let the sum of the number desired be d
prove that it can't have more than 6 digits
if it has n digits and n>6 then the maximum of d is 9*n
(9*n)^3=729*n^3
but for n>6 729*n^3<10^(n-1)
thus there can't be a 7 digit or higher number that is equal
to the cube of the sume of its digits.
now to maximize the number we want it to have as many digits as possible
assume it has 6 digits
then then since d^3 must be in the range of [100000,999999] then
d must be in [47,99] but the maximum digital sum of a 6 digit number is 9*6=54
thus we need only consider d in [47,54]. It is easy to see that in this range
the first digit is always 1. Thus in this range the maximum for d is 1+9*5=1+45=46
but this is less than 47, thus there can not be a 6 digit number that is the cube of
the sum of its digits.
now to try and find a 5 digit number
using similar methods we can narrow d^3 to [10000,99999] thus d in [22,46]
and the max for d is 9*5=45 thus we need only consider [22,45]
now this is a small list of 24 numbers, easily checked by hand to finds that
only 17576 and 19683 work. Of these 2 19683 is the largest.
thus 19683 is the largest base 10 number this is equal to the cube of the sum
of its digits.
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Posted by Daniel
on 2008-10-09 21:28:30 |
analytic solution
