how to??

but the ans given is 3...

solution given:

" 7³ ≡ 343 ≡ 3 (mod 10)

(7³)⁶⁷ ≡ (3)⁶⁷ (mod 10)
therefore the last digit is 3 "

oO??

i agree... but.. that means the solution given in book is either wrong or the answer is indefinite ... ??

The book is 'right' up to ≡3⁶⁷ mod 10. After that to conclude that 3⁶⁷≡3 mod 10 is incorrect.

An easier way using congruences is 7⁴≡1 mod 10 (how did we hit on 4? look up the euler totient function). This is what samagra is saying when he is talking of the last digit being periodic with period 4. The next step elaborates that

So,( 7⁴)⁵⁰ ≡ 7²⁰⁰ ≡ 1 mod 10 and so 7²⁰¹â‰¡7 mod 10.

So the last digit is 7.