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MCII-B Number Theory 2.28, 2.29

 
 
ZhangDaniel的头像
MCII-B Number Theory 2.28, 2.29
ZhangDaniel - 2024年08月14日 Wednesday 20:11
 

I'm struggling to understand question 2.28...it's just really hard for me to understand the question. I kind of get it, but I don't know where I could start proving part a), let alone part b). What should I start with? 

For 2.29, tried expanding bar(abcd) and bar(bcde), giving me 1000a+1100b+110c+11d+e, but I'm not sure how I could use this.

 
LensmireJohn的头像
Re: MCII-B Number Theory 2.28, 2.29
LensmireJohn - 2024年08月15日 Thursday 11:09
 

For 2.28:

  • For (a) you can assume that the fact gcd(b,a) = gcd(b-a, a) is already known. How does this help prove the result of the problem?
  • For (b), the proof will look similar to Example 2.8. Don't stress too much about this one, but if you want to try, the hint is to let d = gcd(b,a) and e = gcd(a,r) and try to prove that d <= e and then e <= d.

For 2.29 you're off to a good start. Could we use something like the divisibility rule for 11 to help here? Do we automatically know some parts of this are divisible by 11?

ZhangDaniel的头像
Re: MCII-B Number Theory 2.28, 2.29
ZhangDaniel - 2024年08月15日 Thursday 13:44
 

For 2.29, 1100b, 110c, and 11d are all already divisible by 11. We need 1000a+e to be a multiple of 11...Oh! I think I know how to do it! Thank you!