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This problem is not as daunting as it might at first blush appear!
Since the results have ‘become very large’, we may further take it that all values of the original function are positive and increasing.
Assume that two results are successive. Because only k is variable for the results, and because the results are taken at increments of 10, the difference between 2 successive results is p((k+10)^n-k^n) + q((k+10)^(n-1)-k^(n-1)) + r((k+10)^(n-2)-k^(n-2))… +C, where p,q,r,… and C are constant for all results. Hence, the second result exceeds the first by a multiple of 10, because 10x is a factor of (k+10)^n-k^n. It follows that the last digit of each result is constant, in this case, 8.
Accordingly, for the purpose of ordering the results, we need to turn to the penultimate digits of the results. Since p,q,r,… and C are constant for all results, and because 10x is a factor of each term of the second result, this is equivalent to saying that we need to look at the change of the penultimate digit under multiplication by the last (nonzero*) digit of x.
The result will be some number from the 10x10 multiplication table; but in base 10 the final-digit pattern of all such products is constrained into one of these different forms:
{0,0,0… } for 10 or
{5,0,5…} for 5 or
{1,2,3,4,5,6,7,8,9,0…} for 1 (and its ‘reverse’ for 9) or
{2,4,6,8,0…} for 2, or
{4,8,2,6,0…} for 4 (and its ‘reverse’ for 8) or
{6,2,8,4,0…} for 6, or
{3,6,9,2,5,8,1,4,7,0…} for 3 (and its ‘reverse’ starting with 7, for 7).
Since the foregoing holds good as between any pair of the given results in their desired original and successive order, we now have a viable means for determining such original order.
Turning to the observed values of the function, we see that each number from 0 to 9 occurs exactly once in the penultimate place. This allows us immediately to discard all those patterns (2,4,5,6,8,10) that fail to cycle through all the numbers, leaving 1,3, and their mirror images 7,9.
Assuming that 454 and 457 are in their correct positions, the pattern is 4xx1, which is consistent with {...4,3,2,1...} so the order is:
453. xxx2478
459. xxx5968
455. xxx5158
454. xxx0748
451. xxx1438
458. xxx9828
457. xxx2618
452. xxx5808
460. xxx6098
456. xxx0088
Assuming that 451 and 456 are correct, we get 3xxxx8, and 2 possible sequences: {3,6,9,2,5,8,1,4,7,0} or {1,2,3,4,5,6,7,8,9,0}; the corresponding orders are:
451. xxx1438
454. xxx0748, or 459. xxx5968
455. xxx5158, or 460. xxx6098
459. xxx5968, or 458. xxx9828
453. xxx2478, or 455. xxx5158
456. xxx0088
460. xxx6098, or 457. xxx2618
452. xxx5808, or 454. xxx0748
457. xxx2618, or 453. xxx2478
458. xxx9828, or 452. xxx5808
Lastly, assuming the maximum possible number of results are correct yields:
451. xxx1438
452. xxx5808
453. xxx2478
454. xxx0748
457. xxx2618
456. xxx0088
455. xxx5158
458. xxx9828
460. xxx6098
459. xxx5968
with just two pairs of results switched.
* We can see from the results in this case, that this is the last digit of x, for otherwise there would be a 2-digit constant at the end of each result. |
