Question

2^{51} mod 22 in words, two to the power of fifty-one mod twenty-two

Answer 1

Since 2⁵ = 32, and

2⁵ ≡ 32 ≡ 10 (mod 22),

we have

2⁵¹ ≡ 2 • 2⁵⁰ ≡ 2 • (2⁵)¹⁰ ≡ 2 • 10¹⁰ (mod 22)

Now consider 10¹⁰ (mod 22):

10 = 2 • 5

10¹⁰ ≡ 2¹⁰ • 5¹⁰ ≡ (2⁵)² • 5¹⁰ ≡ 10² • 5¹⁰ ≡ 2² • 5¹² (mod 22)

so that

2⁵¹ ≡ 2³ • 5¹² (mod 22)

Now consider 5¹² (mod 22):

5 and 22 are coprime, and ɸ(22) = 10 (where ɸ(n) is the Euler totient function). By Euler's theorem,

5¹² ≡ 5² • 5¹⁰ ≡ 5² • 1 ≡ 25 ≡ 3 (mod 22)

and so

2⁵¹ ≡ 2³ • 3 ≡ 24 ≡ 2 (mod 22)

Another, more tedious method: Start with smaller powers of 2 and look for a pattern.

2 ≡ 2 (mod 22)

2² ≡ 4 (mod 22)

2³ ≡ 8 (mod 22)

2⁴ ≡ 16 (mod 22)

2⁵ ≡ 32 ≡ 10 (mod 22)

2⁶ ≡ 2 • 32 ≡ 2 • 10 ≡ 20 (mod 22)

2⁷ ≡ 2 • 20 ≡ 40 ≡ 18 (mod 22)

2⁸ ≡ 2 • 18 ≡ 36 ≡ 14 (mod 22)

2⁹ ≡ 2 • 14 ≡ 28 ≡ 6 (mod 22)

2¹⁰ ≡ 2 • 6 ≡ 12 (mod 22)

2¹¹ ≡ 2 • 12 ≡ 24 ≡ 2 (mod 22)

2¹² ≡ 2 • 2 ≡ 4 (mod 22)

and so on, with a cyclic pattern of length 10. That is, $2^{10k+1}\equiv2\pmod{22}$ for any integer k ≥ 0. So 2⁵¹ ≡ 2 (mod 22).