alex.gellatly wrote:If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
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To determine the remainder when 3^(8n+3) + 2 is divided by 5, we need to know the UNITS DIGIT of 3^(8n+3) + 2.
Let n=1.
Then 3^(8n+3) = 3^11.
To determine the units digit of an integer raised to a GREAT power, examine the resulting units digits when the integer is raised to SMALLER powers.
3^1 = 3.
3^2 = 9.
3^3 = 27.
3^4 = 81.
3^5 = 243.
The units digits repeat in a CYCLE OF 4:
3,9,7,1...3,9,7,1...
Thus, when 3 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, the units digit of 3^12 is 1.
Since 3^11 is the preceding value in the cycle, the units digit of 3^11 is 7.
Thus, the units digit of 3^(8n+3) + 2 is 9.
When an integer with a units digit of 9 is divided by 5, the remainder is 4:
19/5 = 3 R4.
29/5 = 5 R4.
39/5 = 7 R4.
The correct answer is
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