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sofibalerina
@sofibalerina
August 2022
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Помогите написать двоичное, четверичное и восьмеричное кодирование даты рождения 22.03.2003
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petyaGavrikov
Verified answer
22(10) = 16+4+2 = 2^4+2^2+2^1 = 10 110(2)
03(10) = 2+1 = 2^1+2^0 = 11(2)
2003 = 1024+512+256+128+64+16+2+1 =
= 2^10+2^9+2^8+2^7+2^6+2^4+2^1+2^0 = 11 111 010 011(2)
22(10) = 16+4+2 = 4^2+4^1+2*4^0 = 112(4)
3(10) = 3*4^0 = 3(4)
2003(10) = 1024+3*256+3*64+1*16+3 =
= 1*4^5+3*4^4+3*4^3+1*4^2+3*4^0 = 133 103(4)
22(10) = 16+6 = 2*8^1+6 = 26(8)
3(10) = 3(8)
2003(10) = 3*512+7*64+2*8+3 = 3*8^3+7*8^2+2*8^1+3 = 3723(8)
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Answers & Comments
Verified answer
22(10) = 16+4+2 = 2^4+2^2+2^1 = 10 110(2)03(10) = 2+1 = 2^1+2^0 = 11(2)
2003 = 1024+512+256+128+64+16+2+1 =
= 2^10+2^9+2^8+2^7+2^6+2^4+2^1+2^0 = 11 111 010 011(2)
22(10) = 16+4+2 = 4^2+4^1+2*4^0 = 112(4)
3(10) = 3*4^0 = 3(4)
2003(10) = 1024+3*256+3*64+1*16+3 =
= 1*4^5+3*4^4+3*4^3+1*4^2+3*4^0 = 133 103(4)
22(10) = 16+6 = 2*8^1+6 = 26(8)
3(10) = 3(8)
2003(10) = 3*512+7*64+2*8+3 = 3*8^3+7*8^2+2*8^1+3 = 3723(8)