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May 16th, 2018, 08:37 AM   #1
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Exclamation Can you solve these sequences?

Hi

Can you solve these?

1) 12345, 21295, 218143, 2812793, ?

2) 110, 108, 99, 81, ?

Best regards.
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May 16th, 2018, 09:24 AM   #2
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Quote:
Originally Posted by bishop View Post
Hi

Can you solve these?

1) 12345, 21295, 218143, 2812793, ?

2) 110, 108, 99, 81, ?

Best regards.
For the first one ... 7
For the second one ... 13.2

But I'm guessing that my formula is probably different from yours.
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May 16th, 2018, 10:43 AM   #3
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for #2

$\begin{align*}
&100-108=2\\ \\
&108-99=9\\ \\
&99-81=18=2\cdot 9\\ \\
&81-(9\cdot 18) = -81
\end{align*}$
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May 16th, 2018, 10:45 AM   #4
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Quote:
Originally Posted by mrtwhs View Post
For the first one ... 7
For the second one ... 13.2

But I'm guessing that my formula is probably different from yours.
I'd love to see these two answers explained.
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May 16th, 2018, 12:00 PM   #5
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I'd love to see these two answers explained.
I assume you are kidding. Take (1,110), (2,10, (3,99), (4,81), (5,13.2) and fit a 4th degree polynomial. Any finite sequence of $\displaystyle n$ terms can have an $\displaystyle (n+1)^{\text{st}}$ term that can take on any value.

Consider this sequence 1, 4, 9, 16, 25. Then the next term is not 36. It is 156. Use this: $\displaystyle a_n=n^2 + (n-1)(n-2)(n-3)(n-4)(n-5)$
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May 16th, 2018, 01:53 PM   #6
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Quote:
Originally Posted by bishop View Post
2) 110, 108, 99, 81, ?
Difference between 2 consecutive terms = sum of digits of left term!

110 - 108 = 2 = 1 + 1 + 0

108 - 99 = 9 = 1 + 0 + 8

99 - 81 = 18 = 9 + 9

81 - 72 = 9 = 8 + 1

72 - 63 = 9 = 7 + 2
.....

3 cheers for me?
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