Online Course Discussion Forum
MCIII NT 5.25-5.29
I haven't solved any of these problems, however I do have some thoughts.
5.25: I used the law of cosines, getting a relationship of m and n which is to the power of 4, so I guess I'm probably on the wrong track.
5.26: I experimented with a few numbers, finding that a4=33, which works. I discovered that an+3 + (n+2)an = (n+3)an+2. I'm not quite sure how to proceed.
5.27: Since a b c d are all distinct, x-a and such are all distinct. meaning we need to find 4 distinct integers that multiply to 4. I believe it is 1, -1 , 2, -2. However I'm confused about how to represent x, and are abcd random numbers
5.28: This works for all n=1 I think. But I'm confused on how to continue. divisible can be written as 0 mod sth but I'm not sure what to do next
5.29: n=2 works, then I'm pretty much stuck...
Any hints would be appreciated!
5.25: Law of Cosines on angle A can give you some relation about $m$ and $n$. Remember they are integers, and if they appear on a denominator, they should be a factor of the numerator.
5.26: Try to figure out $a_{n+2}-a_{n+1}$.
5.27: If you have got the values $1, -1, 2, -2$, and don't know which is which, one thing to try is to add them up, because the sum would be the same.
5.28: This is a practice on factoring formulas. $n$ can be any value, but not all pairs of $(m,n)$ can work.
5.29: Consider a prime factor $p$ of $2n-1$. $p$ must also be a factor of either $n-2$ or $n+1$, but note that $(n-2)+(n+1)=2n-1$, and what would that say about $p$?
5.25: I'm a little confused-- I used the law of cosines on angle B and got 2 equations. dividing out cosB, I found \( \frac{33^2+m^2-21^2}{(33-n)^2+(m-n)^2+n^2}=\frac{33m}{(33-n)(m-n)} \), which is really complicated. When I tried using the law of cosines on A, I found 21^2+33^2-m^2=2x21x33xcosA, which doesn't seem to form any relationships between n and m.
5.26: an+2−an+1=(n+2)(an+1-an) I'm not quite sure what to do/ how to solve this equation though. BTW, where will these equations be covered in math comp learning? I seem to be really bad at solving equations like these.
5.27: I still didn't quite understand. If I know the roots are 1, -1, 2, -2, wouldn't I know that {a,b,c,d}={1,2,3,4}? Then abcd would be determined. Regarding adding the roots together, we would get abc+abd+acd+bcd=0, using Veita's Theorom. However I'm not quite sure what to do next.
5.28: I think I understand the question now. Could you please give me another hint?
5.29: Since p is a factor of 2n-1, and is also a factor of either n-2 or n+1, then p must be factors of both n-2 and n+1 , which means p<=3. Since 2n-1 is odd, p isn't 1, leaving us with 3 and 1. 2n-1=3, n=2
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