No, it's wrong. If the order was 9 1 8 5 2 4 7 6 3 10, the order of the exit of the first 3 card is 1-2-7!
Not according to the algorithm clarification that I posted. If that's not the correct algorithm then correct it and/or clarify your original post. However, according to the process and order that I posted, here's what happens:
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the 9 (hidden) is moved to the bottom of the pile which now has: 1 8 5 2 4 7 6 3 10 9
The 1 is turned over and removed. The pile is now: 8 5 2 4 7 6 3 10 9
The 8 and 5 (hidden) are moved to the bottom and the 2 is turned over and removed. The pile is now: 4 7 6 3 10 9 8 5.
The 4 7 an 6 (hidden) are moved to the bottom and the 3 is turned over and removed. The pile is now 10 9 8 5 4 7 6.
The 10 9 8 and 5 (hidden) are moved to the bottom and the 4 is turned over and removed. The pile is now 7 6 10 9 8 5.
The 7, 6, 10, 9, and 8 (hidden) are moved to the bottom and the 5 is turned over and removed. The pile is now 7 6 10 9 8.
The 7, 6, 10, 9, 8, and 7 (again since we have to move 6 cards and only 5 remain) are then moved to the bottom in order. The 6 is then turned over and removed leaving 10 9 8 7.
The 10, 9, 8, 7, 10, 9, and 8 are moved to the bottom leaving the 7 at the top which turned over and removed. Leaving 10 9 8. Then the 10, 9, 8, 10, 9, 8, 10, 9 are moved to the bottom in order, the 8 is turned over and removed leaving 10 9.
The 10, 9, 10, 9, 10, 9, 10, 9, 10 are moved to the bottom, the 9 is turned over and removed leaving 10.
Finally, the 10 is moved to the bottom 10 times, then turned over to reveal 10.
number 5: how about
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10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 100.
Unless I'm not understanding the problem again.
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[Last edited by stigant at 01-31-2008 03:39 PM]