Number Theory Problem

Show that 199319^{93} - 139913^{99} is a positive integer divisible by 162.

NOTE:Actually I could not understand how to proceed. Can anyone show me the path ?

Note by Nishant Sharma
6 years, 4 months ago

No vote yet
3 votes

  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

What I had done is that I wrote it as (18+1)93(12+1)99(18+1)^{93}-(12+1)^{99} and now it is easy to see that a very few terms are left, after the binomial expansion, that are not divisible by 162162,i.e., 3423^4*2. And then it is very easy to show that the remaining terms add up to be a multiple of 162162 as well.

Shourya Pandey - 6 years, 4 months ago

Log in to reply

latex not loading proprly pls type it

superman son - 6 years, 4 months ago

Log in to reply

Another method is to note that 1327(mod162)13^2 \equiv 7 \pmod {162}.

Shourya Pandey - 6 years, 4 months ago

Log in to reply

We will have to prove that its divisible by 8181 and 22. But its obviously divisible by 22 and if we consider both numbers mod 9, it turns out that its divisible by 99.

Vikram Waradpande - 6 years, 4 months ago

Log in to reply

Got that it is divisible by 2 and 9. But we need to prove that it is divisible by 162. Anyways, I could not think of approaching it this way.

Nishant Sharma - 6 years, 4 months ago

Log in to reply

it goes for 81 also

superman son - 6 years, 4 months ago

Log in to reply

hey vikram bhaiya please mujhe yei problem latex chor kar likh do mere computer par load nehi ho raha latex

superman son - 6 years, 4 months ago

Log in to reply

show that 19^93 - 13^99 is a positive integer divisible by 162

Anoopam Mishra - 6 years, 4 months ago

Log in to reply

@Anoopam Mishra thanks

superman son - 6 years, 4 months ago

Log in to reply

right but first prove 19^93>13^99

superman son - 6 years, 4 months ago

Log in to reply

not needed to show the greater than sign.

Shourya Pandey - 6 years, 4 months ago

Log in to reply

@Shourya Pandey yes if we do bionomial then not but otherwise Show that 19^93 - 13^99 is a positive integer

superman son - 6 years, 4 months ago

Log in to reply

@Superman Son inequality korchhis bole sob jaegae inequality? positivity or negativity is not at all needed as the problem is asking to prove divisibility.

Soham Chanda - 6 years, 4 months ago

Log in to reply

@Soham Chanda hmm thikie bolecho go sohomda

superman son - 6 years, 4 months ago

Log in to reply

@Superman Son But why is it required?

Shourya Pandey - 6 years, 4 months ago

Log in to reply

@Shourya Pandey yeah i got my mistake

superman son - 6 years, 4 months ago

Log in to reply

A way to reduce the bulk of this problem is to use Euler's totient function. Also, it is inevitably evident that the given number is divisible by 22, so we have to worry about 8181 only.

After applying Euler's totient function, we only need to show that: 1939c(mod81)19^{39} \equiv c \pmod{81} and 1345c(mod81)13^{45} \equiv c \pmod{81} i.e 193919^{39} and 134513^{45} leave the same remainder namely "c""c" on division by 8181.

Aditya Parson - 6 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...