Unofficial Maths237 thread (1 Viewer)

[flyin']

If you have time, try 2003 D1, Q2, c).

If a, b, c are natural numbers and 31|(5a + 7b + 11c),
prove that 31|(6a + 27b + 7c).

 

[...]

If you have time, try 2003 D1, Q2, c).

If a, b, c are natural numbers and 31|(5a + 7b + 11c),
prove that 31|(6a + 27b + 7c).

erk, no i can't do it
theres 3 variables with only 2 "conditions"

hmm
any hints

and oh, have you done one of the recurrence relation q from the past paper (forgot what yr)

q: q_n+2 - a_n+1 - 2 a_n = 1 + 5(3)ⁿ

my a(c)n = c_1 (2)^2 + c_2(-1)^n

a(p)n = 1 + 5(3)^n

 

[flyin']

Can do the question I posted. A friend asked me about it.

a_c(n) is fine. But a_p(n) = -1/2 + (5/4)*(3^n)

 

[...]

Can do the question I posted. A friend asked me about it.

a_c(n) is fine. But a_p(n) = -1/2 + (5/4)*(3^n)

err..wtf bbq

how did you get your a_p(n)

don;t you start off as A_1 + A_2(3)^n