It is known that ЖЖ + Ж = МЁД. What is the last digit of the product: В $\cdot И \cdot H \cdot H \cdot U \cdot \Pi \cdot У \cdot X$ (different letters represent different digits, the same letters represent the same digits)?

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Since a two-digit number ЖЖ was added to a one-digit number Ж to get a three-digit number, then Ж $=9$, and МЁД = 108. Four digits have already been used. In the product $\mathrm{B} \cdot \mathrm{U}^{\prime} \cdot \mathrm{H} \cdot \mathrm{H} \cdot И \cdot П \cdot У \cdot \mathrm{X}$, six other digits are used.

Therefore, among them, there must be the digits 2 and 5, so this product ends in 0.

## Answer

Ends in 0.