Question 1197923
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Recall the divisibility by 8 rule says that we only need to look at the last 3 digits of the number to see if the number is a multiple of 8 or not.


This means numbers like 
800
1800
11800
111800
are all multiples of 8 since 800, in the last three digits, is a multiple of 8.


Then note how 8*16 = 128
So numbers like 
128
4128
54128
754128
are all multiples of 8 because 128 is a multiple of 8.


Furthermore, adding 5 more to each number causes us to get a remainder of 5 when dividing by 8.


128+5 = 133
133/8 = something remainder 5


4128+5 = 4133
4133/8 = something remainder 5


54128+5 = 54133
54133/8 = something remainder 5


and so on. 


In short, the last three digits of 133 indicate we'll get a remainder of 5 when dividing it over 8.


So even if the number 567288133 is really massive, we only need to care about the 133 at the end.
133/8 = something remainder 5
567288133/8 = something remainder 5
We don't need to worry about the quotient, so I just labeled it as "something".


This leads 567288133^6 to reduce to 5^6 mod 8.


Let's look at powers of 5 mod 8.
Meaning that I'll divide numbers of the form 5^k over 8 to determine the remainder
5^1 = 5 (mod 8)


5^2 = 5*5 (mod 8)
5^2 = 25 (mod 8)
5^2 = 1 (mod 8) since 25 is 1 more than 24 = 3*8


5^3 = 5*5^2 (mod 8)
5^3 = 5*25 (mod 8)
5^3 = 5*1 (mod 8) .... replace 25 with 1 mod 8
5^3 = 5 (mod 8)
This tells us that dividing 5^3 over 8 gets us a remainder of 5.


5^4 = 5*5^3 (mod 8)
5^4 = 5*5 (mod 8) .... replace 5^3 with 5 mod 8
5^4 = 25 (mod 8)
5^4 = 1 (mod 8)


As you can see, we get this pattern of remainders: 5, 1, 5, 1, ...
It cycles forever like this.


If the exponent k is odd, then 5^k = 5 (mod 8)
If the exponent k is even, then 5^k = 1 (mod 8)


Therefore, 5^6 = 1 (mod 8)  because of the even number exponent
i.e. (5^6)/8 = something remainder 1


Check:
5^6 = 15625
15625/8 = 1953.125 = 1953 remainder 1


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