple of 3.

Because the number has remainder 3 when it is divided by 4 then it must also be 1 less than an exact multiple of 4.

Finally because the number has remainder 4 when it is divided by 5 then it must also be 1 less than an exact multiple of 5.

So overall the number must be 1 less than an exact multiple of 3, 4 and 5.

The most straight-forward multiple of 3, 4 and 5 is
3 x 4 x 5 = 60

1 less than 60 is 59

59 is a whole number less than 100 so must be the right answer. :)


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 Post subject: Re: CLSB Group 2 HELP
PostPosted: Fri Nov 08, 2013 1:32 pm 
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Now that's a nice explanation which makes no leaps in the dark.


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