1997 AJHSME Problems/Problem 25

Revision as of 17:57, 31 March 2023 by Skatingkitty (talk | contribs) (Solution)

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

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

Simplifiying the ins384)^{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$ (Error compiling LaTeX. Unknown error_msg)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}$$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)\{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$\boxed{D}$

See Also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Question
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All AJHSME/AMC 8 Problems and Solutions

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