Online Course Discussion Forum

MCIII NT 5.19-5.24

 
 
Picture of Katrina Liu
MCIII NT 5.19-5.24
by Katrina Liu - Saturday, April 22, 2023, 9:08 AM
 

For 5.19, I'm not sure about where to start: I don't know how to use Bezout's Identity
For 5.20: I'm guessing that I should factor it out? But how could I find its factorization?
For 5.21: 
First prove that there are at least 2 2's, we do this by trying out evens and odds-- if all are even, all of there differences are even as well; one odd, the three evens have three even differences; 2 odd, evens and odds have even differences respectively; 3 odd, three even differences; all odd, all even differences. Therefore, there exists at least 2^2 in the polynomial.
Then we prove that there is at least one factor of three: first, to not have a difference divisible by 3, every difference must be different mod 3. If there is for example, b-a and c-a =1 mod 3, then b-c=0 mod 3. Therefore, all six differences must be different mod 3 for there to not exist a factor of 3. However, a number can only = 0, 1, 2 mod3. Therefore, there is at least one factor of 3
Therefore 12 divides the polynomial

For 5.22: I don't know where to start
For 5.23:  I don't know where to start
For 5.24: 13 is not a very random number, the divisibility rule tells us that a number is divisible by 13 if the last three digits minus the first three digits is divisible by 13. I know how to count the total amount of numbers, however I do not know how to proceed.

Any hints would be appreciated!

 
Picture of Dr. Kevin Wang
Re: MCIII NT 5.19-5.24
by Dr. Kevin Wang - Tuesday, April 25, 2023, 12:59 PM
 

Here are some hints.

5.19: If a number $a$ is divisible by $k$, then you can write $a=xk$ where x is an integer. Bezout's identity is not required for this one.

5.20: To find out about 169, probably you can start with 13.

5.21 is okay, write it clearly in your homework.

5.22: There are at least two approaches that can work. First method: when powers of a number is involved in divisibility or remainder questions, Fermat's Little Theorem is often useful.  Note that $30=2\times3\times5$.  Second method: can you factor $a^5-a$, and then use the same method as in 5.21?

5.23: Consider pairing.

5.24: Do not just memorizing the rules without understanding why.  The divisibility rule for $13$ in the textbook is based on the fact that $13$ is a factor of $1001$.  Consider pairing again.