8.2. Vasya replaced identical digits in two numbers with identical letters, and different digits with different letters. It turned out that the number ZARAZA is divisible by 4, and ALMAZ is divisible by 28. Find the last two digits of the sum ZARAZA + ALMAZ.
Answer: 32.

Solution. From the condition, it follows that the numbers ZARAZA and ALMAZ are divisible by 4. By the property of divisibility by 4, the numbers ZA and AZ are also multiples of 4. In particular, it follows from this that both digits 3 and A are even. But a number divisible by 4, in which the second-to-last digit is even, must end in a digit that is a multiple of 4. Therefore, both digits 3 and A are multiples of 4. Since they are non-zero (they start the numbers ZARAZA and ALMAZ), one of them is 4, and the other is 8.

The specified sum ends with the same 2 digits as $3 \mathrm{~A}+\mathrm{A} 3=48+84=132$, that is, with 32.