What are the required steps to convert base 10 decimal system
number 1 520 324 131 218 to base 2 unsigned binary equivalent?
- A number written in base ten, or a decimal system number, is a number written using the digits 0 through 9. A number written in base two, or a binary system number, is a number written using only the digits 0 and 1.
1. Divide the number repeatedly by 2:
Keep track of each remainder.
Stop when you get a quotient that is equal to zero.
- division = quotient + remainder;
- 1 520 324 131 218 ÷ 2 = 760 162 065 609 + 0;
- 760 162 065 609 ÷ 2 = 380 081 032 804 + 1;
- 380 081 032 804 ÷ 2 = 190 040 516 402 + 0;
- 190 040 516 402 ÷ 2 = 95 020 258 201 + 0;
- 95 020 258 201 ÷ 2 = 47 510 129 100 + 1;
- 47 510 129 100 ÷ 2 = 23 755 064 550 + 0;
- 23 755 064 550 ÷ 2 = 11 877 532 275 + 0;
- 11 877 532 275 ÷ 2 = 5 938 766 137 + 1;
- 5 938 766 137 ÷ 2 = 2 969 383 068 + 1;
- 2 969 383 068 ÷ 2 = 1 484 691 534 + 0;
- 1 484 691 534 ÷ 2 = 742 345 767 + 0;
- 742 345 767 ÷ 2 = 371 172 883 + 1;
- 371 172 883 ÷ 2 = 185 586 441 + 1;
- 185 586 441 ÷ 2 = 92 793 220 + 1;
- 92 793 220 ÷ 2 = 46 396 610 + 0;
- 46 396 610 ÷ 2 = 23 198 305 + 0;
- 23 198 305 ÷ 2 = 11 599 152 + 1;
- 11 599 152 ÷ 2 = 5 799 576 + 0;
- 5 799 576 ÷ 2 = 2 899 788 + 0;
- 2 899 788 ÷ 2 = 1 449 894 + 0;
- 1 449 894 ÷ 2 = 724 947 + 0;
- 724 947 ÷ 2 = 362 473 + 1;
- 362 473 ÷ 2 = 181 236 + 1;
- 181 236 ÷ 2 = 90 618 + 0;
- 90 618 ÷ 2 = 45 309 + 0;
- 45 309 ÷ 2 = 22 654 + 1;
- 22 654 ÷ 2 = 11 327 + 0;
- 11 327 ÷ 2 = 5 663 + 1;
- 5 663 ÷ 2 = 2 831 + 1;
- 2 831 ÷ 2 = 1 415 + 1;
- 1 415 ÷ 2 = 707 + 1;
- 707 ÷ 2 = 353 + 1;
- 353 ÷ 2 = 176 + 1;
- 176 ÷ 2 = 88 + 0;
- 88 ÷ 2 = 44 + 0;
- 44 ÷ 2 = 22 + 0;
- 22 ÷ 2 = 11 + 0;
- 11 ÷ 2 = 5 + 1;
- 5 ÷ 2 = 2 + 1;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
2. Construct the base 2 representation of the positive number:
Take all the remainders starting from the bottom of the list constructed above.
1 520 324 131 218(10) Base 10 decimal system number converted and written as a base 2 unsigned binary equivalent:
1 520 324 131 218 (base 10) = 1 0110 0001 1111 1010 0110 0001 0011 1001 1001 0010 (base 2)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.