Q: Convert the hexadecimal number A45C to octal and decimal?

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BB895C: Octal = 56704534 Decimal = 12290396

import java.util.Scanner; public class NumberSystem { public void displayConversion() { Scanner input = new Scanner(System.in); System.out.printf("%-20s%-20s%-20s%-20s\n", "Decimal", "Binary", "Octal", "Hexadecimal"); for ( int i = 1; i <= 256; i++ ) { String binary = Integer.toBinaryString(i); String octal = Integer.toOctalString(i); String hexadecimal = Integer.toHexString(i); System.out.format("%-20d%-20s%-20s%-20s\n", i, binary, octal, hexadecimal); } } // returns a string representation of the decimal number in binary public String toBinaryString( int dec ) { String binary = " "; while (dec >= 1 ) { int value = dec % 2; binary = value + binary; dec /= 2; } return binary; } //returns a string representation of the number in octal public String toOctalString( int dec ) { String octal = " "; while ( dec >= 1 ) { int value = dec % 8; octal = value + octal; dec /= 8; } return octal; } public String toHexString( int dec ) { String hexadecimal = " "; while ( dec >= 1 ) { int value = dec % 16; switch (value) { case 10: hexadecimal = "A" + hexadecimal; break; case 11: hexadecimal = "B" + hexadecimal; break; case 12: hexadecimal = "C" + hexadecimal; break; case 13: hexadecimal = "D" + hexadecimal; break; case 14: hexadecimal = "E" + hexadecimal; break; case 15: hexadecimal = "F" + hexadecimal; break; default: hexadecimal = value + hexadecimal; break; } dec /= 16; } return hexadecimal; } public static void main( String args[]) { NumberSystem apps = new NumberSystem(); apps.displayConversion(); } }

public class Dataconversion { public static void main(String[] args) { System.out.println("Data types conversion example!"); int in = 44; System.out.println("Integer: " + in); //integer to binary String by = Integer.toBinaryString(in); System.out.println("Byte: " + by); //integer to hexadecimal String hex = Integer.toHexString(in); System.out.println("Hexa decimal: " + hex); //integer to octal String oct = Integer.toOctalString(in); System.out.println("Octal: " + oct); } }

A binary number system has two states '0' '1' for a long word in bits it can be as follows 101010101010101010101011 intimidating RIGHT? it can be represented in groups of 3 bits in octal 10/010/101/010/101/010/101/011= 22525253 digital or in group of 4 bits as 10/1010/1010/1010/1010/1010 = 2AAAAA 111 =7 octal 1111=f F in hexadecimal numbers 1000 =8 1010 =10 or A

Depends on the encoding. It is probably 0xF (15), but it could be -1 (if 4 bits in 2's complement), or any other value if a non-standard encoding. the coding can be octal or hexadecimal the value in decimal does not change in octal it is 17 in hexadecimal it is F in decimal is 15. ALL of these numbers are right it depends of what code the reference is to

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BB895C: Octal = 56704534 Decimal = 12290396

NA

Octal = 56704534 Decimal = 12290396

221122: Binary = 1000100001000100100010 Octal = 10410442 Decimal = 2232610

Hexadecimal -> BB895Cdecimal -> 12.290.396octal -> 56.704.534

101102 = 2210 = 1616 = 268

The answer depends on what you are converting from: binary, ternary, octal, hexadecimal ...

Octal = 52746757 Binary = 101010111100110111101111

56704534

C65A = 143132

The answer depends on what you are converting from: binary, ternary, octal, hexadecimal ...

Decimal, binary, octal and hexadecimal.