A help here please

How many positive integers nn satisfy the condition 2n+n8n+n 2^n + n \mid 8^n + n?

Note by A Former Brilliant Member
2 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Log in to reply

I tried. cant get a clue to start

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

Since we have a Diophantine equation, which approach can we use?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

through brute force 1,2 come. but what canbe a logical way to do this?

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

i tried dat way too. I believe i am leaving a minute detail. Can ya please send me a solution?

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

You can write up your solution and obtain feedback from the community on how to improve it.

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

@CalvinLin-this was my approach. If 2^n+n devides 8^n+n then it devides 8^n+n-2^n+n.I wanted to eliminate n. Now i arrive at finding 2^n+n devides (2^n)(2^n+1)(2n-1). Here i am stuck to find the answer

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

That's a good start. Actually, 8n 8^n is "more complicated" than nn. How can we "eliminate exponents" to only obtain polynomials?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

i tried that too. I multiplied 2^2n to 2^n+n and subtracted 8^n+n from that and proceeded as above. What i dont get is what to do next.

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

What is the equation that you get? How can we simplify it further?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

(2^2n)n-n|2^n+n

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

Great. So once again, " 22n×n 2^{2n} \times n is more complicated than 2n 2^n ". How can we simplify further?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

(2^n-1)(2^n-1)|2^n+n

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

@Calvin Lin-can you make me understand how to get a polynomial from this?

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

You already have the ideas, but just are not executing on them. You saw how to "simplify" 8n 8^n into other terms, and just need to "simplify" these other terms into polynomials.

IE We have a 22n(×n) 2^{2n} ( \times n) term. How can we get rid of it?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

i did it already, multiplied and subtracted to to get (2^n-1)(2^n+n )|2^n+n. all i dnt know is how to get a polynomial out of it

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

@Calvin Lin-i did it already, multiplied and subtracted to to get (2^n-1)(2^n +n )|2^n+n. all i dnt know is how to get a polynomial out of it

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

Are you sure the RHS is 2^n+n? That seems to be the LHS. At this point in time, I'm not certain what you're doing.

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

@Calvin Lin-2^n+n|(2^n-1)(2^n +n )

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

Can you recheck? You are saying ABA A \mid BA where A=2n+n A = 2^n + n .

This doesn't quite look right to me. I now do not know what you are doing. Please list out your steps carefully, and do not just jump to the final step.

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

@Calvin Lin-sir, please wait because i need to get the answer today. I dnt have anyone here else to help me out.

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

@Calvin Lin-so sorry! it is 2^n+n| 2^n(2^n+1)(2^n-1)

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

That's a good start. Actually, 8n 8^n is "more complicated" than nn. How can we "eliminate exponents" to only obtain polynomials?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

@Calvin Lin-i really have no idea how to go further. Could you please post a solution to this?

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

2n+n8n+n2n+n(2n+n)×4nn×4n+n2n+nn×4nn 2^n + n \mid 8^n + n \Leftrightarrow 2^n + n \mid (2^n + n ) \times 4^n - n \times 4^n + n \Leftrightarrow 2^n + n \mid n \times 4^n - n was what I thought you said you were doing earlier.

So we were able to remove the 8n 8^n and replaced it with a n×4n n \times 4 ^n . How can we remove the 4n 4^n now to further simplify the statement?

Calvin Lin Staff - 2 years, 9 months ago

Log in to reply

i did it already. please post the entire solution.i am quite...disturbed by this probelem

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

@Calvin Lin- could you please give an entire solution to this?

A Former Brilliant Member - 2 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...