| If n^3 = n*10^m + p where p <10^m
then p=qn, so n^2 = 10^m + q where q is small :-)
which suggests using the manual square root method (I'm sure it's in
here somewhere) to successively get square roots of 10,1000,100000 etc
and round them up:
3 1 6 2 2
-- -- -- -- -- -- ...
10 00 00 00 00 00
9
--
6(1) 1 00
61
----
62(6) 39 00
37 56
-----
632(2) 1 44 00
1 26 44
-------
6324(2) 17 56 00
12 64 84
--------
...
Trying each step we get 32^3 = 32768 good
317^3 = 31855013
3163^3 = 31644451747
31623^3 = 31623446801367 good
and no doubt there are more
Dick
|
| Solution by Peter Hurthig, Columbia College, Burnaby, B.C.
We claim that: if the (p+1)st digit to the right of the decimal in
sqrt(10) is greater than or equal to 6, then N = floor(10^p * sqrt(10))
+ 1 has the desired property. [Solutions are shown for p = 1, 4, 5, 6,
7, match the first five solutions in the previous response.]
For N to have the desired property, N^3 = 10^m * N + a where 0 <= a <
10^m. Therefore
0 <= N^3 - 10^m * N < 10^m. (1)
If m = 2p, then N(N^2-10^(2p)) >= 0 implies N >= 10^p where equality
yields the trivial solutions. However if N = 10^p + k for some
positive integer k then
N^3 - 10^m * N = N(N^2-10^(2p)) = (10^p + k)(2k*10^p + k^2) > 10^m,
contradicting (1). Therefore the non-trivial solutions can occur only
if m is an odd number, m = 2p+1. Since N^3 - 10^(2p+1) must be
positive, N > 10^p * sqrt(10). So if there are any solutions of (1)
for m = 2p+1, then N = floor(10^p * sqrt(10)) + 1, the smallest integer
which is greater than 10^p * sqrt(10), must be one of them.
If we let q = 10^p * sqrt(10) and delta = N-q then N(N^2 - 10^m) < 10^m
becomes
(q+delta)(2*q*delta + delta^2) < q^2
or
(1 - 2*delta)q^2 - 3*delta^2*q - delta^3 > 0.
If delta >= 1/2 then the left side is negative and there are no
solutions. [In particular, N = [10^p * sqrt(10)] + 1 is the only
possible choice for N for each value of p, because any larger choice
would mean delta >= 1. -- Ed.] If delta < 1/2 then the inequality is
satisfied if
3*delta^2 + delta*sqrt(delta^2+4*delta)
q > ---------------------------------------.
2 - 4*delta
In particular, if delta <= 2/5 then
3*delta^2 + delta*sqrt(delta^2+4*delta) 6+2*sqrt(11)
--------------------------------------- <= ------------ < 3 < 10^p*sqrt(10)
2 - 4*delta 5
for all p.
Therefore [since delta = N-q is the difference between 10^p * sqrt(10)
and the next highest integer] if the (p+1)st digit after the decimal in
the expansion of sqrt(10) is greater than or equal to 6, then
10^p * sqrt(10) - floor(10^p * sqrt(10)) = 1-delta >= 0.6,
and floor(10^p * sqrt(10)) + 1 will have the desired property.
It would of course be very gratifying to show that sqrt(10) is a normal
number (i.e., all the digits occur in the expected frequency) so that
there would be infinitely many such N.
|