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 Post subject: Help! Divisible questionsPosted: Thu Jan 13, 2011 10:23 pm

Joined: Sun Nov 21, 2010 12:56 am
Posts: 14
How do you do questions like this:

You have numbers 4, 3, 6, 8, 5 on some cards. What is the smallest number you can make which is divisible by 4?

What method is there for these type of questions?

Thanks for any advice, which I'd be grateful for tonight!!!!

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 Post subject: Re: Help! Divisible questionsPosted: Thu Jan 13, 2011 10:27 pm

Joined: Sun Mar 21, 2010 5:06 pm
Posts: 267
Look up HCF and LCM in google

HCF = Highest Common Factor
LCM = Lowest Common Multiple

Eg:
http://www.cimt.plymouth.ac.uk/projects ... k8_2i4.htm

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 8:47 am

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6966
Location: East Kent
is the question from a multiple choice paper? If so use answers to help you. the last digit cannot be 3 or 5 because to be divisible by 4 number must be even, so you can discard those answers straight away.

if standard then quickly write out all possibilities (without 3 or 5 as last digit)

eg 83,456
45, 368 etc to be divisible by 4 you can halve it , then halve again to give a whole number..

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 9:29 am

Joined: Wed Oct 13, 2010 6:48 pm
Posts: 115
Not sure if this is correct method, but using LCM and a bit of logic-
1. Cannot end in 3 or 5 has to be 4 , 6, 8
2. So can have these poss. ... ... ... ... 8
... ... ... ... 6
... ... ... ... 4
3. Now need to decide on 1st no. To be smallest need to start with 3, then use up the other nos. the smallest first, one after the other to build up smallest no. divisible by 4.
i.e. 34568.
4. NB the 8 has to be at the end otherwise putting it anywhere else will make no. bigger. Does that make sense?

Maybe someone has easier way!

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 10:59 am

Joined: Sun Nov 21, 2010 12:56 am
Posts: 14
This is the way I was doing it but sometimes I found it took quite long..Thank You to your replies!

It's my DD's last exam today!! Such a relief for her and me!!

Well done to all children and parents for getting through this stressful time!!!!

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 6:47 pm

Joined: Tue Mar 02, 2010 8:13 pm
Posts: 118
There are some rules abouthow to check if a number is divisible by another number.
If the number is even it is divisible by 2.

If the sum of the numbers adds up to a number that is divisible by 3 then it is divisible by 3, e.g. the sum of the numbers 213609 adds up to 21 which is 3 x 7 so it is divisible by 3.

If the last 2 digits are divisible by 4 then the whole number is divisible by 4.

The smallest number you can make is 34568. The last 2 digits are 68 which is divisible by 4. If it wasn't you'd have to try the next smallest number and so on.

You can easily google a list of divisibility rules.

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 7:55 pm

Joined: Sun Nov 21, 2010 12:56 am
Posts: 14
Thank you Chilled for this very useful information..wish I wrote sooner on this forum.

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 9:15 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3813
Location: Chelmsford and pleased
Other useful tips.

If the digits add to 9 then it is divisible by 9. If it is even and the digits add to a number divisible by three then it is in the 6 times table. Divisible by 8 use the last three digits - if they divide by 8 then the whole divides by 8 (as 1000 is divisible by 8 ).

Divisible by 11 is a case of adding every other digit and then adding the digits that you didn't add the first time and subtracting the second total from the first. If the answer is 0 or divisible by 11 then it is in the 11 times table e.g.

112233

1 + 2 + 3 = 6
6 - (1 + 2 + 3) = 0

or

847

8 + 7 = 15
15 - 4 = 11

I don't know of an easy trick for 7.

Multiplying large numbers by 11. E.g. 21 times 11, write 2_1, then add the 2 and 1 and write the answer in the middle
21 x 11 = 231
17 x 11 = 187
This is only difficult if the digits add to something beyond 9, e.g. 77 x 11 = 7_7 (7+7 = 14)
Keep the 4 and carry the 1 to hundreds column (7+1)47 to give 847.

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 Post subject: Re: Help! Divisible questionsPosted: Fri Jan 14, 2011 11:14 pm

Joined: Sun Nov 21, 2010 12:56 am
Posts: 14
Thanks Moved!! Some more great tips. I'm feeling pretty silly for not asking for help earlier. I underestimated the power of this forum.

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 Post subject: Re: Help! Divisible questionsPosted: Sat Jan 15, 2011 2:46 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3813
Location: Chelmsford and pleased
11plusparent wrote:
Thanks Moved!! Some more great tips. I'm feeling pretty silly for not asking for help earlier. I underestimated the power of this forum.

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