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anonymous
 one year ago
Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?
anonymous
 one year ago
Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Base Case: n=1 8^13^1=5 5/5=1 Induction Hypothesis: 8^k3^k is divisible by 3 8^(k+1)3^(k+1) 8^(2)3^(2) 649=53 53/3

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I think you meant to say `Induction Hypothesis: 8^k3^k is divisible by 5`

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if we assume that claim is true, then we can say 8^k3^k = 5q where q is some integer. Agreed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the significance of 5q??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it just means "5 times some integer"

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if I said "100 is divisible by 5" I can state that as 100 = 5q In this case, q = 20

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, yeah. so that just means divisible by 5

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the goal is to prove that if 8^k3^k = 5q is assumed to be true, then 8^(k+1)3^(k+1) = 5s where q and s are integers

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1here are 2 hints: 1) isolate the `8^k` in `8^k3^k = 5q` to get `8^k = 5q + 3^k` 2) break down `8^(k+1)3^(k+1) ` into `8*8^k3*3^k`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i do not understand.8^(k+1) is just 64, no? not 64^K?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1how are you getting 64? you can't replace k with 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i didn't know i couldn't do that.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you can replace k with 1 if you wanted to prove that specific piece. But we're trying to prove the whole thing in general (for any number k)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1do you see how I turned `8^(k+1)3^(k+1)` into `8*8^k3*3^k` ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, that was my next question. i thought k+1 would turn into ^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1recall that \[\LARGE x^{y+z} = x^y * x^z\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so \[\Large 8^{k+1} = 8^k*8^1 = 8^k*8 = 8*8^k\] make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes. thank you. and now i have to do the same thing for the other side, right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1for the 3^(k+1) part you mean? yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so i just get 3^k*3+5q=6k+5q

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1we have 8*8^k3*3^k now replace the 8^k with 5q + 3^k 8*8^k3*3^k 8*(5q + 3^k)  3*3^k 8*5q + 8*3^k  3*3^k do you see what's next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.040q + 24^k  9^k 40q + 15^k

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.18*3^k isn't equal to 24^k

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.18*5q + 8*3^k  3*3^k 5*8q + 8*3^k  3*3^k 5*8q + (8*3^k  3*3^k) 5*8q + (8  3)*3^k 5*8q + 5*3^k 5*(8q + 3^k) 5*s where s = 8q + 3^k and s is an integer. So this all proves that if 8^k  3^k = 5q, then 8^(k+1)3^(k+1) = 5s q and s are integers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im going to have to study that, tomorrow. can you look at a different kind of problem with me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01^3 + 2^3 + 3^3 + … + n^3 = n^2(n + 1)^2/4 for all positive integers n

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm sure you have the base case already done?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. i just started looking at it. i have not tried solving a sequence kind of problem yet.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you'll use induction here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have less idea about this problem then i did the last :

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the base case is n = 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1prove that is true
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