***urgent. willing to pay high points*** use each of the digits 2,3, 4,6,7 and 8 exactly once to construct two three digit integers, m and n so that m-n is positive ans as small as possible. compute m-n. show how you solved as well. you!

The digits in the hundred should be as close as possible. this is the minimize the hundreds gap. then make the larger top number (number being subtracted) smallest as possible and the bottom number largest as possible. this maximizes the subtractionpossible numbers pairs for first digit: 3 and 2; 4 and 3; 7 and 6; 8 and 7.so now, first case: 2 and 3remaining: 4, 6, 7, 8choose digits that will make the top number smallest as possible, bottom as large as possible.346 - 287 = 59second case: 4 and 3remaining:   2, 6, 7, 8choose digits that will make the top number smallest as possible, bottom as large as possible.426 - 387 = 39third case: 7 and 6remaining:   2, 3, 4, 8choose digits that will make the top number smallest as possible, bottom as large as possible.723 - 684 = 39fourth case: 8 and 7remaining:   2, 3, 4, 6choose digits that will make the top number smallest as possible, bottom as large as possible.823 - 764 = 59conclusioni'm not too sure, but could be either {m = 723, n = 684} or {m = 426, n = 387}.