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**Soroban** Hello, saberteeth!

There are $\displaystyle 6^2=36$ possible outcomes.

The product is odd if *both* numbers are odd.

There are 3 choices of an odd number for the first die: $\displaystyle \{1,3,5\}$

. . and 3 choices of odd number for the second die.

Hence, there are: .$\displaystyle 3\cdot3\:=\:9$ odd-product outcomes.

So there are: .$\displaystyle 36 - 9 \:=\:27$ even-product outcomes.

Therefore: .$\displaystyle P(\text{even product}) \:=\:\frac{27}{36} \;=\;\frac{3}{4}$

Two consecutive integers can have the following pairs of units-digits:

. . $\displaystyle (0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0) $

The unit-digits of their product are: . $\displaystyle 0,\;2,\;6,\;2,\;0,\;0,\;2,\;6,\;2,\;0$

Got it?