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 Post subject: Help please :)
PostPosted: Thu Aug 30, 2012 10:21 am 
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Joined: Tue May 29, 2012 8:16 pm
Posts: 57
Hi everyone, you helped me before with a math sum so was hoping if you could help again.

There are 4 cards with the numbers, 7, 5, 2, 3.

Using each card only once, make the smallest possible multiple of 6.

I think I am missing something obvious but I'm having a thick day :)


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 10:31 am 
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Joined: Tue Jul 21, 2009 9:56 pm
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7 5 2 3

Well you can't make 0x6 so far as I can see, so let's try for 6 ....... how about 7 - (2x3 - 5) = 7 + 5 - 2x3 ........ just saw that by looking as I saw that 2x3 = 6, the difference between 5 and 6 is one, and I needed to get the 7 down by one to get to 6 if you see what I mean.

Is this a bit like that game on TV - Countdown?


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 10:42 am 
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Joined: Mon Aug 22, 2011 8:20 pm
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Location: Warwickshire
Yes I think they must be asking for operators, because it obviously isn't just putting the numbers into sequence.

7+5+2+3 = 17.

17 isn't divisible by 3, so no combination of these numbers (regardless of order) can possibly make a number divisible by 6 since all numbers with 6 as a factor will have 3 as a factor too.


Last edited by Okanagan on Thu Aug 30, 2012 11:09 am, edited 1 time in total.

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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 10:52 am 
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Joined: Tue Oct 27, 2009 8:19 pm
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18 is my guess.

7+5 = 12 / 2 = 6 x 3 = 18


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 11:20 am 
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Joined: Tue May 29, 2012 8:16 pm
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Oh I see - I thought it was asking for each card to be used once. There were 2 other questions using the same cards - 1) make the smallest possible even number and 2) make a number as close as possible to 4000. So I assumed it was using all 4 but couldn't see how??


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 11:24 am 
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Joined: Tue May 29, 2012 8:16 pm
Posts: 57
Okanagan wrote:
Yes I think they must be asking for operators, because it obviously isn't just putting the numbers into sequence.

7+5+2+3 = 17.

17 isn't divisible by 3, so no combination of these numbers (regardless of order) can possibly make a number divisible by 6 since all numbers with 6 as a factor will have 3 as a factor too.


Yes, this is where I got stumped - because the sum of numbers was 17 so impossible. Just a badly worded question then :)


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 11:26 am 
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My answer uses all 4 once.

I make the lowest even number 2

7-5 = 2

3 - 2 (from above) =1

1 (from above ) x 2 = 2

Too much time spent watching countdown I am afraid.


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 11:52 am 
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Joined: Thu May 10, 2012 8:30 am
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mystery wrote:
7 5 2 3

Well you can't make 0x6 so far as I can see


(5 x 2) - (7+ 3) = 0.


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 12:48 pm 
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Now a debate as to whether 0 is an even number or divisible by 6 :lol: :lol: :lol:

Is there are lower even number than 2 or

a lower multiple of 6 than 18?

Really curious now as just love these questions.

Working on the 4000 now whilst sewing on name tapes!


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 Post subject: Re: Help please :)
PostPosted: Thu Aug 30, 2012 1:05 pm 
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Joined: Thu Jun 14, 2012 9:28 am
Posts: 121
smallest possible even number 3 - (7-5)/2

smallest multiple of 6 - 7 - (5-3)/2


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