GO Classes Weekly Quiz 2 | Programming in C | Propositional Logic | Question: 1

Answer 1

$7^2\equiv 9\; \mod \; 10$

$7^3 \equiv 7\ast 7^2\equiv 7\ast 9 \equiv 3 \; \mod \; 10$
and finally

$7^4\equiv 7\ast 7^3 \equiv 7 \ast 3 \equiv 1 \; \mod \; 10$

$7^{19522} = (7^4)^{4880} \times 7^2$

$7^{19522} \equiv 7^2 \equiv 9\; (\mod\; 10)$

Answer 2

$7^{^{19522}} mod 10 = (7^{^{2}})^{9761} mod 10 = (49)^{^{9761}}mod 10=9^{^{9761}}mod10=-1^{9761}mod10=-1 mod 10=9$

So we can say last digit is 9.

Answer 3

$7^0 = 1 , 7^1 = 7 , 7^2 = 49 , 7^3 = 343 , 7^4 = 2401 , 7^5 = 16807$

now if we see only the last digit than we can say for

$7^0 = 1 , 7^1 = 7 , 7^2 = 9 , 7^3 = 3 , 7^4 = 1 , 7^5 = 7$

so we can say that after every forth multiple of 7 , its last digit keeps on repeating in cycle .so if , we get to find last digit of 7 raise to x than we should divide x by 4 and see what is the remainder of it (or) we should do x%4 and we should write the last digit of the respective value.

as  x=19522 hence x%4 = 2  therefore
last digit of 7 raise to 19522 = $7^2 = 9$ and hence the answer will be 9.