Question

refer to the following attachment and find the last digit of the following using modular Arithmetic or congruence modulo and best answer will be marked as the brainliest ​

Answer 1

Answer:

well the answer to this is very interesting

NO COMPUTER IN THE WORLD HAS CALCULATED SUCH VALUE

BUT U CAN FIND THE LAST NO OF IT

Last digit means mod 10.

We use Euler reduction of powers

73^75^64^76 mod 10

E(10) = 1/2 x 4/5 x 10 = 4

So power becomes 75^64^76 mod 4

Again E(4) = 1/2 x 4 = 2

So power becomes 64^76 mod 2

= 0

So problem reduces as

73^75^64^76 mod 10

= {73^ [ 75^64^76 mod 4 ] } mod 10

= {73 ^ [ 75 ^ (64^76 mod 2) mod 4 ] } mod 10

= { 73 ^ [ 75 ^ (0) mod 4 ] } mod 10

= 73^1 mod 10

= 3

Answer 2

We know that,

$\longrightarrow\sf{5^n\equiv5\pmod{10}\quad\forall\,n\in\mathbb{N}}$

This implies the units digit of $\sf{75^{64^{76}}}$ is 5.

But since $\sf{64^{76}}$ is an even number (units digit is even), the tens digit of $\sf{75^{64^{76}}}$ is 2, because,

$\begin{minipage}{11cm}\textit{``The tens digit of the number in the form (10a+5)^b will always be 2, if and only if atleast one among a and b is even. If both are odd then the tens digit will always be 7. }\end{minipage}$

Hence let,

$\longrightarrow\sf{75^{64^{76}}=100n+25}$

Then,

$\longrightarrow\sf{73^{75^{64^{76}}}=73^{100n+25}}$

$\longrightarrow\sf{73^{75^{64^{76}}}=73^{100n+24+1}}$

$\longrightarrow\sf{73^{75^{64^{76}}}=73^{4(25n+6)+1}}$

$\longrightarrow\sf{73^{75^{64^{76}}}=\left(73^{4}\right)^{25n+6}\times73}$

Since $\sf{73^4\equiv3^4\equiv1\pmod{10},}$

$\longrightarrow\sf{73^{75^{64^{76}}}\equiv1^{25n+6}\times73\pmod{10}}$

$\longrightarrow\sf{73^{75^{64^{76}}}\equiv1\times73\pmod{10}}$

$\longrightarrow\sf{73^{75^{64^{76}}}\equiv73\pmod{10}}$

$\longrightarrow\sf{\underline{\underline{73^{75^{64^{76}}}\equiv3\pmod{10}}}}$

Hence 3 is the answer.