Proof #1 , Word problem #1

[mathwannabe]

Hello everybody

I got the problem that says:


1) Prove that for every \(\displaystyle m\in N\) , \(\displaystyle m^5-m\) is divisible by 30.

I got to:

\(\displaystyle m^5-m = m(m^4-1)=m(m^2-1)(m^2+1)=m(m-1)(m+1)(m^2+1)\)

From here it's obvious to me that for every \(\displaystyle m\in N\) , \(\displaystyle m^5-m\) is going to be evenly divided by 30. But, how do I prove it?


2) 65 workers can dig a particular canal for 23 days. After 15 days of work, 13 workers leave the job. How much time do the rest of the workers need to finish the canal?

This is my workflow:

\(\displaystyle 65\) workers, \(\displaystyle 23\) days, \(\displaystyle 1\) canal

\(\displaystyle 65\) workers, \(\displaystyle 1\) day, \(\displaystyle \dfrac{1}{23}\) canal

\(\displaystyle 1\) worker, \(\displaystyle 1\) day, \(\displaystyle \dfrac{\frac{1}{23}}{65}=\dfrac{1}{1495}\) canal

After \(\displaystyle 15\) days, \(\displaystyle \dfrac{15}{23}\) of canal is finished. That leaves the remaining \(\displaystyle 52\) workers to finish the remaining \(\displaystyle \dfrac{8}{23}\) of canal.

So:

\(\displaystyle 52*\dfrac{1}{1495}*x=\dfrac{8}{23}\)

\(\displaystyle \dfrac{52}{1495}x=\dfrac{8}{23}\)

\(\displaystyle x=\dfrac{8}{23}*\dfrac{1495}{52}\)

\(\displaystyle x=10\)

The rest of the workers need 10 days to finish the remainder of the canal. Is this correct? Is there some other way to do this?

 

[mathwannabe]

I suspect that there are multiple ways to prove this.

First, the proposition is obviously true if n = 1.

Second, consider the sequence (m - 1), m, (m + 1), at least one of those numbers must be divisible by 2, and exactly one must be divisible by 3. So m^5 - m is divisible by 6.

\(\displaystyle \exists\ k \in N\ such\ that\ (m - 1) \le 5k \le (m + 3).\)

If 5k = (m - 1) or m or (m + 1), 5 is also a factor of m⁵ - m, which is consequently divisible by 30.

Assume 5k = (m + 2).

\(\displaystyle So\ m = 5k - 2 \implies m^5 - m = (5k - 2)^5 - m = \)

\(\displaystyle \displaystyle \left[\left(\sum_{i = 0}^4\binom{5}{0}2^ik^{(5 - i)}\right) - 32 - (5k - 2)\right]\), which is divisible by 5.

Assume 5k = (m + 3)

\(\displaystyle So\ m = 5k - 3 \implies m^5 - m = (5k - 3)^5 - m = \)

\(\displaystyle \displaystyle \left[\left(\sum_{i = 0}^4\binom{5}{0}3^ik^{(5 - i)}\right) - 243 - (5k - 3)\right]\), which is divisible by 5.

If a number is divisible by both 5 and 6, it is divisible by 30.

Edut: I assume there is a more elegant way to prove this, presumably by induction or reductio ad absurdum, but I just took an ax to it. Very ugly. Subhotosh will probably show you how to do it right.

Anyway, I value this reply highly. In other words, I'm "getting" what the problems of this kind are all about. Thanks Jeff. Also, looking forward to hear from Mr. SK.

 

[mathwannabe]

I am trying not to imagine what kind of mind dreamt this problem up.

By the way, this problem is in the official first year high school math text book in my country. It's in the section discussing rational algebraic expressions. :lol: I guess they advise kids who solve this on their own to switch schools and go to math gymnasium.

 

[mathwannabe]

This problem would cause a riot among students and parents in most schools in the US. It took me quite a while to dope out a route to the proof. I played around with a bunch of ideas before getting out the ax.

When is your exam?

My exam is in 3 months (+- 5 days).

I have progressed immensely in the short amount of time since I started studying (around the same time I joined this forum).

To think I couldn't simplify \(\displaystyle \dfrac{4x+1}{16x^2-1}\)... Speaks a lot about how much progress I have made. I'm getting confident with this stuff. Currently I'm heading on to quadratic equations and exponential equations, after that trigonometry will be my major problem.

 

[lookagain]

1) Prove that for every \(\displaystyle m\in N\) , \(\displaystyle m^5-m\) is divisible by 30.

I got to:

\(\displaystyle m^5-m = m(m^4-1)=m(m^2-1)(m^2+1)=m(m-1)(m+1)(m^2+1)\)

From here it's obvious to me that for every \(\displaystyle m\in N\) , \(\displaystyle m^5-m\) is going to be evenly divided by 30. But, how do I prove it?

.


\(\displaystyle m^5 - m \ =\)

\(\displaystyle m(m^4 - 1) \ = \)

\(\displaystyle m(m^2 - 1)(m^2 + 1) \ = \)

\(\displaystyle m(m - 1)(m + 1)(m^2 + 1) \ = \)

\(\displaystyle (m - 1)(m)(m + 1)[(m^2 + 5m + 6) \ + \ (-5m - 5)] \ = \)

\(\displaystyle (m - 1)(m)(m + 1)[(m + 2)(m + 3) \ - \ 5(m + 1)] \ = \)

\(\displaystyle (m - 1)(m)(m + 1)(m + 2)(m + 3) \ - \ 5(m - 1)(m)(m + 1)^2 \)


Refer to each of these two products as "first" and "second,"
respectively.


The first and second are each divisible by 2, because each is the
product of at least two consecutive integers.


The first and second are each divisible by 3, because each is the
product of at least three consecutive integers.


Therefore, the first and second are each divisible by 6.


The first is divisible by 5, because it is the product of five
consecutive integers.


The second is divisible by 5, because 5 is one of the factors.


Then the first and the second, each, are divisible by 30,
because each is divisible by 6 and by 5. So each is a
multiple of 30.


And then the difference of the first and second is divisible
by 30, because the difference of two multiples of 30 is
itself divisible by 30.


\(\displaystyle Thus, \ \ m^5 - m \ \ is \ \ divisible\ \ by \ \ 30.\)

 

[mathwannabe]

Ha ha, bravo lookagain. Looks like you really did look again XD