Discussion
What is the unit digit of \( 7^{95} \)?
Answer: B
Correct answer: 3
The unit digits of the powers of 7 repeat in a cycle of four: \[ 7^{1} = 7,\quad 7^{2} = 49,\quad 7^{3} = 343,\quad 7^{4} = 2401, \] giving the sequence 7, 9, 3, 1 and then repeating. To find where the 95th power falls, divide the exponent by the cycle length: \( 95 \div 4 \) leaves a remainder of 3. The third term of the cycle is 3. A remainder of 0 would mean the last term of the cycle, not the first - that is the usual trap. The cycle lengths for the other digits are worth memorising: 2, 3, 7 and 8 all cycle in four; 4 and 9 cycle in two; and 0, 1, 5 and 6 always end in themselves. The same technique answers questions about the unit digit of a product or sum of large powers: find each unit digit separately, then combine them.
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