Discussion

Explanation:

Last 2 digits of

7¹ = 07
7² = last 2 digits of (07 × 7) = 49
7³ = last 2 digits of (49 × 7) = 43
7⁴ = last 2 digits of (43 × 7) = 01
7⁵ = last 2 digits of (01 × 7) = 07
7⁶ = last 2 digits of (07 × 7) = 49
7⁷ = last 2 digits of (49 × 7) = 43
7⁸ = last 2 digits of (43 × 7) = 01

As we can see, for every 4^th power of 7, the last two digits are 01.

Since 2008 is divisible by 4, we can conclude that last two digits of 7²⁰⁰⁸ are 01.

Hence, option (c).

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