Okay, time for some math help :)

[Hoeboy]

Problem: Prove that 2^(5n+1) + 5^(n+2) is divisible by 27 for any positive integer n.


I figured that 5^(n+2) is always divisible by 25. So 5^(n+2) mod 27 will always be -2mod7. Now if only I can figure out how to apply that somewhere

 

[Yomicron]

It must be Sunday night.


looks like you should do a proof by mathematical induction.

 

[gopunk]

nm

 

[UglyCasanova]

What math are you taking? (Please don't say Calc 1). I have to take Calc 1 next quarter and I'm not sure how I will do.

 

[Hoeboy]

Okay I forgot to mention that the problem wants me to prove it via arithmitic mod, not induction

dparker, don't worry. Unless you plan to get a math degree or teach math, you won't have to learn this kind of junk. Calc 1 should be easy

 

[Yomicron]

Originally posted by: dparker
What math are you taking? (Please don't say Calc 1). I have to take Calc 1 next quarter and I'm not sure how I will do.

He is probably taking discrete mathematics (generally for CS or Math majors)

 

[Moonbeam]

Are you saying that 2 to the 6th plus 5 to the seventh, where n=1 is divisible by 27 cause I make that at 64 + 78125 = 78189 / 27 = 2895.8888888888888etc. which ain't a whole number. On the other hand any number is dividible by 27 so where's the problem?

 

[gopunk]

Originally posted by: Moonbeam
Are you saying that 2 to the 6th plus 5 to the seventh, where n=1 is divisible by 27 cause I make that at 64 + 78125 = 78189 / 27 = 2895.8888888888888etc. which ain't a whole number. On the other hand any number is dividible by 27 so where's the problem?

why would 5 be to the seventh... should be to the 3rd if 2 is to the 6th

 

[Yomicron]

Originally posted by: Moonbeam
Are you saying that 2 to the 6th plus 5 to the seventh, where n=1 is divisible by 27 cause I make that at 64 + 78125 = 78189 / 27 = 2895.8888888888888etc. which ain't a whole number. On the other hand any number is dividible by 27 so where's the problem?

The question is, if you let x=2^(5n+1) + 5^(n+2) for any n >=1, then x/27 will have a remainder of zero.

 

[aux]

2^(5n+1) = 2*(2^(5n)) = 2*((2^5)^n) = 2*(32^n)
5^(n+2) = 25*(5^n)
now, 25*(5^n) = (-2)*(5^n) (mod 27)
because 25*(5^n) + (2)*(5^n) = 27*(5^n), which is divisible by 27 (reminder: if a+b is divisible by 27, a= (-b) (mod 27))
so the statement you want to prove is equivalent to
(-2)*(5^n) + 2*(32^n) = 0 (mod 27)
which is the same as
2*(32^n - 5^n) = 0 (mod 27)
the above is true because
(32^n - 5^n) = (32 - 5) * (32^(n-1) + .... + 5^(n-1)) = 27* (some integer)
(you should know what (a^n - b^n) is equal to)

 

[hamburglar]

discreet math!!!






























































































is hard

 

This is abstract algebra. Since Aux helped, I won't bother trying it.

You're always helpful, Aux. Is it just 66 times you've ever has the chance to help? Keep it up. Maybe more members will post their math problems more often.

 

[aux]

Originally posted by: luvly
This is abstract algebra. Since Aux helped, I won't bother trying it.

You're always helpful, Aux. Is it just 66 times you've ever has the chance to help? Keep it up. Maybe more members will post their math problems more often.

Not all of my posts are answers of math questions, but significant part are.
I would post more on different topics but I'm too lazy