By O. A. Sarma
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15 moves: (15, 0, 0), (9, 0, 6), (9, 6, 0), (3, 6, 6), (3, 10, 2), (3, 10, 0), (3, 4, 6), (7, 0, 6), (7, 6, 0), (1, 6, 6), (1, 10, 2), (11, 0, 2), (11, 2, 0), (5, 2, 6), (5, 8, 0), (0, 8, 5). E6. The goal is to find the smallest integer values of Ò is a minimum and ¼ By numerical search we find ¿ ½ ¼¼¼ ¾´ · ½µ ½ ¼½ or ¼ and and so that ½ ¼¼ ¼¾ are the smallest values satisfying the above equations. Thus and Ò ¾½¼ transfers is the minimum. E7. One of the terms is ´Ü Üµ, so ½ ¼½. , ¼. E8. The number is 381,654,729.
The Lord GOD hath . . sheep that are even . . beside Eloth, on the . . every man his ËÏÇÊ ËÏÇÊÆ ËÀÇÊÆ, ËÀÇÊ ËÀ Ê , against . . by himself, . . which came . . of the Red . . and his . . ( Micah 4 : 3 ) ( Amos 6 : 8 ) ( Song of Solomon 4 : 2 ) ( 1 Kings 9 : 26 ) ( 1 Samuel 13 : 20 ) There are 13 other possible citations for ËÏÇÊÆ, and 1 Kings 4 : 29 could also be used for ËÀÇÊ , still avoiding forward steps. Puzzle #3. First observe that the word in Luke 17 : 27 must have at least one letter in common with both Æ Ã and ÇÎ Ê.
My argument is to show that large inputs have smaller outputs, thus reducing an infinite universe to a manageable finite one. At that point, an argument by cases, or a computer check of the finitely many cases, suffices. 999, then In the case of the 123 hole, we can argue as follows: If n f (n) n. In other words, the new number that counts the digits is smaller than the original number. ) Thus, starting at 1000 or above eventually pulls one down to under 1000. For n 1000, I’ve personally checked the iterates of f (n) for n = 1 to 999 by a computer program such as the one above.