Find the number of n that can be divided by 7

[songoku]

Homework Statement

Given that U_n=4n⁶+n³+5. Find the number of n, where n = {1, 2, 3,... 2009} so that U_n can be divided by 7

Homework Equations

Don't know

The Attempt at a Solution

Completely blank...

 

[mtayab1994]

I'd say to use induction.

 

[micromass]

Why don't you check a few n (say for n from 1 to 10) to see what you get?

Another thing you should do is factorize your expression to see if you get something nice.

 

[Dick]

Think about the problem mod 7. Check n=0,1,2,3,4,5,6.

 

[songoku]

I'd say to use induction.

I only know mathematical induction is used for proofing. Can it be used to find number of n in this question?

Why don't you check a few n (say for n from 1 to 10) to see what you get?

Another thing you should do is factorize your expression to see if you get something nice.

U₁=10
U₂=269
U₃=2948
U₄=16453
U₅=62630
U₆=186845
U₇=470944
U₈=1049093
U₉=2126498
U₁₀=4001005

I can't see any patterns there that can be used to determine which n can be divided by 7...
I also can't factorize the expression. Or maybe you mean using perfect square?

Think about the problem mod 7. Check n=0,1,2,3,4,5,6.

Actually I don't have any knowledge about mod but I'll give it a try.

Let U_n=a so for this question a = b (mod 7), where b must be 7k (multiplication of 7)

This is as far as I can go

 

[Dick]

Actually I don't have any knowledge about mod but I'll give it a try.

Let U_n=a so for this question a = b (mod 7), where b must be 7k (multiplication of 7)

This is as far as I can go

Ok, let's not use the word 'mod'. Can you show $U_{n+7}-U_n$ is divisible by 7? That means $U_n$ is divisible by 7 iff $U_{n+7}$ is divisible by 7.

 

[songoku]

Ok, let's not use the word 'mod'. Can you show $U_{n+7}-U_n$ is divisible by 7? That means $U_n$ is divisible by 7 iff $U_{n+7}$ is divisible by 7.

I can show that U_n+7 - U_nis divisible by 7 by expanding U_n+7 and after subtraction, the terms left will all have 7 as factor.

I have several things that I still not clear:
1. How "U_n+7 - U_nis divisible by 7" be used to find number of n?
2. Why you use U_n+7 - U_n? Is it because divisible by 7? If the question asking about divisibility by 5 then U_n+5 - U_n?

Thanks

 

[Dick]

I can show that U_n+7 - U_nis divisible by 7 by expanding U_n+7 and after subtraction, the terms left will all have 7 as factor.

I have several things that I still not clear:
1. How "U_n+7 - U_nis divisible by 7" be used to find number of n?
2. Why you use U_n+7 - U_n? Is it because divisible by 7? If the question asking about divisibility by 5 then U_n+5 - U_n?

Thanks

Look. U_0 is not divisible by 7, right? That means U_7, U_14, U_21,... are not divisible by 7. Agree with that?

 

[songoku]

Look. U_0 is not divisible by 7, right? That means U_7, U_14, U_21,... are not divisible by 7. Agree with that?

Yes. So if we know which n produces U_n that is divisible by 7, then U_n+7 will also be divisible by 7 and they will form arithmetic sequence. Do i get your hint correctly?

Then how to find the first n?

Thanks

 

[Curious3141]

Yes. So if we know which n produces U_n that is divisible by 7, then U_n+7 will also be divisible by 7 and they will form arithmetic sequence. Do i get your hint correctly?

Then how to find the first n?

Thanks

Why not simply try testing U₁ to U₇? I did it, and it took me no more than 5 minutes with a calculator.

EDIT: Or U₀ to U₆, as Dick suggested, which entails less work. In fact you can even test U_-3 to U₃, and that would involve the smallest numbers of all, and still give a complete result.

 

[Dick]

Yes. So if we know which n produces U_n that is divisible by 7, then U_n+7 will also be divisible by 7 and they will form arithmetic sequence. Do i get your hint correctly?

Then how to find the first n?

Thanks

There may not be any n. Check each point that might be a start of the arithmetic sequence. Like I said before, try n=0,1,2,3,4,5,6.

 

[songoku]

Why not simply try testing U₁ to U₇? I did it, and it took me no more than 5 minutes with a calculator.

EDIT: Or U₀ to U₆, as Dick suggested, which entails less work. In fact you can even test U_-3 to U₃, and that would involve the smallest numbers of all, and still give a complete result.

There may not be any n. Check each point that might be a start of the arithmetic sequence. Like I said before, try n=0,1,2,3,4,5,6.

Ah, I think I get it. Thanks a lot for the help