yes, you get the remainder 1 from a of them and 0 from (5-a) of them since 3|(5-a) of them, I got this partThere are 5 squares of which 5−a are divisible by 3 and a of them have remainder 1 when divided by one. 1⋅a is the same as just a. From that we know that a has remainder of 1 when divided by 3, and since 0≤a≤5 we know that a can be either 1 or 4.
Does this make sense?
Correction: since we are talking about primes "divisible by 3" means "equal to 3" of course.There are 5 squares of which 5−a are divisible by 3 and a of them have remainder 1 when divided by one. 1⋅a is the same as just a. From that we know that a has remainder of 1 when divided by 3, and since 0≤a≤5 we know that a can be either 1 or 4.
Does this make sense?
Does the rest make sense too?yes, you get the remainder 1 from a of them and 0 from (5-a) of them since 3|(5-a) of them, I got this part
yeah I got it not, thanks for your helpDoes the rest make sense too?
To answer your question about the significance of 3 and 8: there is none, i.e., whatever works![]()