AMC 8 1997 Problem 25

AMC8 1997 年第 25 题

余数与同余★★★☆☆

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

将 2 到 98（含）之间所有不以 0 结尾的偶数相乘。乘积的最右边数字（个位数字）是多少？

正确答案 · Correct Answer

解析 · Solution

Solution 1

All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of

(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot...

There will be 10 groups of 4 numbers. The number now can be rewritten as

(2\cdot 4 \cdot 6 \cdot 8)^{10}

Multiplying, we get

384^{10}

Again, we can disregard the tens and hundreds digit of 384, since we only want the units digit of the number, leaving

4^{10}

Now, we try to find a pattern to the units digit of

4^n

. To compute this quickly, we once again discard all tens digits and higher.

4^1 = 4

4^2 = 4\cdot 4 = 1\underline{6}

, discard the 1.

4^3 = 1 \cdot 4 = \underline{4}

4^4 = 4 \cdot 4 = 1\underline{6}

, discard the 1.

4^5 = 1 \cdot 4 = \underline{4}

Those equalities are, in reality, congruences

\mod {10}

Thus, the pattern of the units digits is

\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}

. The cycle repeats so that term

n

is the same as term n+2. The tenth number in the cycle is 6, giving an answer of

D

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