Consider the following high-level recursive procedure: long long int flong long int n, long long int k long long int b b k+2; if
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Answer: C5₁₆ ⇔ 11000101₂ ⇔ 197⏨
Explanation: 11000101 is a binary number
it's equivalent decimal number is
1x2⁷+ 1x2⁶+0x2⁵ + 0x2⁴+0x2³+1x2²+0x2¹+1x2⁰ = 128+64+4+1 = 197
Hexadecimal system has equivalents in decimal system
0 ⇔0
1 ⇔ 1
2 ⇔ 2
3 ⇔ 3
4 ⇔ 4
5 ⇔ 5
6 ⇔ 6
7 ⇔ 7
8 ⇔ 8
9 ⇔ 9
10 ⇔A
11 ⇔B
12 ⇔C
13 ⇔D
14 ⇔E
15 ⇔F
197 ⇔??
To convert this decimal number in a hexadecimal number we have to divide it by 16
197/16
12,...
the quotient without the decimal part is multiplied by 16 and subtracted from the number
12x16 = 192
197-192 = 5
Then this number can be expressed as 12x16¹ + 5x16⁰ = 192 + 5 = 197
But in the hexadecimal system there are not digits > 9 , so 12 ⇔ C
197⏨ = Cx16¹ + 5x16⁰ = C5₁₆
Answer: C5₁₆ ⇔ 11000101₂ ⇔ 197⏨