10
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Two numbers go at the $?$s, what are they and why?

$733,\space \space 716,\space \space 645,\space \space 565,\space \space 324,\space \space 276,\space \space 77,\space \space 75,\space \space 64,\space \space 56,\space \space ?,\space \space ?$

Hints

As noted in the comments these are the final two numbers in the sequence.

The sequence is strictly decreasing and contains only positive integers
(that narrows them down to $1485$ choices).

We could extend the sequence to the left using the same logic, but there are many ways to do so.

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  • $\begingroup$ Does the sequence continue after the second ? ? $\endgroup$ – humn Feb 14 '17 at 0:08
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    $\begingroup$ They are the final two in this sequence. $\endgroup$ – Jonathan Allan Feb 14 '17 at 0:24
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    $\begingroup$ @JonathanAllan Are there enough clues given to identify the missing numbers? $\endgroup$ – Techidiot Feb 14 '17 at 7:31
  • $\begingroup$ @Techidiot I believe so. $\endgroup$ – Jonathan Allan Feb 14 '17 at 18:57
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    $\begingroup$ Interesting that only the digits 1-7 show up. Maybe there's something there? $\endgroup$ – Scott M Feb 17 '17 at 15:41
15
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The sequence is:

733, 716, 645, 565, 324, 276, 77, 75, 64, 56, 7, 4

Because:

The sequence consists of digits 1-7 and each digit x occurs x times. There is one 1, two 2s, three 3s, three 4s, five 5s, six 6s, six 7s. The next two numbers should have a 7 and a 4 in them to complete the pattern. Since the sequence is strictly decreasing, the last two numbers are 7,4.

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  • $\begingroup$ Oh darn! How bad of me to not see that. +1 $\endgroup$ – Techidiot Feb 18 '17 at 16:50
  • $\begingroup$ Enthusiastic ^vote with a note: Care to mention in the answer how you realized this? $\endgroup$ – humn Feb 19 '17 at 3:15
2
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Since I have spent quiet a few days on this one, needed to post something I finally got working(at least in my own head)

The Mystery Sequence is

$733,\space \space 716,\space \space 645,\space \space 565,\space \space 324,\space \space 276,\space \space 77,\space \space 75,\space \space 64,\space \space 56,\space \space 48,\space \space 38$

Pattern is

We need to take the sum of all the digits which leads to 13 14 15 16 9 15 14 13 10 11 Now, lets place them vertically to see how they look. When done, we can see the following sequence -
.
.
.
.
13 (Set n-2)
14 (Set n-2) $\Leftarrow$ DESCENDING ORDER
15 (Set n-2)
16 (Set n-2)
-> 9 (Set n-2 begins)
15 (Set n-1)
14 (Set n-1) $\Leftarrow$ ASCENDING ORDER
12 (Set n-1)
-> 10 (Set n-1 begins)
11 (Set n)
12 (Set n) $\Leftarrow$ DESCENDING ORDER
-> 11 (Set n begins) BEGIN HERE $\Uparrow$

Which means we need two numbers whose sum gives 12 and 11. For 12, the next number we have is 48 and for 11, the next number we have is 38. Since, I am assuming -> to be the start, there cannot be any numbers beyond that. I am not good at formatting and I suck at mathematical formatting so bear with me on this... Well, I am satisfied with that thing above :)

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    $\begingroup$ Pretty amazing find, but more complicated reasoning than the expected solution. $\endgroup$ – Jonathan Allan Feb 17 '17 at 22:08
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    $\begingroup$ Well, this is amazing, but the sum of digits in 75 is not 13, but 12, and this fact breaks it. $\endgroup$ – elias Feb 18 '17 at 11:23
1
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The two numbers are clearly 37224 and 342283.

This is because these are the values of the 9th-degree polynomial

$$\frac{3923 x^9}{72576} - \frac{53903 x^8}{20160} + \frac{3423661 x^7}{60480} - \frac{321713 x^6}{480} + \frac{84048523 x^5}{17280} - \frac{21394487 x^4}{960} + \frac{5800232051 x^3}{90720} - \frac{552402481 x^2}{5040} + \frac{126557399 x}{1260} - 36003$$

evaluated at x = 1...10. So the next two numbers are this polynomial evaluated at x = 11 and x = 12.

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  • $\begingroup$ Much as I love a polynomial fit with more inputs than the sequence, I'm afraid it's much more logical. You'll know the answer is right when you have it. $\endgroup$ – Jonathan Allan Feb 13 '17 at 23:40
  • $\begingroup$ y = 23.34446 + (723.1129 - 23.34446)/(1 + (x/4.927903)^4.959924),,36 and 32 $\endgroup$ – Amruth A Feb 14 '17 at 5:49
  • $\begingroup$ polynomial answers are not appropriate answers for this kind of questions in my opinion. $\endgroup$ – Oray Feb 14 '17 at 14:37
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    $\begingroup$ I think it was just a joke. $\endgroup$ – noneuclideanisms Feb 16 '17 at 6:00
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    $\begingroup$ Yes, this is primarily a joke answer. But this is the polynomial fit with least degree. This is partially why I hate sequence puzzles. $\endgroup$ – edderiofer Feb 16 '17 at 12:29

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