The last digit of 2^m cdot 5^n (m, n in mathb | vedclass

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(A) We are given the expression $2^m \cdot 5^n$ where $m, n \in \mathbb{N}$.We can rewrite this as $2^m \cdot 5^n = 2^{m-n} \cdot (2^n \cdot 5^n)$ ass

Vedclass · Accepted Answer

$0$

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(A) We are given the expression $2^m \cdot 5^n$ where $m, n \in \mathbb{N}$.We can rewrite this as $2^m \cdot 5^n = 2^{m-n} \cdot (2^n \cdot 5^n)$ assuming $m \ge n$,or $2^m \cdot 5^n = 5^{n-m} \cdot (2^m \cdot 5^m)$ assuming $n > m$.In either case,we can factor out $10^k$ where $k = \min(m, n)$.Since $2 \cdot 5 = 10$,the expression will always contain at least one factor of $10$ because $m, n \ge 1$.Any integer multiplied by $10$ ends in the digit $0$.Therefore,the last digit of $2^m \cdot 5^n$ is $0$.

The last digit of $2^m \cdot 5^n$ $(m, n \in \mathbb{N})$ is . . . . . . .

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