random problem, seems easy but am i missing something

[furyshade]

this was an extra credit on a calculus test, not calc related but supposed to be difficult, seemed to easy, am i missing something? i think the problem went

1&lt;a&lt;b&lt;c&lt;d&lt;e <u>&lt;</u> 9

a, b, c, d, and e are whole numbers of differnet values, what combination of a b c d and e in a two and three digit number will produce the greatest product? i thought
a=5 b=6 c=7 d=8 e=9, 965*87, is there something im missing?

[GMontag]

96*875 is a greater product than 965*87.

EDIT:

But really, the 5 digits are obvious, because if you use any of 2-4, you can always replace it with a larger number to get a larger product. In the same way, the ordering of the digits within the two numbers is also obvious. That only leaves 10 possible combinations of multiplicands. Just try them all and see:

987*65 = 64155
986*75 = 73950
985*76 = 74860
976*85 = 82960
975*86 = 83850
965*87 = 83955
876*95 = 83220
875*96 = 84000
865*97 = 83905
765*98 = 74970

[mindflayer]

[ QUOTE ]

987*65 = 64155
986*75 = 73950
985*76 = 74860
976*85 = 82960
975*86 = 83850
965*87 = 83955
876*95 = 83220
875*96 = 84000
865*97 = 83905
765*98 = 74970


[/ QUOTE ]

there are only 6 combinations
eliminate 1,2,3 and 10
only combinations where 9 and 8 are the initial numeral of each number are allowed
the second number in the multiplication are

85
86
87
95
96
97