remainder

shortest method fir this
No description
22 Replies
iTeachChem Helper
@Apu
iTeachChem Helper
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Opt
Opt3w ago
Oh crap they're not coprime. They share 7 as a common prime factor @SirLancelotDuLac you're the NT expert here.
Opt
Opt3w ago
Oh, one more thing, the modulo 35 for 7, goes in cycles of 5, and 17 in cycles of 3 What's the index for your second line? Modulo what number?
SirLancelotDuLac
mod 5
Opt
Opt3w ago
CRT keeps tripping me up.
hardcoreisdead
hardcoreisdeadOP3w ago
how did mod 5 become mod 35 suddenly all clear upto second last line
SirLancelotDuLac
Ye sorry. we have this number congruent to 0 mod 7 which is 7 mod 7 and congruent to 2 mod 5 which is also 7 mod 5. So by CRT, the number is 7 mod(7*5) (as 7 and 5 are co-prime)
hardcoreisdead
hardcoreisdeadOP3w ago
CRT referes to chineese remainder theorem? can u state it once i am not aware of it
Opt
Opt3w ago
Yup.
Opt
Opt3w ago
No description
generic display name
What does (ni,nj)=1 mean How to interpret it?
Opt
Opt3w ago
It means the GCD is 1. They're coprime.
generic display name
Oh okay
Aetherfly
Aetherfly3w ago
is there some good video on this? they only taught us the bt method
SirLancelotDuLac
Yeah, in jee syllabus only binomial method is there. But if you are familiar with congruent arithmetic, you can look up books like pathfinder (Or some less detailed material if you want to see theory only for jee) or lectures to refer.
Aetherfly
Aetherfly3w ago
alright thanks
hardcoreisdead
hardcoreisdeadOP3w ago
how does the imply we can just multiply mod ni and mod nj values
SirLancelotDuLac
No, we can't multiply the values but if you can some some value a such that a1=a2=...an=k then you can say that x is congruent k modulo n1n2....
hardcoreisdead
hardcoreisdeadOP3w ago
so given number when divided by 7 gives remainder 7 , when divided by 5 gives 7 , so upon division by 7*5 is must give 7 too ?? x=a (mod n1 ) x=a (mod n2) then x=a (mod n1n2) given n1 and n2 are coprime correct?
Opt
Opt3w ago
Yup

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