# 1997 AJHSME Problems/Problem 25

## Problem

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? $\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

## 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 $\boxed{D}$.

## Solution 2

Again, the value we seek is equal to $(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$. We can use equivalence to simplify. $(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$ $\equiv (2\cdot 4\cdot (-4)\cdot (-2))^{10} \mod 10$ $\equiv 64^{10} \mod 10$ $\equiv 4^{10} \mod 10$ $\equiv 2^{20} \mod 10$ $\equiv 8^6\cdot 2^2 \mod 10$ $\equiv (-2)^6\cdot 2^2 \mod 10$ $\equiv 2^8 \mod 10$ $\equiv 256 \mod 10$ $\equiv 6 \mod 10$

Thus our answer is $\boxed{D}$.

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