Why is this wrong? Why can’t we solve separately and then add?

I did:

4^35/5, R = 4

3^35/5, R= 2

R=4+2

R=6

Can a number have remainder 6 when divided by 5?

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Oh right - so if it’s a number greater than what’s in the denominator, we divide again by the denominator to get the remainder? correct?

Better to think of it like this: what’s the remainder when 4^{35} + 3^{35} is divided by 5?

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