CAT 1998 — QA Question 29

Since 2 has a cyclicity of 4,
i.e. 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64 ..., the last digits (2, 4, 8, 6) are in four cycles.
∴ On dividing $\frac{51}{4}$, we get the remainder as 3.

∴ The last digits has tobe 2³ = 8

Shortcut:
Since cyclicity of the power of 2 is 4, so 2⁵¹ can be written in 2^{4(12) + 3} or unit digit will be 2³ = 8.