Mathematica Proving 2^(2^n) - 6 is a Multiple of 10: Mathematical Induction Method

[dglee]

Show taht for every natural numbers n>=2 n \geq 2 the number 2^2^n - 6 2^{2^n} -6 is a multiple of 10. Using mathematical induction.
Okay i got no clue how to start this question. Ahhh Is there a series where the x^n is a series? Well this stuff really sucks.

 

[Tide]

HINT:

2^{2^{n+1}} = \left( 2^{2^n}\right)^2

 

[dglee]

HINT:
2^{2^{n+1}} = \left( 2^{2^n}\right)^2

Oh WOw... that helped a lot.. now i can show that its inductive hmm but how would i show its a multiple of 10?
hmm maybe if i said 10^2 is 100 so that's a multiple of 10. 10*n\leq2^{2^n}-6 if i proved that.. hmm i wonder if that would be right... 10*(n+1)\leq2^{2^(n+1)} hmm if i proved that would i have solved the question?
wow if hmm well ahaha thanks a LOT! YOUR AWSOME. that little hint helped a lot but i got no clue if I am actually doing it right.

 

[pizzasky]

For Mathematical Induction, you assume that P_{k} is true, in this case for n greater or equal to 2. With this, you can immediately say that 2^(2^k) - 6 is equal to 10a, for some a, where a is a positive integer.
Can you then use this fact, and the hint provided, to prove the desired result for 2^(2^(k+1)) - 6?

 

[dglee]

hmm should could i say that

10(a+x)\leq (\left2^{2^n}\right )^2 - 6 where x is some positive number and should that it is inductive to prove that 10 is a multiple?
soo
10(a)\leq2^{2^n} - 6 then
10(a+x)\leq(\left 2^{2^n}\right)^2 - 6
then show that
10(a+x)\leq2^{2^{n+1}} -6 ahhh I am confusd now.. ahh

 

[Tide]

HINT: If M - 6 is a multiple of 10 then M ends in the digit 6! :)

 

[pizzasky]

Hmmm... Why are there so many inequalitites in your working? From what I know, the only inequality to appear in your solution should be the fact that n is greater than or equal to 2, but this is just a specification, and should not appear in your proof.

 

[pizzasky]

Steps in Mathematical Induction
1) Let P_{n} be the statement 2^{2^n}-6 is divisible by 10, for n\geq 2.
2) Check that the result you want to prove is valid for n=2, so P_{2} is true.
3) Assume P_{k} is true, for some n \geq 2. So, 2^{2^k}-6 = 10a, for some a, which is a positive integer.
4) Using this result, you must somehow prove that 2^{2^{k+1}}-6 is a multiple of 10. How would you go around doing it? Look at the first hint provided and observe... What has been done to the term 2^{2^n} ? USE BOTH THE RESULT FROM STEP 3 AND THE FIRST HINT

5) Once you have proven step 4, give a conclusion. "Since P_{2} is true, and for some n\geq2, P_{k} is true \Longrightarrow P_{k+1} is true. By Mathematical Induction, P_{n} is true for all n\geq2."

 

[dglee]

Wow thanks a lot for your help. I will try to figure this out. You helped a lot pizzasky and Tide.