Answer :

If the number 12n, for any natural number n, ends with the digit 0 or 5, then it is divisible by 5.

That is, the prime factorization of 12n contains the prime 5.

This is not possible because prime factorisation of

12n = (2^{2} × 3)^{n} = 2^{2n} × 3^{n}

So, the only primes in the factorisation of 12^{n} are 2 and 3

And according to the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 12^{n}. So, there is no natural number n for which 12^{n} ends with the digit zero.

**OR**

Let √2+√5 be a rational number.

A rational number can be written in the form of p/q where p,q are integers.

√2+√5 = p/q

Squaring on both sides,

(√2+√5)^{2} = (p/q)^{2}

√2^{2}+√5^{2}+2(√5)(√2) = p^{2}/q^{2}

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