What are the possible values of m in the equation 3m + 3n - 8m - 4n! = 680?

[timetraveller123]

Homework Statement

this problem came out in the math olympiad i took today and i got completely wrecked by this

consider the following equation where m and n are positive integers:

3^m + 3ⁿ - 8m - 4n! = 680
determine the sum all possible values of m:

Homework Equations

not sure which

The Attempt at a Solution

no attempt my mind is completely blank give me some hints please

 

[scottdave]

Try moving everything with m to one side and n to the other side. What can you say about the values of the m-function and n function ?

 

[BvU]

Were you allowed to use a calculator ?

    n: 1 2 3 4 5 6
  m 1 -6 -4 -2 -20 -242 -2156
    2 -8 -6 -4 -22 -244 -2158
    3 2 4 6 -12 -234 -2148
    4 48 50 52 34 -188 -2102
    5 202 204 206 188 -34 -1948
    6 680 682 684 666 444 -1470

 

[timetraveller123]

@BvU
no calculator

@scottdave
that m is bigger than n?

 

[BvU]

@BvU
no calculator

Just kidding. But from the brute force results you should be able to extract some restrictions on the (m,n) pairs to check. Especially in combination with a list of powers of 3:
3
9
27
81
243
729

and a list of 4n! helps a lot too!
4
8
24
96
480
2880
These don't take long to set up, even without a calculator.

subsequent hypotheses:

• one of the two has to be 6
• m has to be 6, n has to be smaller than 5

I think the hard part is to find the one (6,1) combination that does yield 680. Then 'conclude' that there can be no others.

 

[timetraveller123]

@BvU
not to offend you but i am not really fan of brute force approaches

and why do you say n has to be 6

based on what scottdave said

i inferred m > n is this true
and if it is does it help to write m = k + n

 

[Buffu]

Homework Statement

this problem came out in the math olympiad i took today and i got completely wrecked by this

consider the following equation where m and n are positive integers:

3^m + 3ⁿ - 8m - 4n! = 680
determine the sum all possible values of m:

Homework Equations

not sure which

The Attempt at a Solution

no attempt my mind is completely blank give me some hints please

$3^m + 3^n = 680 + 8m + 4n!$ or $3^m + 3^n = 4(170 + 2m + n!)$

Since $170 + 2m + n!$ is divisible by $3$ what does this tell you about parity of $170 + 2m + n!$ ? and for what values of $n$ it is possible ?

 

[BvU]

No offence taken. I'm just interested, and to be sure: I don't know the proper way to deal with this one, so I just did trial and error. 4n! goes real fast as you can see, so n can't be 6. You need at least one of the two with a value of 6 to get over 680 .

I hadn't pulled out the factor 4 Buffu mentions. Hats off !

 

[timetraveller123]

oh wow that is interesting so 2m + n! ≡ 1 mod3

if n≥3 then m must be multiple of 3

if n <3 then 2m ≡ 2 mod 3

am i getting somewhere

 

[timetraveller123]

@BvU
and more over i am a really careless boy i am sure to make lots of mistakes i probably lost 8 questions just by careless today so that's another reason i did not opt for your method without a calculator so you anyways trial and error to estimate bounds and restrictions is always uselful

 

[Buffu]

oh wow that is interesting so 2m + n! ≡ 1 mod3

if n≥3 then m must be multiple of 3

if n <3 then 2m ≡ 2 mod 3

am i getting somewhere

You can do what you are doing but an easy way would be to check if $170 + 2m + n!$ is odd or even ?

 

[timetraveller123]

oh is n't that always even for n bigger than 2

but what does that tell you the left hand side is also always even

 

[Buffu]

oh is n't that always even for n bigger than 2

but what does that tell you the left hand side is also always even

But we want it odd, so what values of $n$ is available to us ?

 

[timetraveller123]

why do we want it odd? please i am beginner in olympiads

but anyways the answer to your question is n =1

edit :

oh sorry now i get why it must be odd because it is divisible by 3 sorry for being so dumb

so n= 1 to give 170 + 8m + 1! to be odd

3^m = 680 + 8m + 1

so m must be at least 6 and so is it only m = 6 solves it

that is very brilliant method there how did you get to it did you just play around with the expression until you got something useful

 

[Buffu]

oh sorry now i get why it must be odd because it is divisible by 3 sorry for being so dumb

Nooo, It is wrong, since even multiples of 3 are there :).

I said that expression is odd because $3^n + 3^m$ is not divisible by 8 for any value of $m,n$ * and if $170 + 8m + n!$ is even, then $3^m + 3^n$ must be divisible by $8$.

* $3^n \mod 8 = 1, 3$ using power rule for modulus and the fact that $3^n = 1$ for $n = 0$.

So $3^n + 3^m \mod 8 = 2, 4, 6$ hence never divisible by 8.

 

[Buffu]

that is very brilliant method there how did you get to it did you just play around with the expression until you got something useful

I remember some tricks from olympiads when I used to give those.

 

[timetraveller123]

oh okay now i get it thanks for the help