The last digit of number obtained by multiplying numbers 81 × 82 × 83 × 84 × 86 × 87 × 88 × 89 will be

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.

$$\eqalign{ & \frac{{\left( {81 \times 82 \times 83 \times 84 \times 86 \times 87 \times 88 \times 89} \right)}}{{10}} \cr & {\text{We}}\,{\text{take}}\,{\text{individual}}\,{\text{remainder}}\,{\text{of}}\,{\text{each}}\,{\text{digit,}} \cr & \frac{{\left( {1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9} \right)}}{{10}} \cr & {\text{Numbers}}\,{\text{multiplied}}, \cr & \frac{{\left( {24 \times 42 \times 72} \right)}}{{10}} \cr & {\text{Individual}}\,{\text{Remainder}}\,{\text{has}}\,{\text{been}}\,{\text{taken}}, \cr & \frac{{\left( {4 \times 2 \times 2} \right)}}{{10}} \cr & Or,\,\frac{{\left( {16} \right)}}{{10}} \cr & Or,\,6 \cr & {\text{Remainder}} = 6 \cr & {\text{So,}}\,{\text{the}}\,{\text{last}}\,{\text{digit}}\,{\text{will}}\,{\text{be}}\,6 \cr} $$