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Sir Col
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(A)symmetrical Sums
« on: Dec 29th, 2004, 4:49pm » |
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Given that x,y,k [in] [bbz], find the number of solutions to each of the inequalities.
[easy]
(1) x+y [le] k, where x,y [ge] 0.
(2) |x|+|y| [le] k.
[harder]
(3) x+y [le] k, where 0 [le] x [le] y.
(4) |x|+|y| [le] k, where x [le] y.
Note: |x| is the absolute, or unsigned, value of x.
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John_Gaughan
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Re: (A)symmetrical Sums
« Reply #1 on: Dec 29th, 2004, 6:38pm » |
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1. n = k + 1
[forall] x, 0 <= x <= k, [exists] y [therefore] x + y = k
There are k + 1 such x's.
2. n = 4(k + 1)
Same as 1, except that for each value of x there are four possibilities: positive and negative x and y.
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towr
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Re: (A)symmetrical Sums
« Reply #2 on: Dec 30th, 2004, 1:23am » |
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close but no cigar
'[le]' [ne] '=' 
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| « Last Edit: Dec 30th, 2004, 1:24am by towr » |
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John_Gaughan
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Re: (A)symmetrical Sums
« Reply #3 on: Dec 30th, 2004, 6:15am » |
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Ah crap, you're right. I guess this should work then:
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1.
n = [sum]a=0ka+1
n = ((0+1) + (1+1) + (2+1) + ... + (k+1))
n = ((0+1+2+3+...+k) + (1+1+1+1+...+1))
n = k(k+1)/2 + (k+1)
2.
n = 4(k(k+1)/2 + (k+1))
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towr
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Re: (A)symmetrical Sums
« Reply #4 on: Dec 30th, 2004, 7:23am » |
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I'm sceptical about the second. You shouldn't count the points for (0,y) and (x,0) 4 times, nor (0,0) even twice.
You can easily draw it for small k in the integer-plane. The first is a triangle, the second two overlaid squares.
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1) 1/2 (k+1)(k+2)
2) (k+1)2 + k2
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| « Last Edit: Dec 30th, 2004, 7:43am by towr » |
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towr
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Re: (A)symmetrical Sums
« Reply #5 on: Dec 30th, 2004, 7:42am » |
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3)
(k+2)2/4 for even(k)
(k+1)(k+3)/4 for odd(k)
4)
[(k+2)2 + k2]/2 - 1 for even(k)
(k+1)2 - 1 for odd(k)
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| « Last Edit: Dec 30th, 2004, 7:51am by towr » |
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Sir Col
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Re: (A)symmetrical Sums
« Reply #6 on: Dec 30th, 2004, 9:09am » |
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You're good, towr! I liked your explanation about the two overlapping squares for (2).
In fact, it is possible to obtain a compact single function, in terms of k, for each of (3) and (4), by making use of the integer part function.
I'm intrigued to know your approach for (3) and (4). When I was playing with these puzzle ideas it was (4) that I started with and after taking an algebraic approach I made the geometrical connection with the graph |x|+|y|[le]k (shaded square region). For efficiency in approach I used a geometrical approach with them all. For (4) I had two overlapping rectangular regions: only counting values below or on the line y=x.
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towr
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Re: (A)symmetrical Sums
« Reply #7 on: Dec 30th, 2004, 10:34am » |
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For 3) and 4) I did the same as for 1) and 2), I drew the case k=4 and k=5
It's easy to generalize from there.
for the odd case of 4) it took me a while to recognize it was twice two overlaid squares (with sides (k+1)/2), minus of course the origin (which would otherwise be counted twice)
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