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use the index arithmetic to find the remainder when $46^{88}$ times $5^{400}$ is divided by $29$.

I have solved this using typical modulo operations

first by reducing $46$ to $17$ then finding the order of $17$ and dividing $88$ by the order. which in this case is $4$ so it leaves me with $1$ times $5^{400}$ (mod $29$)

I then find the order of $5$ (mod $29$) which is $14$ and leaves me with

1 times $5^8$ (mod $29$). So i get the answer of $3$(mod $29$)

BUT my professor wants me to use the index arithmetic notation and I am confused on how to go through these steps using that notation.

Any help would be appreciated.

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So I first find a primitive root of $29$ which I used $3$. Let $\phi$ = Euler function$(n)$.

$$3^{x} \equiv a \mod{29} \equiv ( \text{ ind$3(a)$ = $x$ })$$ $$\text{ind}3(46^{88} \cdot 5^{400}) (\text{ mod }\phi(29))$$ $$\text{ind}3(46^{88}) + \text{ind}3(5^{400}) (\text{mod} \phi(29))$$ $$88(\text{ind}3(46)) + 400(\text{ind}3(5)) (\text{mod} \phi(29))$$ $$4(\text{ind}3(17)) + 8(\text{ind}3(5)) (\text{mod} \phi(29))$$ $$4(21) + 8 (10) (\text{mod} \phi(29))$$ $$0 + 24 (\text{mod} \phi(29))$$ and finally $24 \mod{29}$ is the remainder.

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