**What are the last two digits of 7**

^{2008}(1) 21 (2) 61 (3) 01 (4) 41 (5) 81

Solution follows here:

Solution:

Let us observe the pattern of last two
digits for lower powers of 7:

7
^{1} = 07 |
7
^{5} = 16807 |

7
^{2} = 49 |
7
^{6} = 117649 |

7
^{3} = 343 |
7
^{7} = 823543 |

7
^{4} = 2401 |
7
^{8} = 5764801 |

Observe
that it is following a pattern--- 07,49,43,01, this repeats for every 4 power
increments.

Given
exponent 2008, which is multiple of 4 and hence the last two digits are “01”

**Answer(3)**

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