This was a difficult puzzle, though I continue to be amazed how so many people solve them with such speed…my hats off to all the ones who solved this one. That includes Saurabh Aggarwal, Abhinav Jain, Akshay Joshi, Suman Saraf and Amit Mittal.
I could not find any good method to find the answer, though the answer is 679. the sequence goes as follows:
I do not have any proof that this is indeed the smallest way of doing this. I do not know which of the Martin Gardner books this is from, once I find, I will make another post on this.
Thanks to Amit Mittal who gave the most elaborate answer, I am reproducing this here (though I do not fully understand the first part of his answer):
Congratulations for being able to carry on with kth powers of prime numbers in digits. The easiest persistent number with persistence as k is usually a series of numbers with k in the digit and the series of 1s in the digits from 1 1 to n 1s
2 12 21 121 112
3 13 31 113 131 311
5 15 51 115 511
Our answer as is widely available has a persistence digital root of 6
further traversing up to discover the method of the madness
6 – 32 and the roots of 32 are 22222 or 48
6-32-48 we can pad 1 to get 148 or ( from an admitted hindsight bias) 168
roots of 148 (factors) are 74*2 or 37*4 which goes nowhere in constructing the number as we see below we can get the same 148 from 168 which uses the 7 prime factor to up the product persistence
while roots of 168 are 21*8 or 378
we now have
the roots of 378 are 189*2 or 3*3*3 * 7 * 2 that gives us 6 * 7 * 9
which gives us 679 which is probably the smallest as we use a lot of 2s in here.
I do know to construct numbers with persistence of 1, 2, 3, 4, 5 is harder even knowing you can arrange prime factors of the number
237 or 732 give the same persistence of two 732-42-8
Thus the smallest number with persistence of n will be arranged in increasing order of digits or like 77 have the same digits
679 – 378 – 168 – 48 – 32 – 6
Hope you all enjoyed the puzzle!