sod(n) denotes the sum of the digits of a
base ten positive integer n, and:
R
_{t} = 11...11 (the digit 1 repeated precisely t times.)
Determine the values of t for which:
sod(R_{t}^{2}) = (sod(R_{t}))^{2}
(In reply to
computer exploration by Charlie)
we can determine a hard upper limit as follows:
r(t)=(10^t1)/9 and thus has t digits
r(t)^2=(10^t1)^2/81=(10^(2t)2*10^t1)/81 and thus has 2t1 digits
now since r(t) is comprised of all 1's sod(r(t))=t
thus the RHS is sod(r(t))^2=t^2
the LHS is
sod(r(t)^2)
since r(t)^2 has 2t1 digits the upper limit on the sod is 9*(2t1)=18t9
thus we are certain not to find any more solutions once
18t9<t^2
t^218t+9>0
t^218t+81>72
(t9)^2>72
t9>sqrt(72) or t9<sqrt(72)
t>9+sqrt(72) or t<9sqrt(72)
t>17.49 or t<0.51
thus we need only check up to t=17

Posted by Daniel
on 20131002 10:36:11 