If 5 to the power of 501 is devided by 126, find the remainder?

Answer 1

(5 to the 501st power)/(5 squared + 1)


1

Answer 2

since 501 can be devided by 3, 5 to the power of 501 can be represented as 125 to the power of 501 / 3 = 125 to the power of 167.
125 can be represented as (126 - 1) therefore

(1) 125^167 = (126 - 1)^167

Now this can be represented using the n-th power of the sum of two numbers a and b meaning that

(2) (a + b)^n = sum{k:0..n}o(n, k) a^(n-k) b^k

where o(n, k) represents n over k
In our case n is 167, a is 126 and b is -1
if we take that into consideration we see that (1) comes down to

(3) 125^167 = sum{k:0..167} o(167, k) 126^(167 - k) (-1)^k

we can see that for every k except 167 the k-th element of this sum can be devided by 126 therefore the only element thet cannot be divided by 126 is

(4) 126^0 (-1)^167 = -1

And that would be our remainder if it were positive, but since it is not in order to get the remainder we should subtract -1 from 126
Therefore the remainder is 125.

Answer 3

5^501 ends in 5 and is evenly divisible by 125.

5^501 = 125^(501/3) = 125^167

The remainders of 125^n/126 are 125 if n is odd and 1 if it is even. So the remainder is 125.

Answer 4

5^n ≡1 (mod)126 if n ≡ 0 (mod 6)
5^n ≡5 (mod)126 if n ≡ 1 (mod 6)
5^n ≡25 (mod)126 if n ≡ 2 (mod 6)
5^n ≡125 (mod)126 if n ≡ 3 (mod 6)
5^n ≡121 (mod)126 if n ≡ 4 (mod 6)
5^n ≡101 (mod)126 if n ≡ 5 (mod 6)

501 ≡3 (mod 6)

So 5^(501) ≡125 (mod)126

Answer 5

5^501=5^3(167)=125^167=(126-1)^167

now expand using binomial expantion all terms but one will contain 126 hence ramainder depends upon that
that will be (-1) hence remainder = -1+126=125

Answer 6

antilog(501*log5-log126)