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Question

What are the last two digits of 72008?   (CAT 2008)
  1. 21
  2. 61
  3. 1
  4. 41


Solution

The correct option is C 1
The last two digits of a number is nothing but the remainder obtained when the number is divided by 100. The number can be rewritten as 2401502.
When 2401 is divided by 100, the remainder is 1. So, the last two digits of 72008 will be 1502 = 01.

Alternatively, :

Using the last 2 digit rule for 7 - Unit digit will be 1 (power cycle of 7). Ten's digit will be - 0×8 = 0. Hence, last two digits will be 01.

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