6? can end with the digit 0 for any natural

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Q.
Assertion (A): $6^{n}$ can end with the digit 0 for any natural number n.

Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

1 Both A and R are true, and R is the correct explanation of A.

2 Both A and R are true, but R is not the correct explanation of A

3 A is true, but R is false.

4 A is false, but R is true.

Ans.
(4)

As we know that any number whose prime factors consist of 2? × 5?, where m and n are natural numbers, can end with the digit 0, but $6^{2}$ has only prime factors as 2 and 3. So $6^{2}$ cannot end with the digit 0.

Solution

Q.
Assertion (A): $6^{n}$ can end with the digit 0 for any natural number n.

Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

1 Both A and R are true, and R is the correct explanation of A.

2 Both A and R are true, but R is not the correct explanation of A

3 A is true, but R is false.

4 A is false, but R is true.

Ans.
(4)

As we know that any number whose prime factors consist of 2? × 5?, where m and n are natural numbers, can end with the digit 0, but $6^{2}$ has only prime factors as 2 and 3. So $6^{2}$ cannot end with the digit 0.

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Solution

Q. Assertion (A): 6n can end with the digit 0 for any natural number n. Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

1 Both A and R are true, and R is the correct explanation of A.

2 Both A and R are true, but R is not the correct explanation of A

3 A is true, but R is false.

4 A is false, but R is true.

Ans. (4) As we know that any number whose prime factors consist of 2? × 5?, where m and n are natural numbers, can end with the digit 0, but 62 has only prime factors as 2 and 3. So 62 cannot end with the digit 0.

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Q.
Assertion (A): $6^{n}$ can end with the digit 0 for any natural number n.

Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

Ans.
(4)

As we know that any number whose prime factors consist of 2? × 5?, where m and n are natural numbers, can end with the digit 0, but $6^{2}$ has only prime factors as 2 and 3. So $6^{2}$ cannot end with the digit 0.