AMC8 1997

AMC8 1997 · Q25

AMC8 1997 · Q25. It mainly tests Remainders & modular arithmetic, Powers & residues.

All of the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?

将从 2 到 98（包含）的所有偶数相乘，但排除那些末位为 0 的数。乘积的最右边数字（个位数）是多少？

(A) 0 0

(B) 2 2

(D) 6 6

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): If the numbers 2, 4, 6, and 8 are multiplied, the product is 384, so 4 is the final digit of the product of a set of numbers ending in 2, 4, 6, and 8. Since there are ten such sets of numbers, the final digit of the overall product is the same as the final digit of $4^{10}$. Now, $4^{10}=(4^2)^5=16^5$. Next, consider $6^5$. Since any number of 6’s multiply to give 6 as the final digit, the final digit of the required product is 6.

答案（D）：如果把 2、4、6 和 8 相乘，积为 384，因此以 2、4、6、8 结尾的一组数的乘积末位是 4。由于有十组这样的数，整体乘积的末位与 $4^{10}$ 的末位相同。现在，$4^{10}=(4^2)^5=16^5$。接着考虑 $6^5$。因为任意多个以 6 结尾的数相乘，其末位仍为 6，所以所求乘积的末位是 6。

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