What is the last digit of the number 537^1279089028?

This question is about cyclicity of the powers of 7.

Units place of 7^1 = 7,

Units place of 7^2 = 7* 7 = 9,

Units place of 7^3 = 9 * 7 = 3,

Units place of 7^4 = 3 * 7 = 1

Units place of 7^5 = 1 * 7 = 7.

Hence the units place is repeating after 4 powers. Hence cyclicity of powers of 7 is 4, and the units place will be (7,9,3,1) depending on whether the remainder of the power when divided by 4 is (1,2,3,0) .

Now we have to find the remainder when 1279089028 is divided by 4. That will be equal to the remainder when 28 is divided by 4. = 0. Hence the units place is 1.

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