Use all the digits 1 to 9 create three 3-digit numbers, which, when multiplied together creates the highest value possible.
(In reply to answer
by K Sengupta)
Let the three 3-digit numbers be given by ABC, DEF and GHI.
At the outset for the product to be maximum, (A, D, G) (- (9, 8, 7).
In other words, the triplet (A, D, G) must correspond to a perm of (7, 8, 9).
Any other choice would force a product less than 504*10^6, which is a contradiction.
It can be easily shown that if x> y, with (x+b)(y+a) > (x+a)(y+b), then a>b, and vice versa. So proceeding to determining the optimal tens digit, we observe that choices are now limited to the set (1,2,3,4,5,6).
Thus, in accordance with the foregoing, (B, E, H) = (4, 5, 6) must be the optimal troika of tens digits.
It now remains to select the unit digit triplet (C, F, I) out of the
six perms of (7,8,9). In view of the above inequality it is now
trivial to observe that (C, F, I) = (1, 2, 3) so that the required
maximal product is given by:
941*852*763 = 611,721,516