Q. 37

# Show that (12)<su

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 = (22 × 3)n = 22n × 3n

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

And according to the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 12n. So, there is no natural number n for which 12n 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

√22+√52+2(√5)(√2) = p2/q2

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<span lang="EN-USRS Aggarwal - Mathematics