Question

Prove that for all integers n>0, 5^(2n)-2^(5n) is divisible by 7.

Answer 1

sounds like some sort of induction process to me

Answer 2

It is an induction problem. I am typing out the work I have down, now.

Answer 3

Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7.
5^(2)*5-2^(5)*2=61
61|7

Inductive Hypothesis: n=k
5^((2)(k+1))-2^((5)(k+1))
5^((2)(k)*5-2^((5)(k))*2

Answer 4

i think your induction is off

Answer 5

I am not sure how to proceed from here. I know that I need to get 5^((2)(k))-2^((5)(k) because we have assumed that is divisible. Then the rest needs to be proved | by 7.

Answer 6

5^(2(k+1)) - 2^(5(k+1))

5^(2k +2) - 2^(5k +5)

5^(2k) 5^2 - 2^(5k) 2^5

Answer 7

Okay, let me try to work that out. I'll post what I have in couple moments.

Answer 8

5^(2n) - 2^(5n)

[5^2]^n - [2^5]^n

this might be a better format to play with

Answer 9

@AlexADB definitely got the right idea

Answer 10

5^(2(k+1)) - 2^(5(k+1))

[5^2]^(k+1) -[2^5]^(k+1)

Answer 11

but yeah, ive got no clear direction for this in my head yet

Answer 12

well, that was annoying. i had a problem posting for a second. okay, still trying to figure this out.

Answer 13

take out the 25 might work

Answer 14

\[25\times 5^{2k}-32\times 2^{5k}\]

Answer 15

\[25(5^{2k}- 2^{5k})-7\times 2^{5k}\]

Answer 16

seen it before is how i know

Answer 17

i was trying to divide it by 5^k - 2^k to see where it led :)

Answer 18

i am trying to piece together the steps you guys are throwing around. i'm not good at math so i'm having an issue keeping up :X

Answer 19

32 2^(5k) = 25 2^(5k) +7 2^(5k)

Answer 20

25 5^(2k) - (32 2^(5k))

25 5^(2k) - (25 2^(5k) +7 2^(5k))

25 5^(2k) - 25 2^(5k) -7 2^(5k)

[25 5^(2k) - 25 2^(5k)] -7 2^(5k)

etc ...

it was not obvious to me at first either

Answer 21

both sides are factorable by 7, and therefore ...

Answer 22

both *terms* are factorable by 7 that is

Answer 23

does it make sense yet?

Answer 24

im copying that onto paper. i need another 2 minutes to look at it.

Answer 25

25x + 32y

32y = 25y + 7y

25x + 25y + 7y

25(x + y) + 7y

but it was assumed that 7|x+y

25*7s + 7y

7(25s + y)

Answer 26

not sure if thats the proper method, but its proof enough to me

Answer 27

Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7. 5^(2)*5-2^(5)*2=61 61|7

Inductive Hypothesis: n=k 5^((2)(k+1))-2^((5)(k+1)) 5^((2)(k)*5-2^((5)(k))*2

Okay, I kind of need to see this from the beginning. First, is my base case and inductive hypothesis right?

Answer 28

your induction is off again ...

Answer 29

might help to get rid of the 'bases' and use the integer representations

Answer 30

Which part of my induction is wrong? Is it the base case? Is it the Hypothesis? What do you mean by getting rid of the 'bases?'

Answer 31

for example ...

5^((2)(k+1)) is not 5^(2k) * 5

Answer 32

5^(2k) is a base of 5, to a power if 2k ...

5^(2k) = 25^k

Answer 33

i didn't know i couldn't do that. 5^2 is the same as 5x5. I just reach up and grab a 5 from the exponent when I need more stuff to make the equation work, no?

Answer 34

5^(2(k+1)) = 5^(2k+2) = 5^(2k) * 5^2

you are not distributing correctly and it is throwing off your setup

Answer 35

Alternatively, as you did it:
\(S(n)=5^{2n}-2^{5n}\)
=>
Base case: n=1
\( S(1)=5^{2\times 1}-2^{5\times 1}=25-32=-7\) => 7|S(1)

Answer 36

@mathmate
Thank you. I like your notation mathmate.

Next, my Inductive Hypothesis would be:
5^(2(k+1))-2^(5(k+1)) I don't think there is any trouble there.

Inductive Step:
5^(2k)*5-2^(5k)*2

After this, I am not quite sure what to do. I want 5^2k-2^5k because that assumption is being treated as true. But, what should be done with the 5 and 2 that are in the way?

Answer 37

A lot of your misery is in mis-distribution. Once that's fixed, you can go along with what @Amistre64 suggested.
For the Inductive Hypothesis, it's just for the general case "n", or
\( 7~|~(5^{2n}-2^{5n})\), or simply 7|S(n)

For the Induction step,
You have to show that 7|S(n+1), \(given\) that 7|S(n).
Re-read @Amistre64's last few posts for hints on this step.

Answer 38

@satellite73
hey man, I still don't understand what is going on in this problem.
I understand 5^2k * 5^2 - 2^5k * 2^5; so, doing 25 and 32 i understand. but, getting here 25(5^2k−2^5k)−7×2^5k i don't see it. If you have time that would be awesome.

Answer 39

Lets just clean up the expression
\[\Huge 25(\underbrace{5^{2k}}_{\color{red}{=x}})-32(\underbrace{2^{5k}}_{\color{green}{=y}})\]

Now, we have 25x - 32y ... and we want to factor out (x-y)

we can just as easily add zero to this setup,say (7x-7x)

25x + (0) - 32y

25x + (7x-7x) - 32y

25x + 7x -7x - 32y

(25x + 7x) -7x - 32y

32x -7x - 32y

32x - 32y - 7x

(32x - 32y) - 7x

32(x - y) - 7x

Answer 40

it is assumed, or given, that (x-y) is divisible by 7 ... so it is some value 7s.

Answer 41

waow that's some pretty LaTeX :) lol
underbrace? Hmm never seen that before :3