What is the remainder when 324 is divided by 5 ?
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Remainder (3^24/5) = Remainder (81^6/5) = 1 Correct Option: B Another way: 3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243 Hence the cyclicity of 3 = 4 Therefore 3^24 will have the last digit = 1 Remainder when divided by 5 = 1
3的幂次方个位数为1379循环
3^24=81^6 个位数1六次方还是为1，那么除以5，余数为1。【若问3^23，同理余数为3；若问3^86，3^86=9*81^21，因为81^个位数始终为1，乘以9，个位数始终为9，所以余数为9-5=4】。
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Remainder (3^24/5) = Remainder (81^6/5) = 1
Correct Option: B
Another way:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
Hence the cyclicity of 3 = 4
Therefore 3^24 will have the last digit = 1
Remainder when divided by 5 = 1
3的幂次方个位数为1379循环
3^24=81^6 个位数1六次方还是为1，那么除以5，余数为1。【若问3^23，同理余数为3；若问3^86，3^86=9*81^21，因为81^个位数始终为1，乘以9，个位数始终为9，所以余数为9-5=4】。
3^24=81^6 个位数1六次方还是为1，那么除以5，余数为1。【若问3^23，同理余数为3；若问3^86，3^86=9*81^21，因为81^个位数始终为1，乘以9，个位数始终为9，所以余数为9-5=4】。