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Show that $45^n$ can not end with the digit $0$, $n$ being a natural number. Write the prime number '$a$' which on multiplying with $45^n$ makes the product end with the digit $0$.
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Solution: $45^n = (3 \times 3)^n \times 5^n$ [1 mark]
To end with digit $0$, $45^n$ should have prime factors $2$ and $5$ both. So it cannot end with digit $0$. [1/2 mark]
$45^n$ should be multiplied by $2 \Rightarrow a = 2$ [1/2 mark]
To end with digit $0$, $45^n$ should have prime factors $2$ and $5$ both. So it cannot end with digit $0$. [1/2 mark]
$45^n$ should be multiplied by $2 \Rightarrow a = 2$ [1/2 mark]