1. May 30, 2017

vishnu 73

• Member warned that some attempt must be shown
1. The problem statement, all variables and given/known data
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:

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

2. Relevant equations
not sure which

3. The attempt at a solution
no attempt my mind is completely blank give me some hints please

2. May 30, 2017

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 ?

3. May 30, 2017

BvU

Were you allowed to use a calculator ?
Code (Text):

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

4. May 30, 2017

vishnu 73

5. May 30, 2017

BvU

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.

6. May 30, 2017

vishnu 73

@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

7. May 30, 2017

Buffu

$3^m + 3^n = 680 + 8m + 4n!$ 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 ?

8. May 30, 2017

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 !

9. May 30, 2017

vishnu 73

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

10. May 30, 2017

vishnu 73

@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 ya anyways trial and error to estimate bounds and restrictions is always uselful

11. May 30, 2017

Buffu

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

12. May 30, 2017

vishnu 73

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

13. May 30, 2017

Buffu

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

14. May 30, 2017

vishnu 73

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

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

3m = 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

Last edited: May 30, 2017
15. May 30, 2017

Buffu

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.

16. May 30, 2017

Buffu

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

17. May 31, 2017

vishnu 73

oh okay now i get it thanks for the help