2x < 10 + 5

2x < 15

x < 15/2

x < 7.5

x/3 > 6

x > 6(3)

x > 18

write down the integer value of n which satisfy the inequality -2<n1

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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