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, 07,49,43,01so, 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 ]