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[Salvadoramerican]

The U-Knee-Cue tribe uses a special base to solve their math
problem. Using their special bases they found that 16x -70=0 and 13x-60=0. What base are they using
? ans 15

Larry, Moe, Curly and 4 other rascals are randomly arrange in the 1st row of seven desks. What is the probability that Larry, Moe and Curly will be seated in no specific order next to each other?

I moved to the old forum

[nsguy1350]

I solved a similar problem to the base question, but I don't see how the method would work with these numbers.

For the Larry/Moe/Curly problem, you have 7 desks. You need 3 consecutive desks. There are 5040 (7!) arrangements that the three could be in. There are 5 ways, excluding order, that the three could be seated when next to each other (123,234,345,456,567). If you multiply that by 6, for the numebr of total permutations for each position, you get 30 possiblities. This gives 30/5040, or 1/168 probability.

Is that correct? I have my doubts, but that's what comes to mind.

[Salvadoramerican]

I think i figure it out
We have 16x-70=0 and 13x-60=0, where 16, 70, 13, and 60 are numbers in
some unknown base. If we replace each
number with an expression in terms of the base, b, we have a system of
equations in two variables,

(b + 6)x - 7b = 0
(b + 3)x - 6b = 0

In order to eliminate x, we multiply the first equation by (b+3) and
the second by (b+6), and subtract:

(b + 3)(b + 6)x - 7b(b + 3) = 0
(b + 3)(b + 6)x - 6b(b + 6) = 0

6b(b + 6) - 7b(b + 3) = 0

This simplifies to

6b^2 + 36b - 7b^2 - 21b = 0
-b^2+15=0
then
-b(b-15)=0
the base has to be 15 not 0 if we find the zeros
If anyone denies it tell me

[Salvadoramerican]

also no its says 14.3%

[nsguy1350]

Oh, I was treating 16, 70, 60, and 13 as base 10. How did I not see that? Also, if my solution was wrong for the other problem, I don't know.