A big number N becomes 6*N if you take its last digit and place it as its first like transforming 123466 into 612346.
REM: Big big number!
Find it.
We seek (a,b) such that (10a+b)*6=(10^n)b+a where
a=all but the units digit of N
b=the units digit of N.
Rewrite as
a=(10^n-6)b/59
Since 59 is prime, (10^n-6) must be divisible by 59.
If n=1, the remainder = r(1)=4
Increasing n by 1 is in the expression (10^n-6) is equivalent to multiplying the previous number by 10 and adding 54.
Using the remainder function on my calculator
remainder(Ans*10+54,59)
gives the sequence: 4,35,50,23,48,3,...
keeping count eventually we reach the n=57 term which is 0.
So (10^57-6) is divisible by 59.
We just need b to be at least 6 so it can move to the front. The first four numbers then for b={6,7,8,9} are
60(10^57-6)/59+6 = 101694915...6
70(10^57-6)/59+7 = 118644067...7
80(10^57-6)/59+8 = 135593220...8
90(10^57-6)/59+9 = 152542372...9
These numbers are 58 digits long. The next four probably have 117 digits.
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Posted by Jer
on 2022-03-24 14:55:22 |
Non-computer solution
