What are the last 2 digits of 7 to the power of 2018. This question was on a past exam paper.

49

Step-by-step explanation:

Try finding pattern of last two digits of 7 to the power of some numbers.

7^1=07

7^2=49

7^3=343

7^4 = 2301

7^5 = 16807

so, the pattern is, 07,49,43,01 so, if we start writing all numbers from 1 to 2018, four number in each line, 2018 will fall in the second column. so it will have 49 as the last 2 digits [by seeing this pattern : 07,49,43,01 ]