Puzzle

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stimpsonslostson
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Puzzle

Post by stimpsonslostson »

i got sent this...

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:

1. The fifth number plus the third number equals fourteen.

2. The fourth number is one more than the second number.

3. The first number is one less than twice the second number.

4. The second number plus the third number equals ten.

5. The sum of all five numbers is 30.

What were the five numbers and in what order?

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BarcelonAl
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Re: Puzzle

Post by BarcelonAl »

So he's that mentally challenged that he can't remember 5 numbers, but can remember a cryptic set of clues? Pah! :evil:

I'm always rubbish at these things, might try to work it out later when I'm not swearing at Micro$oft for their browser being rubbish and incompatible with standards...
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Mike
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Re: Puzzle

Post by Mike »

Phil you are a git. I have been sitting here for my entire lunch hour trying to figure that one out. The really frustrating thing is I know I can do it but I have forgotten how to do equations properly!

PS my first solution seems so right but comes to 31 and does not meet the text above. How stupid do I feel right now? Very.
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Mike
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Re: Puzzle

Post by Mike »

Jo did it in 5 minutes. That is really frustrating!!!!
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stimpsonslostson
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Re: Puzzle

Post by stimpsonslostson »

I confess I got it in about that length of time too.
I've no idea how you got 31 tho. I got 18 22 26 34 and obviously 30 as the possible totals.
How should we share the solution?

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Mike
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Re: Puzzle

Post by Mike »

I wanted to have another go yet to see if I can crack it, now I have had the Jo tips. . . . how about we post solutions tomorrow morning - either scan your workings or type em out. :-D
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johnriley1uk
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Re: Puzzle

Post by johnriley1uk »

The obvious answer is that he phoned a colleaugue on his mobile and asked what the number was. He then gained immediate access to the building without causing any overheating of the brain... 8)
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Re: Puzzle

Post by Mike »

Well, shamed, this is how far I got which is really poor. I know it should be a set of equations that you arrange so that you only have one term for example x in the a + b + c + d + e = 30 final formula.

This is how far I got
1 st number is a
2 nd number is b
3 rd number is c
4 th number is d
5 th number is e

therefore
a = 2b-1
b + c = 10 or b = 10 - c
c + e = 14
d = b + 1
e = 14 - c

this is where i come unstuck. i know i need to get the common letter to replace all the terms in the a + b + c + d + e = 30 but cant get there!

edit
actually i have just realised that b can be replaced in all of the equations by 10-c therefore the equation should read like
a = 2(10-c) -1 or a = 19 - 2c
d = 10 - c + 1 or d = 11 - c
therefore
(19 - 2c) + (10 - c) + c + (11 - c) + (14 - c) = 30
54 - 4c = 30
4c = 24
c = 6

therefore
a = 2(4) -1 = 7
b = 10 - 6 = 4
c = 6
d = 4 + 1 = 5
e = 8

7 + 4 + 6 + 5 + 8 = 30

phew (thanks jo, i was forgetting about the final forumula) it was only when I posted the above i realised!
Last edited by Mike on Fri Apr 18, 2008 8:42 am, edited 2 times in total.
Reason: final solution realised whilst posting, doh!
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stimpsonslostson
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Re: Puzzle

Post by stimpsonslostson »

2 solutions! Written in blue to stop you reading accidentally- highlight text below to read!

Algebra... From the clues we know that
Clue1) C+E=14 Clue2)D=B-1 Clue3)2B-1=A Clue4)B+C=10 Clue5)A+B+C+D+E=30 so...
A+B+D=16 (Clue5-Clue1) so,
A+B+B+1=16 (above+Clue2) so A+2B=15, (add clue3 to give) 2B-1+2B=15 therefore B=4
Return to A+2B=15 gives A=7
B+C=10 (Clue4) B=4 so C=6
D-B=1 (Clue2) so D=5
C+E=14 C=6 so E=8
solution is: A=7 B=4 C=6 D=5 E=8

The other way was to write out the possible soultions for clue1;
Pairs are (C=5 E=9), (C=6 E=8), (C=7 E=7), (C=8 E=6), (C=9 E=5) then to work out the other numbers according to these rules- less mentally taxing, but more time consuming.

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Mike
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Re: Puzzle

Post by Mike »

lol, it is nice to know we got the same answer through different methods!
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mr_e
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Re: Puzzle

Post by mr_e »

I'm slightly ashamed to say that I "brute-forced" this, as I was eating food and didn't want to stop eating to pick up a pen or type. Once I realised it was actually a maths problem and not one of those weird logic problems ("He used the intercom and had his colleagues let him in!" Twat) I started off with the second number as one, then gradually raised it, as you can work out everything pretty easily in your head from that number.

Nice work with the equations though guys! If the numbers had been larger I would've been shafted, so equations are the way to go, really.
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stimpsonslostson
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Re: Puzzle

Post by stimpsonslostson »

John, good effort! I like the minimalist approach to problem solving- parsimony is what we science types look for. Sometimes just crunching thru the alternatives is the only way (although algebra is elegant) :-D
If I stumble across any more I'll post em.
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Mike
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Re: Puzzle

Post by Mike »

Here is another one for you.

Using the following numbers make 100.
1, 7, 7, 7, 7 - as well as the numbers you can use (, +, -, x, ÷.

For example (7+1) + (7x7) which in this case equals 112 but you get the picture.

The question is from www.mathsisfun.com
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mr_e
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Re: Puzzle

Post by mr_e »

God, that is hard. I eventually got it though (in about 20-30 minutes), but almost gave up in frustration. It's excrutiatingly simple once you realise what the only way to solve it is... I ended up working backwards from the actual answer.
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stimpsonslostson
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Re: Puzzle

Post by stimpsonslostson »

got it... took 15mins, gimme algebra any day! :roll:

(7+7) × (7+(1 ÷ 7)) = 100

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ps... thought I'd mention that my first attempt didn't really go to plan- I thought I'd got it, ((7x7)+1)+((7x7)+1)=100 then I realized that its using that I only had 5 digits. :oops: so I went back to it.
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