Areteem: MCII-B Number Theory 2.28, 2.29

Online Course Discussion Forum

MCII-B Number Theory 2.28, 2.29

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.

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?

Show parent

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!

Show parent