If n = 67 then find the unit digit of [3^n+2^n]
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I think the answer is 15
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3^67
3^4 gives unit digit =1
Hence, (3^4)^16 = unit digit--1
3^67 = (3^4)^16 * 3^3 = (...1)* 27 = ...7
Hence, unit digit of 3^67 = 7
2^67
2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 =(1)6.... 2^5=_2... 2^6=_4..2^7 = _8...2^8= _6
Hence,the unit digit repeats after a cycle of '4'
2^67 = (2^4)^16 * 2^3 = (...6)* 8 = ....8
Hence unit digit of 2^67 = 8
Unit digit of (3^67 + 2^67) =...7 + ....8 = ....5
Hence unit digit = 5
Hope it helps
3^4 gives unit digit =1
Hence, (3^4)^16 = unit digit--1
3^67 = (3^4)^16 * 3^3 = (...1)* 27 = ...7
Hence, unit digit of 3^67 = 7
2^67
2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 =(1)6.... 2^5=_2... 2^6=_4..2^7 = _8...2^8= _6
Hence,the unit digit repeats after a cycle of '4'
2^67 = (2^4)^16 * 2^3 = (...6)* 8 = ....8
Hence unit digit of 2^67 = 8
Unit digit of (3^67 + 2^67) =...7 + ....8 = ....5
Hence unit digit = 5
Hope it helps
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