24^4/8 find remainder problem

GRE Quant Problem Solving
1

R) if we apply the formula discussed in prep swift arithmetic section column 2 video 15th
q) 24^4/8

step 1 find last digit 4, 6, 4, 6, 4, 6
so 4 follows this pattern

step 2 now power is 4 which is even
resulting unit digit for 24^4 as 6

step 3 now 6/8 so as per this remainder should be 6

step 4 but if you check with calculator remainder is 0, or you can think in little more depth and understand like if last two digits of number is xxx56 so number is still ending with 6 but this time there is a chance remainder is going to be zero depending upon number before 56 maybe xxx456

please let me know either if there’s a gap in my understanding or whether i am violating any given considerations or instructions for this formula.
but so far i am concerned with implementation or working of this formula

2

If you need to find the remainder, you need to find the pattern of the Remainder as well (not last digit)

Also, logically if you think, if 24 is divisible by 8, then any power of 24 will also be divisible by 8.
So, essentially, you don’t even need to find any pattern for this.

3

thankyou Jovyn got it, we need to focus on denominator for eg when we have 2 as Denominator remainder is either 0 or1, for 10 in denominator find unit digit for denominator, similarly 5,6,7.. each digit has its own nature and specific way to easily find the remainder when numerator power is too big.

4

Yes exactly.