Answer:
Step-by-step explanation:
basicallyi you want to find a common denominator before you add them up. the common denominator in this case would be 20, since it's the first common multiple all of them share
50/20 plus 72/20 plus 55/20
177/20 would be your answer
Recall Euler's theorem: if
, then
![a^{\phi(n)} \equiv 1 \pmod n](https://tex.z-dn.net/?f=a%5E%7B%5Cphi%28n%29%7D%20%5Cequiv%201%20%5Cpmod%20n)
where
is Euler's totient function.
We have
- in fact,
for any
since
and
share no common divisors - as well as
.
Now,
![37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6](https://tex.z-dn.net/?f=37%5E%7B32%7D%20%3D%20%281%20%2B%2036%29%5E%7B32%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20%2B%2036c_1%20%2B%2036%5E2c_2%20%2B%2036%5E3c_3%2B%5Ccdots%2B36%5E%7B32%7Dc_%7B32%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20%2B%206%20%5Cleft%286c_1%20%2B%206%5E3c_2%20%2B%206%5E5c_3%20%2B%20%5Ccdots%20%2B%206%5E%7B63%7Dc_%7B32%7D%5Cright%29%20%5C%5C%5C%5C%20%5Cimplies%2032%5E%7B37%5E%7B32%7D%7D%20%3D%2032%5E%7B1%20%2B%206%28%5Ccdots%29%7D%20%3D%20%2032%5Ccdot%5Cleft%2832%5E%7B%28%5Ccdots%29%7D%5Cright%29%5E6)
where the
are positive integer coefficients from the binomial expansion. By Euler's theorem,
![\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9](https://tex.z-dn.net/?f=%5Cleft%2832%5E%7B%28%5Ccdots%29%5Cright%29%5E6%20%5Cequiv%201%20%5Cpmod9)
so that
![32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9](https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%7D%20%5Cequiv%2032%5Ccdot1%20%5Cequiv%20%5Cboxed%7B5%7D%20%5Cpmod9)
Yesyesyesyes yesyeso online
6^9 x 6^-7 = 6^2
x= -7
Hope this helps :)