What is the remainder when (3^84)/26

No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion.

For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method,
I would start by listing,
\(\frac{2^{1}}{7}\)yields remainder of 2

\(\frac{2^{2}}{7}\)yields remainder of 4

\(\frac{2^{3}}{7}\)yields remainder of 1

\(\frac{2^{4}}{7}\)yields remainder of 2

There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4

Alternately, using binomial expansion method:

\(\frac{2^{86}}{7}\) --> \(\frac{(2^3)^{28}*2^2}{7}\) --> \(\frac{(7+1)^{28}* 2^2}{7}\)

Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer.

This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(23-18)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(2-4)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor.
If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5.

Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4.

Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did.