The box (Posted on 2009-01-02)

	Submitted by pcbouhid
Rating: 5.0000 (1 votes)
Solution:	(Hide)
(a+2)(b+2)(c+2) = 2abc c(a+2)(b+2) + 2(a+2)(b+2) = 2abc c[2ab - (a+2)(b+2)] = 2(a+2)(b+2) c = 2(a+2)(b+2)/[2ab-(a+2)(b+2)] c = 2(a+2)(b+2)/(ab-2a-2b-4) c = (2ab + 4a + 4b + 8)/(ab-2a-2b-4) c = [(2ab-4a-4b-8) + (8a+8b+16)]/(ab-2a-2b-4) c = 2 + (8a+8b+16)/(ab-2a-2b-4). (c-2)/8 = (a+b+2)/(ab-2a-2b-4). Making c'=c-2, b'=b-2, a'=a-2, we have: c'/8 = (a'+b'+6)/(a'b'-8). We cannot have a or b = 1 or 2, because (1+2)/1 and (2+2)/2 are already ≥ 2 (so multiplying by (c+2)/c gives > 2). Then a, b ≥3, so a', b' ≥ 1, (a'+b') ≤ (a'b'+1) and (a´+ b´+ 6) ≤ (a´b´+7). Hence c'/8 ≤ (a'b'+7)/(a'b'-8) = 1 + 15/(a'b'-8) ≤ 1 + 15. Hence c´ ≤ 128, and finally c ≤ 130.

	Subject	Author	Date
	spoiler--all solutions	xdog	2009-01-05 00:45:29
	re: Simple stupid way?	Charlie	2009-01-03 12:12:13
	Simple stupid way?	elementofsurprize	2009-01-03 08:24:22
	somewhat analytical solution	Daniel	2009-01-02 21:41:48
	computer exploration (spoiler)	Charlie	2009-01-02 18:33:28
	re: Solution?	Sing4TheDay	2009-01-02 16:52:11
	Solution?	Sing4TheDay	2009-01-02 16:02:23