If n-m must be multiple of 100 (i.e. n-m≥100) and also m≥1, doesn't that mean that min(n+m) is for n=101 and m=1, thus min(n+m)=102? And obviously 101=n≥3, so that also checks out in this case, so isn't the answer 102 instead of 106 or am I missing something?
This factorization 1978^m(1978^n-m -1) made the solution so simple!
Good question, well explained. Thanks prof.
Explained nicely
Really great video
This was awesome
I learnt new things today
Thnx
10:53 nice
nice.
If n-m must be multiple of 100 (i.e. n-m≥100) and also m≥1, doesn't that mean that min(n+m) is for n=101 and m=1, thus min(n+m)=102? And obviously 101=n≥3, so that also checks out in this case, so isn't the answer 102 instead of 106 or am I missing something?
👏👏👏
106?
day 6969 of waiting for official discord server
nice.
Ow do you calculate what is congruent to 1978^4 in modulus 125? How’s you get (-22)^4
How did you set k = to -22^4?
How did you know that (1978)^4 is congruent to (-22)^4 mod (125) can you explain?
How to divide by zero
Your content is great but you have to work on your writing skills
Or you can just type and animate like @mind your decision