Solution (By Examveda Team)
The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10.
$$\eqalign{
& \frac{{\left( {81 \times 82 \times 83 \times 84 \times 86 \times 87 \times 88 \times 89} \right)}}{{10}} \cr
& {\text{We}}\,{\text{take}}\,{\text{individual}}\,{\text{remainder}}\,{\text{of}}\,{\text{each}}\,{\text{digit,}} \cr
& \frac{{\left( {1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9} \right)}}{{10}} \cr
& {\text{Numbers}}\,{\text{multiplied}}, \cr
& \frac{{\left( {24 \times 42 \times 72} \right)}}{{10}} \cr
& {\text{Individual}}\,{\text{Remainder}}\,{\text{has}}\,{\text{been}}\,{\text{taken}}, \cr
& \frac{{\left( {4 \times 2 \times 2} \right)}}{{10}} \cr
& Or,\,\frac{{\left( {16} \right)}}{{10}} \cr
& Or,\,6 \cr
& {\text{Remainder}} = 6 \cr
& {\text{So,}}\,{\text{the}}\,{\text{last}}\,{\text{digit}}\,{\text{will}}\,{\text{be}}\,6 \cr} $$
You mean this expression,
81*82*83*84*85*86*87*88*89
Then, Remainder would be 0.
See, there 82 and 85, if you multiply these two, you will find 10 at the unit digit which will remain till the end.
And when a number containing 0 at unit digit divide by 10, will give remainder 0.
If we will add 85 so what will be answer please update