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Since distributivity is the only ring axiom to mention both addition and multiplication, a ring without distributivity is nothing more than an abelian group and a semigroup that happen to be defined on the same set.
(sin(a) + sin(b) + sin(c))^2 + (cos(a) + cos(b) + cos(c))^2
= sin^2(a) + sin^2(b) + sin^2(c) + cos^2(a) + cos^2(b) + cos^2(c)
+ 2sin(a)sin(b) + 2sin(a)sin(c) + 2sin(b)sin(c)
+ 2cos(a)cos(b) + 2cos(a)cos(c) + 2cos(b)cos(c)
= 3 + 2(cos(a-b) + cos(b-c) + cos(c-a))
= 0
Assuming a,b,c are real we must have
sin(a) + sin(b) + sin(c) = cos(a) + cos(b) + cos(c) = 0
and so the answer is (iv).
Any sequence of the form
In Professor Umbuguio's forumula we have
Since
are all integers, we can use the recurrence relation to prove by induction that is divisible by 1946 for allSurely we should have:
Don't you just love it when two mistakes cancel each other out.
I have found this to be a nice problem, with what I consider to be a pretty solution.
Let g(n) be the number of cuboids that can be made from n cubes.
Write n as a product of powers of distinct primes, say
Then we have