CAT Quant Practice Problems

Question: What are the last two digits of \(7^{2008}\)?
  1. 21
  2. 61
  3. 1
  4. 41
  5. 81

Correct Option:3

$7^4$ = 2401 = 2400+1
So, any multiple of $7^4$ will always end in 01
Since 2008 is a multiple of 4, $7^{2008}$ will also end in 01


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