✔ 最佳答案
1)
What you want is to calculate (2^2011)/13 or the remainder of it?
If it is a matter of remainder,
find a form to represent 2^2011 in (13A + 1)^n or (13A - 1)^n
where A is an integer, and with the help of binomial theorem,
the remainder could be found easily.
(here A could be 5, 315 ... etc)
2^2011
= 2^[6(335) + 1]
= 2(2^6)^335
= 2(64)^335
= 2(65 - 1)^335 <- easy to express?
= 2{ (65)^335 + (335C1)[(65)^334](-1)^1 + (335C2)[(65)^333](-1)^2 + ... + (-1)^335 } ...(i)
In (i), all the elements are divisible by 13 except the last term
since 65 = 13*5 is divisible by 13.
i.e.
(i) = 2(13M + (-1)^335] , where M is an integer
= (13)(2M) - 2
= (13)(2M - 1) + 13 - 2
= (13)(2M - 1) + 11
i.e. remainder = 11
---
in case that A = 315:
2^2011
= 2^[12(167) + 7]
= (2^7)(2^12)^167
= 128(4096)^167
= 128(4095 + 1)^167 <- here's what we want!
= 128{ 4095^167 + (167C1)(4095^166) + (167C2)(4095^165) + ... + 1)} ...(ii)
Similarly, 4095 = 13*315 is divisible by 13.
(ii) = 128(13N + 1) , where N is an integer
= 13(128N) + 128
= 13(128N) + 13(9) + 11
= 13(128N + 9) + 11
i.e. remainder = 11
- - - - - -
2)
x^2 + y^2 = 1^2
Change it into cylindrical system
i.e. [ρcos(φ)]^2 + [ρsin(φ)]^2 = ρ^2
we have ρ = 1
x = ρcosφ = cosφ
y = ρsinφ = sinφ
2x + 5y^2
= 2cosφ + 5sin^2(φ)
= 2cosφ + 5[1 - cos^2(φ)]
= -5[cos^2(φ) - (2/5)cosφ - 1]
= -5[(cosφ - 1/5)^2 - (1/5)^2 - 1]
= -5[(cosφ - 1/5)^2 - 26/25]
= -5(cosφ - 1/5)^2 + 26/5
i.e.
maximum of 2x + 5y^2
= 0 + 26/5
= 26/5
Easily seen that max of -5(cosφ - 1/5)^2 is zero (when cosφ = 1/5)
If not, see here:
-1 < cosφ < 1
-6/5 < cosφ - 1/5 < 4/5
0 < (cosφ - 1/5)^2 < 36/25
0 < 5(cosφ - 1/5)^2 < 36/5
-36/5 < -5(cosφ - 1/5)^2 < 0
-2 < -5(cosφ - 1/5)^2 + 26/5 < 26/5
