Remainder theorem

[Guest]

Find the remainder of 21^84 / 324 using remainder theorem.

[shyamjayakannan]

[tex]\frac{21^{84}}{324}=\frac{3^{84}\times7^{84}}{2^2\times3^4}=\frac{3^{80}\times7^{84}}{4}[/tex]. Now, [tex]3^n[/tex] and [tex]7^n[/tex] alternate between numbers of the form [tex]4k+1[/tex] and [tex]4k+3[/tex], where [tex]k,n\in\mathbb N[/tex].

[tex]3^n=4k+3[/tex] when [tex]n[/tex] is odd and [tex]4k+1[/tex] otherwise and it is the same for 7.

So, [tex]\frac{3^{80}\times7^{84}}{4}=\frac{(4k_1+1)(4k_2+1)}{4}=\frac{16k_1k_2+4k_1+4k_2+1}{4}=\frac{4k_3+1}{4}[/tex], where [tex]k_3=4k_1k_2+k_1+k_2[/tex].

Hence, the remainder is 1.