Yes, good work. I almost failed an arithmetic exam at school. Did go on to study applied maths at university for two years though! Had a wonderful professor called R B Dingle, much missed.
Though we should get going again with an incredibly simple one. (the maths symbols are meant to be a Plus Sign (+); a Minus sign (-); and a Multiplication sign (x).
Apparently I can’t just write E. It isn’t a complete sentence.
Yes ! Well done Eoink !
If you had just written “E” I wouldn’t have known whether you were quoting the correct answer, or squawking in despair “eeeeh !” at a seemingly impossible brain teaser.
Apologies for that one.
The next one might be a trifle more challenging (well, it was for me !) but i’ll post it later this evening when the guys “Down Under” have a chance to be up and running tomorrow morning.
Cheers
Don
Actually as my first name is Eoin I often sign myself as E on emails, so it would have been definitely ambiguous to people who know that.
12 letter word
You need to figure out the starting point, then move from adjoining square to adjoining square, (horizontally or vertically but not diagonally), to spell a 12-letter word, using each letter once and only once.
Of course, you need to fill in the missing letters as well.
Do you want the short answer to the 12 letter word?
Hi Steve,
I’ll keep my answer brief …
You got it
Well done !
I’m lost TBH.
The penny has dropped. Now I can go enjoy some extended music, rather than having a brief session!
… but only for a very short period of time …
And I like the way you condensed the words To Be Honest
Well done !
Ah, obvious when you see’t.
Three across and one down
Arrange the 16 letters so as to form three words across, and one word down the middle of the 16 circles.
It won’t let me post without a sentence first, and doesn’t keep my formatting that aligned neatly!:
W
L I D
V A L I D
Q U A L I T Y
The word down middle is “will”
Nicely done IB.
No worries about the presentation of your answer, you got the correct three words across and the right word down.
Cheers
Don
Seating Arrangements
At the engineering and science conference on cable direction in hifi systems the gentlemen sat at desks numbered 1-5 whilst the ladies sat opposite them at desks numbered 6-10.
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The lady sitting next to the lady opposite No. 1 is Freda
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Freda is three desks away from Geraldine
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Henrietta is opposite Chris
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Eric is opposite the lady next to Henrietta
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If Chris is not central then Adam is.
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Dominic is next to Bruce
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Bruce is three desks away from Chris
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If Freda is not central the Irene is
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Henrietta is three desks away from Josephine
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Dominic is opposite Geraldine
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The lady sitting next to the lady opposite Adam is Josephine
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Chris is not at desk No. 5
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Josephine is not at desk No. 10
Who is sitting at which desk ? ie what is the seating arrangement ?
1 Chris
2 Eric
3 Adam
4 Bruce
5 Dominic
6 Henrietta
7 Freda
8 Irene
9 Josephine
10 Geraldine
Assuming 1 faces 6, 2 faces 7 etc.
Nicely done Eoink.
Yes, the people sit at the desks as per the picture and as you described.
Well done.
The Baron’s Treasure
Baron Von Richenstein kept his gold treasure in a treasure house. In each room of the treasure house were as many chests as there were rooms. In each chest there were as many gold coins as there were chests in that room. All the gold coins were of the same size and value.
When he died, the Baron’s will was that his favourite Barber should receive one chest of gold coins. The remaining coins were to be divided equally between the Baron’s three sons.
The three sons were proud and fierce men who would definitely resort to bloodshed if the coins could not be divided equally.
The question is simply: -
a) Was there blood shed.
b) Was there no blood shed.
c) Is there no way of knowing whether there was blood shed or not.
PS. The proof is the real requirement, not simply a one in three guess.
B I think. If there are n rooms, each room has n chests, thus there are n * n chests. Each room has n chests, thus in each room there are n * n coins, with there being n rooms we have a total of n * n *n coins.
The barber gets one chest, which holds n coins. So the three brothers have n^3-n or n(n^2-1) or n(n-1)(n+1) coins to share. The next bit of logic is easier to see than explain. If n or n-1 or n+1 is a multiple of 3, then there is no bloodshed because the number of coins is the product of those 3 and can therefore be evenly divided by one of them. Quick rough induction proof that one of n-1, n, n+1 is divisible by 3. Starting with n=1, then we have 0,1,2. N=2, 1,2,3 N=3, 2,3,4. Each subsequent number is one of these plus 3x where x is an integer, so in each case there must be a multiple of 3 in the factors.