Hw question answer is 21 We have to find N(mod100)=>N(mod25) and N(mod4) Therefore, N is congruent to 1(mod4)=>N=4a+1. And N is congruent to 21(mod25)=>N=25b+21. Equating the values of N we get, 4a+1=25b+21. Taking (mod4) both sides we get, b is congruent to 0(mod4)=>b=4k putting this value on N=25b+21 we get, N=100k+21 Therefore, the the last two digit of N is 21.
Hw question answer is 21
We have to find N(mod100)=>N(mod25) and N(mod4)
Therefore, N is congruent to 1(mod4)=>N=4a+1.
And N is congruent to 21(mod25)=>N=25b+21.
Equating the values of N we get, 4a+1=25b+21. Taking (mod4) both sides we get, b is congruent to 0(mod4)=>b=4k putting this value on N=25b+21 we get, N=100k+21
Therefore, the the last two digit of N is 21.
Sir please give more assignment or share some ebooks from where u give assignment to the prime batch students
Ans kya he hw question ka 21??
Sir one video that tells about IOQM for class 12
Sir apki book kab aa rahi hai...
a gayii4
7:28
Sorry to ask this sir !!
But why are you coughing all the time during the class u should be taking care of your throat right?
Sir wrong pdf please check