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Mikhail's Sequence (Posted on 2004-06-13) Difficulty: 3 of 5
Mikhail, a great mathematics teacher, used to always give hard and complex sequences to his sons. After much thought, the brilliant mathematician thought that his sequences were a little too hard. So, he made another one that was easier. He showed it to his sons later that day:

6, 10, 4, 9, 6, 11

Then he asked what would be the next number in this sequence. Because there were many possibilities, the sons were stumped. So, Mikhail said, "This sequence cannot continue once you have the next number." After hearing this, the sons figured out the answer. What was the last number?

No Solution Yet Submitted by Victor Zapana    
Rating: 4.1111 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution based on Tim C's idea | Comment 10 of 15 |

I managed to construct a rational function that does the trick:

p(n) = ((35/12)x^5 - (209/4)x^4 + (4223/12)x^3 - (4387/4)x^2 + (9229/6)x - 702)/(8-n)

The seventh term is 600.

There is no eighth term due to division by zero.  (Tim C's idea)

The ninth term is -10266 but this is not a continuance.

Of course, I doubt this is VZ's original plan.


  Posted by Jer on 2004-06-15 09:54:14
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