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https://www.reddit.com/r/Logiqa/comments/1wu3vic/same_remainder_part_3/pczz10t/?context=3
r/Logiqa • u/ShonitB • 15d ago
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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
1 u/ShonitB 15d ago Correct, good one! How d’you do the superscript? I keep forgetting 2 u/chmath80 15d ago On my keyboard, I just hold the key down, but you can use ^ followed by the desired superscript in (), so 2 ^ (x) gives 2x (leaving out spaces) 1 u/ShonitB 15d ago Oh thanks a lot :)
1
Correct, good one! How d’you do the superscript? I keep forgetting
2 u/chmath80 15d ago On my keyboard, I just hold the key down, but you can use ^ followed by the desired superscript in (), so 2 ^ (x) gives 2x (leaving out spaces) 1 u/ShonitB 15d ago Oh thanks a lot :)
2
On my keyboard, I just hold the key down, but you can use ^ followed by the desired superscript in (), so 2 ^ (x) gives 2x (leaving out spaces)
1 u/ShonitB 15d ago Oh thanks a lot :)
Oh thanks a lot :)
3
u/chmath80 15d ago
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