2026-04-26

Problem — 2026-04-26

Easy

Easynumber_theory

What is the remainder when 21002^{100} is divided by 77?

Enter an integer (0 to 6)

Show solution
By Fermat's little theorem, 26≡1(mod7)2^6 \equiv 1 \pmod{7}. Since 100=6⋅16+4100 = 6 \cdot 16 + 4, we have 2100=(26)16⋅24≡116⋅16≡16≡2(mod7)2^{100} = (2^6)^{16} \cdot 2^4 \equiv 1^{16} \cdot 16 \equiv 16 \equiv 2 \pmod{7}.