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        riddles >> easy >> 1-10 and Aa to Ee
        
(Message started by: Noke Lieu on Dec 2^nd, 2007, 5:47pm)

Title: 1-10 and Aa to Ee
Post by Noke Lieu on Dec 2^nd, 2007, 5:47pm

another one of these little things. A bit rusty, slightly ill... apologies if it's not up to muster.

As a recap: two of the numbers 1-10 are allocated a letter, either a or A.
(a<A... n<N you get the picture)

Whilst all the totals (a+A etc) are 5n, only one is 3n
c and C are prime
b is factor of B
a is even
c^b = D
what are they?

Title: Re: 1-10 and Aa to Ee
Post by FiBsTeR on Dec 2^nd, 2007, 6:22pm

Fun puzzle! I find:

[hide]
a = 4
A = 6
b = 2
B = 8
c = 3
C = 7
d = 1
D = 9
e = 5
E = 10
[/hide]

Hint: [hide]Attack b and B first. Since b|B, 5|b+B, c^b=D, and B>b, what are the possible pairs for (b,B)? Many values lock into place from these observations, and I found only 2 cases to consider.[/hide]

I was going to put my solution in a table, but apparently they can't be hidden. :(

Title: Re: 1-10 and Aa to Ee
Post by Joe Fendel on Dec 4^th, 2007, 4:30pm

Cute. Actually, when I solved, I didn't even use two of the facts. That b | B, or that only one pair adds to a multiple of 3. I wonder which other combinations of information were unnecessary for the solution.

[hideb]c and C are primes which add to a multiple of 5. Our options are thus 2+3 or 3+7. Next we see that c^b = D. Clearly b is not 1, since then c = D. If c=2 and C=3 then those values are out for b, so b > 3 which makes D > 10. Thus c=3, C=7, and c^b = D can only be fulfilled with b=2 and D=9. Thus B = 8 is the only value which can be added to b to get a multiple of 5. Next use the fact that a is even, but it is not 2 or 8 (already used) or 10 (too big). If a=6, then the only number larger which adds to a multiple of 5 is A=9, but this can't be because D=9. Thus a=4, and so A=6 (because 1 is too small). We're then left with 1 as the only option for d which can add to 9 to get a multple of 5, and so d=1, and conclude that e=5 and E=10.[/hideb]

Title: Re: 1-10 and Aa to Ee
Post by Noke Lieu on Dec 5^th, 2007, 2:37pm

yup.
Minimum number of clues is a mystery for me.

I realised that the c^b=D was a hugely powerful clue, but like I say- a little rusty at puzzling. And about to beachward.