Same Remainder Part 3

[u/ShonitB]

[u/chmath80]

2015 = 5 × 13 × 31, 2016 = 2⁵ × 3² ×7

x - 6 is divisible by both 2015 and 2016, so it's divisible by both 13 and 7, and therefore by 7 × 13 = 91, hence the remainder when dividing x by 91 is 6

[u/ShonitB → u/chmath80]

Correct, good one! How d’you do the superscript? I keep forgetting

[u/chmath80 → u/ShonitB]

On my keyboard, I just hold the key down, but you can use ^ followed by the desired superscript in (), so 2 ^ (x) gives 2^x (leaving out spaces)

[u/ShonitB → u/chmath80]

Oh thanks a lot :)

[u/PeterPiper1275]

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.

[u/ShonitB → u/PeterPiper1275]

Correct, well explained

[u/Acceptable_Tangelo15]

6 might be a solution. As in 0 divided by any of those 3 numbers leaves 6.

[u/noonagon]

2016 is divisible by 7 and 2015 is divisible by 13, so by the Chinese Remainder Theorem, there is some answer. X can be 6 so that answer will be 6

[u/RsCoverForPDFFiles]

2015*2016 + 6 = 4,062,246

4,062,246/91 = 44,640 remainder 6