The last digit of 2^{m} cdot 5^{n} (where m, | vedclass

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

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$0$

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(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$.We can rewrite this as $2^{m-n} \cdot (2 \cdot 5)^{n}$ if $m \ge n$,or $5^{n-m} \cdot (2 \cdot 5)^{m}$ if $n > m$.In either case,the expression contains a factor of $(2 \cdot 5) = 10$.Any positive integer multiplied by $10$ will result in a number ending in $0$.For example,if $m=1, n=1$,then $2^1 \cdot 5^1 = 10$ (ends in $0$).If $m=2, n=1$,then $2^2 \cdot 5^1 = 4 \cdot 5 = 20$ (ends in $0$).If $m=1, n=2$,then $2^1 \cdot 5^2 = 2 \cdot 25 = 50$ (ends in $0$).Therefore,the last digit is always $0$.

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

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