Question

The sides \(a\gt b\gt c\) of a triangle are integers such that \(3^a, ~3^b, ~3^c\) leave the same remainder when divided by \(10000\).

Find the minimum value of the perimeter of such a triangle.

Answer 1

made expressions for 3^a= 10000e+z
similar for rest of them using the same denominator z
then applied AM>GM i.e. (10000(e+f+g)+3z)/3 > 3^(a+b+c)/3
and minimum value of e+f+g should be zero when all are simultaneous zero => z=0 so equations simplifies to 0>3^(a+b+c)/3=> a+b+c<3

Answer 2

Thats a very interesting way to use AM>GM inequality ! I'm thinking when \(a\) is an integer, how can the remainder can be \(0\) ? because we have \(10000 \nmid 3^a\) for all \(a\)...

Answer 3

if e+f+g=0 then z/3 > 3^(a+b+c)/3 and least value of any remainder is 0

Answer 4

Ahh yes i still dont see how can we take advantage of that fact..

Answer 5

in GM term will form like a+b+c/3 (perimeter)

Answer 6

*

Answer 7

I guess an alternate way to phrase the question is: What are the three smallest powers of 3 to share the same first 4 digits?

Answer 8

Well, they will also have to obey the triangle inequality, so you can apply one more condition on top of that to check in case the numbers are far enough apart that this might fail \[\Large 3^a < 3^b+3^c\]

Answer 9

Correction: I meant to say last 4 digits in my first post there, I think you probably already knew that though. Throwing some ideas out there: Applying a few things and praying that a is even we can manipulate the inequality to look like this: \[\Large (3^ \frac{a-c}{2}+1)(3^ \frac{a-c}{2}-1) < 3^{b-c}\]
Another thing is that since the last 4 digits are the same, then the last digit is the same, so mod 3 means we cycle through 1,3, 9, and 7 so each exponent a,b,and c have the form \[\Large a=4\alpha + n \\ \Large b = 4 \beta + n \\ \Large c = 4 \gamma + n\] where they share the same n, some value from 0 to 3.

Answer 10

that looks like a good start ! but i think the second condition requires a bit tweaking as
\(a \lt b+c \iff 3^a\lt 3^b+3^c \) does not hold always

Answer 11

No, there's not a relationship between those. All I'm saying is we know \[\Large a>b>c \implies 3^a>3^b>3^c\] and since these are the sides of a triangle, it has to obey this inequality If \[\Large 3^a > 3^b+3^c\] then it would look like this

Answer 12

Using this information and plugging it into the inequality \[\Large 3^a < 3^b+3^c\]\[\Large a=4\alpha + n \\ \Large b = 4 \beta + n \\ \Large c = 4 \gamma + n\] we get a new equation \[\Large 81^\alpha < 81^\beta+81^\gamma\] which sort of might or might not help us hmm...

Answer 13

the sides are \(a,b,c\) not \(3^a,3^b,3^c\) right

Answer 14

oh you're right haha guess I better back track

Answer 15

I'm going to take a quick break to clear my brain since everything I just did was pretty much wrong but I can still use some of it, like we still know a>b+c it's just everything is backwards kinda... well not quite. Be back in 15 minutes.

Answer 16

Yes these are still valid
\[\Large a=4\alpha + n \\ \Large b = 4 \beta + n \\ \Large c = 4 \gamma + n\]

Answer 17

eh i miss something here :\

Answer 18

Perfect! thats very much similar to the solution I have. The most hard part of the problem is in finding the least power of \(3\) that yields \(1\) in mod \(10000\)
\[3^n\equiv 1\pmod{10000}\]