3^{15*16} mod 619 can be solved without any calculator :-

3^{15*16} mod 619 = 3^{240} mod 619 = (3^{103} * 3^{103} * 3^{34}) mod 619

Since , **(a*b) mod n = ((a mod n )*(b mod n) ) mod n**

So, (3^{103} * 3^{103} * 3^{34}) mod 619 = [(3^{103 }mod 619)*(3^{103 }mod 619)*(3^{34 }mod 619)] mod 619

Now , let 3^{103 }mod 619 = x

Now , take power of 6 both sides or write x , 6 times and multiply them ,

So, we get , 3^{6*103} mod 619 = x^{6}

3^{618} mod 619 = x^{6}

Now, apply , **Fermat's Little Theorem here , ie a**^{p-1} ≡ 1 (mod p) (or) a^{p-1} mod p = 1 mod p

So, 3^{618} mod 619 = 1 mod 619 = 1....So, 1 = x^{6 }...So , x =1 ...

Now , [(3^{103 }mod 619)*(3^{103 }mod 619)*(3^{34 }mod 619)] mod 619 = [1*1* (3^{34 }mod 619)] mod 619 = 3^{34} mod 619...

So, whole problem reduces to 3^{34} mod 619..now it can be easily solved

3^{34} mod 619 = (3^{2})^{17} mod 619 = 9^{17}mod 619 = (9^{5*3}9^{2}) mod 619 = (((9^{3})^{5} mod 619) * (81 mod 619)) mod 619 = ((729)^{5}mod 619 *(81 mod 619)) mod 619

here (729 mod 619)^{5 }mod 619 = 110^{5}mod 619 = (110^{2*2} *110) mod 619 = [(110^{2} mod 619)^{2}*110] mod 619

((339^{2} mod 619)*110 ) mod 619 = 406*110 mod 619 = 92...

So, overall , we get ,92*81 mod 619 = 24..