Freeman Dyson Problem V2

Freeman Dyson Problem This shows that the problem that appears in the NYT Magazine Article on Freeman Dyson has an 18-digit solution as claimed by Dyson. Suppose n is an r-digit numeral and we apply a transformation f that moves a, the last digit of r, to the front of the numeral. We want to find n such that f (n ) = 2n . Then n−a f (n ) = a ⋅ 10r + , and the equation f (n ) = 2n can be solved for n to give 10 a (10 r +1 − 1) n= 19 It follows from Fermat’s Little Theorem that the smallest value of r for which 10r+1 − 1 is divisible by 19 is 18. The smallest value of n that provides a solution is generated by r = 18, a = 2, namely n = 105263157894736842. (Thank you, Maple). Here is a problem which sounds almost the same, but has no solution: Suppose n is an r-digit numeral and we apply a transformation f that moves a, the first digit of r, to the end of the numeral. We want to find n such that f (n ) = 2n . This problem is easier, and the solution is at the bottom of the page:

We can write f (n ) = 10 ( n − a ⋅ 10r ) + a , and the equation f (n ) = 2n reduces to (1)

(10

r +1

− 1) a = 8n

Since 10r+1 − 1 is odd, and a is a one-digit number, a = 8. Then divide both sides of (1) by 8 to get n = 10 r +1 − 1 . Since all digits of n are 9, a = 9, a contradiction.