Correct me if I am wrong

[Jarvis Bull Dawg]

Prove that of 25 is subtrated from the square of an odd integer greater than 5, the resulting number is always divisible by 8.

Solution:

Let S (n) = n2 – 25
For n = 1, S (1) = (1 )2 – 25 = -24 which is clearly divisible by 8.
Thus the first condition is satisfied as S(1) is true.

Let us assume that S(n) is true for n = 2k+1 Belonging to Odd integers greater than 5, that is,

S(2k+1) = (2k+1)2 - 25 is divisble by 8
= 4k2 + 4k + 1 -25
= 4k2 + 4k - 24
= 4(k2 + k - 6)
= 4(k-2)(k+3) ------ (A)

We can see that (A) is clearly divisible by 8, condition being that k is an odd integer which is bigger than 5. Since both conditions are satisfied, hence by mathematical induction we have proven that S(n) is divisble by 8 for all integers bigger than 5.

 

[ComputerGeek]

umm, you are not exactly correct. you are setting n = to 2k + 1. while that is denoting an odd integer, that is what n should be and you have not shows for n+1 yet which should be 2k + 3.

however, induction doe snot require that you use that form. you can simply say n for all n = 2k+1, now showing for n+1 and just use n+1 to show it rather than 2k+3

 

[shmoe]

Let S (n) = n2 – 25
For n = 1, S (1) = (1 )2 – 25 = -24 which is clearly divisible by 8.
Thus the first condition is satisfied as S(1) is true.

Why is your base case n=1 when you're trying to prove this for odd integers greater than 5?

Let us assume that S(n) is true for n = 2k+1 Belonging to Odd integers greater than 5, that is,

S(2k+1) = (2k+1)2 - 25 is divisble by 8
= 4k2 + 4k + 1 -25
= 4k2 + 4k - 24
= 4(k2 + k - 6)
= 4(k-2)(k+3) ------ (A)

You don't use your induction hypothesis here, and you don't need to. Induction is not necessary.

We can see that (A) is clearly divisible by 8, condition being that k is an odd integer which is bigger than 5.

n>5, not k.

 

[Jarvis Bull Dawg]

Why is your base case n=1 when you're trying to prove this for odd integers greater than 5?

True. Thank you

You don't use your induction hypothesis here, and you don't need to. Induction is not necessary.

What should I use?

Next thing, so should I use k+1 rather than 2k+1? Advice?

 

[shmoe]

What should I use?

Next thing, so should I use k+1 rather than 2k+1? Advice?

2k+1 is good, n is an odd integer after all. You don't know if k is even or odd though.

use what you've got:

n^2-25
=(2k+1)^2 - 25
= 4k^2 + 4k + 1 -25
= 4k^2 + 4k - 24
= 4(k^2 + k - 6)
= 4(k-2)(k+3) ------ (A)

Show A is divisible by 8. 2 cases if you like, k even or k odd.

Actually the n>5 bit isn't necessary, it's just to keep n^2-25 positive, the concept of divisibility still makes sense for negative integers.

By the way you can use Latex to make exponents n^{2} <-click to see how, or please use a ^ to denote exponentiation as I've done above, as "n2" looks like another variable or the variable n times 2.

 

[rachmaninoff]

Odd integers greater than 5: 7, 9, 11...
(5+2), (5+4), (5+6)...

general form: (5+2n), n=1,2,3...


So, what number are you looking at? (5+2n)^2 minus something? Do some algebra here...

Until you're done. (hint: factor out '2's when you can...)

There's no need for induction. If you must, start with n=7, not n=1 (factoring negative integers is uncool).

 

[Mithal]

I think it is easier solved this way

(n^2)-25 divides 8 is equivalent to

(n^2)-1 divides 8 because 24 divides 8

now we have

(n-1) (n+1) divides 8

n= 2 m + 1 ( n is odd as suggested in the problem)

(2 m ) (2 m + 2 ) = 4 m (m+1)

if m is even we are done , if m is odd then (m+1) is even .Both cases prove

divisibility by 8

 

[Jarvis Bull Dawg]

Here is another one, but seriously I have no idea what this is.

Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

 

[shmoe]

what are the last two digits of 7^6? 7^7? Notice a pattern?

Suggestion:7^{4}=100k+1 for some integer k. Use this to show that the last two digits of 7^{4n} are 01 for any natural number n.

 

[Jarvis Bull Dawg]

what are the last two digits of 7^6? 7^7? Notice a pattern?

Suggestion:7^{4}=100k+1 for some integer k. Use this to show that the last two digits of 7^{4n} are 01 for any natural number n.

Yes I do notice the recurring pattern, but I didn't get your suggestion.

 

[shmoe]

Knowing 7^{4}=100k+1 you should be able to prove you can write 7^{4n}=100m+1 where m is some integer (for example use the binomial theorem).

This will prove that the last two digits of 7^{200} are 01. (Since you're only interested in this case you could really take n=50 above, but it doesn't really simplify things)

By the way, you're not familiar with modular arithmetic are you? While not necessary to do problems like this, it does simplify things.

 

[Jarvis Bull Dawg]

Edited for errors

 

[Jarvis Bull Dawg]

Hope yo get it. I solved it for the 201 term. By B-Theorem

 

[Mithal]

Here is another one, but seriously I have no idea what this is.

Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

I tried to do it this way
By binomial theorem I find the last two terms are
(10-3) ^ 201 = (201) (10) (3 ^ 200 )- (3) ^ 201
or
(3 ^ 200) (2010 - 3)
or
(9^100) (2000+7) =((10-1)^100 )( 7) since we are concerned with the last two digits
or
7(1000 t +1)
7000 t +7
This proves the last two digits are 07

 

[Werg22]

You're solutions are perhaps correct but you fail to proove the generality of this statement (for any therm). The true solution is much simpler...

Any odd number greater than 5 can be writtent under the form 5+x, where x is even. Following the statement:

(5 + x)^2 - 25

25 + x^2 + 2(5)x - 25

x^2 + 2(5)x.

Since x is even, let x=2y

(2y)^2 + 2(2)(5)y

4y^2 + 4(5y)

4(y^2 + 5y)

y^2 + 5y is even for any integrer. So let y^2 + 5y=2a

4(2a)

8a

This is the proof that the solution for (5 + x)^2 - 25, where x is even is always a factor of 8.