CBSE Class 10 Mathematics Basic 2025: question 21

Hints

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Step-by-step solution

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1. Prime factorisation: $45=3^2\times 5$, so $45^n=(3\times 3{)}^n\times 5^n=3^{2n}\times 5^n$.

   1 mark

2. A number ends with the digit 0 only if it is divisible by 10, i.e. its prime factorisation contains both 2 and 5. The prime factorisation of $45^n$ has no 2. By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), $45^n$ cannot end with the digit 0 for any natural number $n$.

   ½ mark

3. Multiplying by the prime 2 gives $2\times 45^n=2\times 3^{2n}\times 5^n$. This contains both 2 and 5, so the product ends with 0. Hence $a=2$.

   ½ mark

Answer: $45^n=3^{2n}\times 5^n$ has no factor 2, so it cannot end with 0; $a=2$

Revise the chapter

• Chapter: Real Numbers (notes, formulas, where marks are lost and every past question)
• Topic: Fundamental Theorem of Arithmetic