Va rog sa ma ajutati urgent cu exercitiul 9!
Răspunsuri la întrebare
In baza de numerotatie k, avem cifrele (semnele grafice) 0;1;...;k-1
1a4₍₅₎=1b3₍₆₎ => a>b pentru ca 4>3
1a4₍₅₎=(1*5²+a*5¹+4*5⁰)₍₁₀₎-
=(25+5a+4)₍₁₀₎=
=(29+5a)₍₁₀₎
1b3₍₆₎ = (1*6²+b*6¹+3*6⁰)₍₁₀₎=
=(36+6b+3)₍₁₀₎=
=(39+6b)₍₁₀₎
39+6b=29+5a
39-29=5a-6b
10=5a-6b
a=(10-6b):5
a=10:5
a=2 => b=0 1a4₍₅₎=124₍₅₎ si 1b3₍₆₎=103₍₆₎ => 124₍₅₎=103₍₆₎
a=8 si b=5 => 1a4₍₅₎=184₍₅₍ si 1b3₍₆₎=153₍₆₎ => 184₍₅₎=153₍₆₎
b) 1ab₍₃₎=11a0₍₂₎
1ab₍₃₎=(1*3²+a*3¹+b*3⁰)₍₁₀₎=
=(9+3a+b)₍₁₀₎
11a0₍₂₎=(1*2³+1*2²+a*2¹+0*2⁰)₍₁₀₎=
=(8+4+2a+0)₍₁₀₎=
=(12+2a)₍₁₀₎
(9+3a+b)₍₁₀₎=(12+2a)₍₁₀₎
3a+b=12+2a-9
3a+b=2a+12-9
3a+b=2a+3
3a-2a=3-b
a=3-b
a+b=3
,,a,, poate fi {3;2 si 1}
,,b,,poate fi: {0;1 si 2}
=> 130₍₃₎=1130₍₂₎
121₍₃₎=1120₍₂₎
111₍₃₎=1110₍₂₎
c)
ab₍₂₎+ba₍₂₎=6 => a> 0 si b>0
ab₍₂₎=(a*2¹+b*2⁰)₍₁₀₎=
=(2a+b)₍₁₀₎
ba₍₂₎=(b*2¹+a*2⁰)₍₁₀₎=
=(2b+a)₍₁₀₎
(2a+b)₍₁₀₎=(2b+a)₍₁₀₎
2a=2b+a-b
2a=b+a
2a-a=b
a=b => a=1 si b=1 => ab₍₂₎=(1*2¹+1*2⁰)₍₁₀₎=
=(2+1)₍₁₀₎=
=3₍₁₀₎
ba₍₂₎=(1*2₁+1*2⁰)₍₁₀₎=
=3₍₁₀₎
11₍₂₎+11₍₂₎=3₍₁₀₎+3₍₁₀₎=6