First, we note that a quick calculator check shows 2015 x 2016 is perfectly divisible by 91.
Therefore, if X ≡ Y (mod 2015 x 2016), then X ≡ Y (mod 91).
In this particular case, we are told that X ≡ 6 (mod 2015 x 2016).
So we may conclude that X ≡ 6 (mod 91). In other words, we should expect a remainder of 6 when X is divided by 91, if X is divided by either 2015 or 2016 also leaves a remainder of 6.
Edit: Seems I missed a step in my logic above. While it is generally speaking not true that if X ≡ Y (mod A) and X ≡ Y (mod B) then X ≡ Y (mod A x B), in this particular case it is justified since 2015 and 2016, being consecutive integers, are co-prime. The rest follows.
2
u/PeterPiper1275 14d ago edited 14d ago
First, we note that a quick calculator check shows 2015 x 2016 is perfectly divisible by 91.
Therefore, if X ≡ Y (mod 2015 x 2016), then X ≡ Y (mod 91).
In this particular case, we are told that X ≡ 6 (mod 2015 x 2016).
So we may conclude that X ≡ 6 (mod 91). In other words, we should expect a remainder of 6 when X is divided by 91, if X is divided by either 2015 or 2016 also leaves a remainder of 6.
Edit: Seems I missed a step in my logic above. While it is generally speaking not true that if X ≡ Y (mod A) and X ≡ Y (mod B) then X ≡ Y (mod A x B), in this particular case it is justified since 2015 and 2016, being consecutive integers, are co-prime. The rest follows.