5
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There are 4 missing members in this small series...labeled ???.

Hint 1;

When you need assistance, seek help from higher authority.

Hint 2:

Higher authority can grant you higher powers!

Complete the series.

$191$, $426$, $931$, $???$, $???$, $646$, $971$, $???$, $???$

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3
  • $\begingroup$ If nobody gets it by tomorrow, I will drop a Hint. $\endgroup$
    – Uvc
    Jun 9 '19 at 17:01
  • $\begingroup$ Going to provide the first hint now. $\endgroup$
    – Uvc
    Jun 10 '19 at 10:01
  • $\begingroup$ Second hint is being given. $\endgroup$
    – Uvc
    Jun 11 '19 at 2:00
1
$\begingroup$

The sequence is:

191, 426, 931, 666, 555, 646, 971, 486, 111

Explanation:

For the nth number:

The first digit is: $n^2 \bmod 10$.
The second digit is: $10 - (n^3 \bmod 10)$.
The third digit is: $n^4 \bmod 10$.

Example: eighth number

First digit is: $8^2 \bmod 10 = 64 \bmod 10 = 4$
Second digit is: $10 - (8^3 \bmod 10) = 10 - (512 \bmod 10) = 10 - 2 = 8$
Third digit is: $8^4 \bmod 10 = 4096 \bmod 10 = 6$
Final number: $486$

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1
  • $\begingroup$ Got it......... $\endgroup$
    – Uvc
    Jun 11 '19 at 2:35
5
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Complete answer.

191, 426, 931, 626, 591, 646, 971, 486, 191

because

First digits are the square numbers modulo 10 (1, 4, 9, 6, 5, 6, 9, 4, 1) and last digits alternate between 1 and 6. Respectively, the sum of all the digits add up to 11, 12, 13, 14, 15, 16, 17, 18 and cycle back to 11 since the last term is the same as the starting one.

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3
  • 1
    $\begingroup$ nice progress and approach, +1! $\endgroup$ Jun 9 '19 at 9:09
  • $\begingroup$ any ideas? mine wont work... $\endgroup$ Jun 9 '19 at 9:23
  • $\begingroup$ All I can say is that you are on the right track partially. $\endgroup$
    – Uvc
    Jun 9 '19 at 9:28
1
$\begingroup$

191, 426, 931, ???, ???, 646, 971, ???, ???

After looking for few minute I found pattern as

1.Sum of digit in series as 11,12,13, ???, ???,16,17,???, ???

2.the first digit: square of series 1,4,9,1(6),2(5),3(6),4(9),6(4),8(1)

3.the last digit is flipping between 1& 6

So using 1&2&3 pattern :626,591,486,991

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1
  • $\begingroup$ Not right..will post a hint tomorrow if nobody gets it by that time $\endgroup$
    – Uvc
    Jun 9 '19 at 18:14

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