A. Nastia and nearly Good Numbers | Codeforces Round

[@manthanvats9809]

bro what about 12 and 6....B is not equal to 1 but every no. that is divisible by 12 is divisible by 6, so we will end up having 2 good and 1 nearly good....

[@sKSama]

to be "good" it should divisible by A*B , not A,B separately

[@nemesis_rc]

please increase sound of your videos

[@sushyyyyyyyy]

bro, pls btao aapne ye socha kese??? mera dimag to chala hi nhi ese!!!! 😥😥😥😥😥 thoda tips do is direction mein sochne ke lie!!!!

[@sKSama]

Just played with the equation x+y=z on a paper, replaced X,y,z with different values X aur Y ke good aur nearly good values dala, dekha kaise kaam kar raha ussi se samaj aya Last mein A(1+B) equation bana tph samaj gaya fir

[@sushyyyyyyyy]

@@sKSama 🤩🤗 thanks bro...👍🏻

[@freecoder522]

Ok sir . I am with you ......

[@raunakkumar3073]

ye konsa IDE h plz batoo....

[@sKSama]

Vs code

[@sKSama]

kzbin.info/www/bejne/r2maoKNjjc-Bp6c

[@vijaypatil873]

What if A is divisble by B?

[@sKSama]

Dosent matter, read pinned comment once

[@MALIWAL1000]

Question thoda aur ache se frame karna chahiye tha. Ab change kiya hai author ne..

[@koushikbiswas8289]

Wrong answer on test 2

[@anweshakumar7838]

👌👌👌👌👌👌👌

[@sKSama]

PLEASE READ FULL- To be nearly good- it should be divisible by A, it can be divisible by B also, but not by A*B suppose A=4, B=2, 12 is nearly good as it is not divisible by A*B=8 ALSO, My ans- A.B, A, (A.B)+A When B is greater than 1 A.B< (A.B)+A < 2*A.B Therefore it it will be never be divisible by A.B as it lies btw 1*A.B and 2*A.B