x/3 > 6
x > 6(3)
x > 18
solve these inequalities: 2x - 5 < 10
x/3 > 6
]]>That's really all it is, experience, nothing else irspow.
Oh, and I meant to say integer larger or equal to 0 in my post, since this was about integers and nothing else.
Was this a question that a teacher/professor gave you? It should be worded 9^p and 25^p for p ≥ 1, aka, natural numbers.
It's not wrong the way it is, just...weird.
]]>I need to learn those tricks with adding more terms, very clever.
]]>If p is an integer, then p + 1 would also be an integer the equation would only produce a series like;
(27n + 25n)/4 = 54n/4 = 13n, which of course would all be integers.
Let n = 1:
So you are saying there is an integer p ≥ 1 such that (3*9^(p+1)+25^(p+1))/4 = 13?
Furthermore, you are saying that 9^(p+1) = 25^(p+1) for all p, since n = n
]]>(27n + 25n)/4 = 54n/4 = 13n, which of course would all be integers.
This is obviously not a proof, because I mistakenly used an arithmetic series example instead of a geometric series, but I think that it can be shown to be similar since the relationship of multiples still holds. I will try it that way.
]]>Glad you responded, I wasn't sure if you knew what an inductive proof was, so that saved me a lot of typing.
Inductive assumption:
Inductive assumption, 216, and 600 are all divisible by 4.
]]>I guess it can be prove that 3*9^(p+1)+25^(p+1) is always a multiple of 4.
You could try an inductive proof:
1) prove the base case (p=1),
2) prove that whenever you add 1 to p, the result will still be a multiple of 4 (use p=n → p=n+1)
k = (3*9^(p+1)+25^(p+1))/4
p is any number larger or equal to 0. How can I be 100% sure that k is an integer? I've tried different values to test it, and it looks like it is an integer, but I don't know how to actually proof it.
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