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 $m, n \in N$ (natural numbers).We can rewrite the expression as $2^{m} \cdot 5^{n} = 2^{m-n}

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(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$ (natural numbers).We can rewrite the expression as $2^{m} \cdot 5^{n} = 2^{m-n} \cdot 2^{n} \cdot 5^{n}$ (if $m \ge n$) or $2^{m} \cdot 5^{m} \cdot 5^{n-m}$ (if $n > m$).Case $1$: If $m = n$,then $2^{m} \cdot 5^{m} = (2 \cdot 5)^{m} = 10^{m}$. The last digit of $10^{m}$ is $0$.Case $2$: If $m > n$,then $2^{m} \cdot 5^{n} = 2^{m-n} \cdot (2 \cdot 5)^{n} = 2^{m-n} \cdot 10^{n}$. Since $2^{m-n}$ is an even number and $10^{n}$ ends in $0$,the product will end in $0$.Case $3$: If $n > m$,then $2^{m} \cdot 5^{n} = (2 \cdot 5)^{m} \cdot 5^{n-m} = 10^{m} \cdot 5^{n-m}$. Since $5^{n-m}$ always ends in $5$ for $n > m$ and $10^{m}$ ends in $0$,the product will end in $0$.In all cases,the last digit is $0$.

$Q.1.$ $HCF$ of $8$ and $6$	$A. 49$
$Q.2.$ $LCM$ of $(26, 91)$	$B. 182$
	$C. 2$

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

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$C. 2$

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