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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 9 of 10 · 32 worked questions. Predict your result before expanding the solution.

These worked traces are meant to build fluency through controlled variation. If several answers feel identical, try the next difficult value and look for the point where a carry, sign, branch, or selected memory address changes the outcome.

Foundations · Problem 0257

Express decimal 576 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 576.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 576 is at least 512; write bit 1; new remainder 64.
  5. Weight 256: skip it because remaining 64 is below 256; write bit 0; new remainder 64.
  6. Weight 128: skip it because remaining 64 is below 128; write bit 0; new remainder 64.
  7. Weight 64: take it because remaining 64 is at least 64; write bit 1; new remainder 0.
  8. Weight 32: skip it because remaining 0 is below 32; write bit 0; new remainder 0.
  9. Weight 16: skip it because remaining 0 is below 16; write bit 0; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000000.
  16. Confirm the bit weights sum back to 576.
  17. Answer: 576₁₀ = 1001000000₂.

Final answer: 1001000000

Foundations · Problem 0258

Express decimal 577 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 577.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 577 is at least 512; write bit 1; new remainder 65.
  5. Weight 256: skip it because remaining 65 is below 256; write bit 0; new remainder 65.
  6. Weight 128: skip it because remaining 65 is below 128; write bit 0; new remainder 65.
  7. Weight 64: take it because remaining 65 is at least 64; write bit 1; new remainder 1.
  8. Weight 32: skip it because remaining 1 is below 32; write bit 0; new remainder 1.
  9. Weight 16: skip it because remaining 1 is below 16; write bit 0; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000001.
  16. Confirm the bit weights sum back to 577.
  17. Answer: 577₁₀ = 1001000001₂.

Final answer: 1001000001

Foundations · Problem 0259

Express decimal 578 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 578.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 578 is at least 512; write bit 1; new remainder 66.
  5. Weight 256: skip it because remaining 66 is below 256; write bit 0; new remainder 66.
  6. Weight 128: skip it because remaining 66 is below 128; write bit 0; new remainder 66.
  7. Weight 64: take it because remaining 66 is at least 64; write bit 1; new remainder 2.
  8. Weight 32: skip it because remaining 2 is below 32; write bit 0; new remainder 2.
  9. Weight 16: skip it because remaining 2 is below 16; write bit 0; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000010.
  16. Confirm the bit weights sum back to 578.
  17. Answer: 578₁₀ = 1001000010₂.

Final answer: 1001000010

Foundations · Problem 0260

Express decimal 579 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 579.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 579 is at least 512; write bit 1; new remainder 67.
  5. Weight 256: skip it because remaining 67 is below 256; write bit 0; new remainder 67.
  6. Weight 128: skip it because remaining 67 is below 128; write bit 0; new remainder 67.
  7. Weight 64: take it because remaining 67 is at least 64; write bit 1; new remainder 3.
  8. Weight 32: skip it because remaining 3 is below 32; write bit 0; new remainder 3.
  9. Weight 16: skip it because remaining 3 is below 16; write bit 0; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000011.
  16. Confirm the bit weights sum back to 579.
  17. Answer: 579₁₀ = 1001000011₂.

Final answer: 1001000011

Foundations · Problem 0261

Express decimal 580 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 580.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 580 is at least 512; write bit 1; new remainder 68.
  5. Weight 256: skip it because remaining 68 is below 256; write bit 0; new remainder 68.
  6. Weight 128: skip it because remaining 68 is below 128; write bit 0; new remainder 68.
  7. Weight 64: take it because remaining 68 is at least 64; write bit 1; new remainder 4.
  8. Weight 32: skip it because remaining 4 is below 32; write bit 0; new remainder 4.
  9. Weight 16: skip it because remaining 4 is below 16; write bit 0; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000100.
  16. Confirm the bit weights sum back to 580.
  17. Answer: 580₁₀ = 1001000100₂.

Final answer: 1001000100

Foundations · Problem 0262

Express decimal 581 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 581.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 581 is at least 512; write bit 1; new remainder 69.
  5. Weight 256: skip it because remaining 69 is below 256; write bit 0; new remainder 69.
  6. Weight 128: skip it because remaining 69 is below 128; write bit 0; new remainder 69.
  7. Weight 64: take it because remaining 69 is at least 64; write bit 1; new remainder 5.
  8. Weight 32: skip it because remaining 5 is below 32; write bit 0; new remainder 5.
  9. Weight 16: skip it because remaining 5 is below 16; write bit 0; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000101.
  16. Confirm the bit weights sum back to 581.
  17. Answer: 581₁₀ = 1001000101₂.

Final answer: 1001000101

Foundations · Problem 0263

Express decimal 582 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 582.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 582 is at least 512; write bit 1; new remainder 70.
  5. Weight 256: skip it because remaining 70 is below 256; write bit 0; new remainder 70.
  6. Weight 128: skip it because remaining 70 is below 128; write bit 0; new remainder 70.
  7. Weight 64: take it because remaining 70 is at least 64; write bit 1; new remainder 6.
  8. Weight 32: skip it because remaining 6 is below 32; write bit 0; new remainder 6.
  9. Weight 16: skip it because remaining 6 is below 16; write bit 0; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000110.
  16. Confirm the bit weights sum back to 582.
  17. Answer: 582₁₀ = 1001000110₂.

Final answer: 1001000110

Foundations · Problem 0264

Express decimal 583 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 583.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 583 is at least 512; write bit 1; new remainder 71.
  5. Weight 256: skip it because remaining 71 is below 256; write bit 0; new remainder 71.
  6. Weight 128: skip it because remaining 71 is below 128; write bit 0; new remainder 71.
  7. Weight 64: take it because remaining 71 is at least 64; write bit 1; new remainder 7.
  8. Weight 32: skip it because remaining 7 is below 32; write bit 0; new remainder 7.
  9. Weight 16: skip it because remaining 7 is below 16; write bit 0; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001000111.
  16. Confirm the bit weights sum back to 583.
  17. Answer: 583₁₀ = 1001000111₂.

Final answer: 1001000111

Foundations · Problem 0265

Express decimal 584 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 584.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 584 is at least 512; write bit 1; new remainder 72.
  5. Weight 256: skip it because remaining 72 is below 256; write bit 0; new remainder 72.
  6. Weight 128: skip it because remaining 72 is below 128; write bit 0; new remainder 72.
  7. Weight 64: take it because remaining 72 is at least 64; write bit 1; new remainder 8.
  8. Weight 32: skip it because remaining 8 is below 32; write bit 0; new remainder 8.
  9. Weight 16: skip it because remaining 8 is below 16; write bit 0; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001000.
  16. Confirm the bit weights sum back to 584.
  17. Answer: 584₁₀ = 1001001000₂.

Final answer: 1001001000

Foundations · Problem 0266

Express decimal 585 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 585.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 585 is at least 512; write bit 1; new remainder 73.
  5. Weight 256: skip it because remaining 73 is below 256; write bit 0; new remainder 73.
  6. Weight 128: skip it because remaining 73 is below 128; write bit 0; new remainder 73.
  7. Weight 64: take it because remaining 73 is at least 64; write bit 1; new remainder 9.
  8. Weight 32: skip it because remaining 9 is below 32; write bit 0; new remainder 9.
  9. Weight 16: skip it because remaining 9 is below 16; write bit 0; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001001.
  16. Confirm the bit weights sum back to 585.
  17. Answer: 585₁₀ = 1001001001₂.

Final answer: 1001001001

Foundations · Problem 0267

Express decimal 586 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 586.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 586 is at least 512; write bit 1; new remainder 74.
  5. Weight 256: skip it because remaining 74 is below 256; write bit 0; new remainder 74.
  6. Weight 128: skip it because remaining 74 is below 128; write bit 0; new remainder 74.
  7. Weight 64: take it because remaining 74 is at least 64; write bit 1; new remainder 10.
  8. Weight 32: skip it because remaining 10 is below 32; write bit 0; new remainder 10.
  9. Weight 16: skip it because remaining 10 is below 16; write bit 0; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001010.
  16. Confirm the bit weights sum back to 586.
  17. Answer: 586₁₀ = 1001001010₂.

Final answer: 1001001010

Foundations · Problem 0268

Express decimal 587 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 587.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 587 is at least 512; write bit 1; new remainder 75.
  5. Weight 256: skip it because remaining 75 is below 256; write bit 0; new remainder 75.
  6. Weight 128: skip it because remaining 75 is below 128; write bit 0; new remainder 75.
  7. Weight 64: take it because remaining 75 is at least 64; write bit 1; new remainder 11.
  8. Weight 32: skip it because remaining 11 is below 32; write bit 0; new remainder 11.
  9. Weight 16: skip it because remaining 11 is below 16; write bit 0; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001011.
  16. Confirm the bit weights sum back to 587.
  17. Answer: 587₁₀ = 1001001011₂.

Final answer: 1001001011

Foundations · Problem 0269

Express decimal 588 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 588.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 588 is at least 512; write bit 1; new remainder 76.
  5. Weight 256: skip it because remaining 76 is below 256; write bit 0; new remainder 76.
  6. Weight 128: skip it because remaining 76 is below 128; write bit 0; new remainder 76.
  7. Weight 64: take it because remaining 76 is at least 64; write bit 1; new remainder 12.
  8. Weight 32: skip it because remaining 12 is below 32; write bit 0; new remainder 12.
  9. Weight 16: skip it because remaining 12 is below 16; write bit 0; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001100.
  16. Confirm the bit weights sum back to 588.
  17. Answer: 588₁₀ = 1001001100₂.

Final answer: 1001001100

Foundations · Problem 0270

Express decimal 589 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 589.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 589 is at least 512; write bit 1; new remainder 77.
  5. Weight 256: skip it because remaining 77 is below 256; write bit 0; new remainder 77.
  6. Weight 128: skip it because remaining 77 is below 128; write bit 0; new remainder 77.
  7. Weight 64: take it because remaining 77 is at least 64; write bit 1; new remainder 13.
  8. Weight 32: skip it because remaining 13 is below 32; write bit 0; new remainder 13.
  9. Weight 16: skip it because remaining 13 is below 16; write bit 0; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001101.
  16. Confirm the bit weights sum back to 589.
  17. Answer: 589₁₀ = 1001001101₂.

Final answer: 1001001101

Foundations · Problem 0271

Express decimal 590 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 590.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 590 is at least 512; write bit 1; new remainder 78.
  5. Weight 256: skip it because remaining 78 is below 256; write bit 0; new remainder 78.
  6. Weight 128: skip it because remaining 78 is below 128; write bit 0; new remainder 78.
  7. Weight 64: take it because remaining 78 is at least 64; write bit 1; new remainder 14.
  8. Weight 32: skip it because remaining 14 is below 32; write bit 0; new remainder 14.
  9. Weight 16: skip it because remaining 14 is below 16; write bit 0; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001110.
  16. Confirm the bit weights sum back to 590.
  17. Answer: 590₁₀ = 1001001110₂.

Final answer: 1001001110

Foundations · Problem 0272

Express decimal 591 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 591.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 591 is at least 512; write bit 1; new remainder 79.
  5. Weight 256: skip it because remaining 79 is below 256; write bit 0; new remainder 79.
  6. Weight 128: skip it because remaining 79 is below 128; write bit 0; new remainder 79.
  7. Weight 64: take it because remaining 79 is at least 64; write bit 1; new remainder 15.
  8. Weight 32: skip it because remaining 15 is below 32; write bit 0; new remainder 15.
  9. Weight 16: skip it because remaining 15 is below 16; write bit 0; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001001111.
  16. Confirm the bit weights sum back to 591.
  17. Answer: 591₁₀ = 1001001111₂.

Final answer: 1001001111

Foundations · Problem 0273

Express decimal 592 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 592.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 592 is at least 512; write bit 1; new remainder 80.
  5. Weight 256: skip it because remaining 80 is below 256; write bit 0; new remainder 80.
  6. Weight 128: skip it because remaining 80 is below 128; write bit 0; new remainder 80.
  7. Weight 64: take it because remaining 80 is at least 64; write bit 1; new remainder 16.
  8. Weight 32: skip it because remaining 16 is below 32; write bit 0; new remainder 16.
  9. Weight 16: take it because remaining 16 is at least 16; write bit 1; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010000.
  16. Confirm the bit weights sum back to 592.
  17. Answer: 592₁₀ = 1001010000₂.

Final answer: 1001010000

Foundations · Problem 0274

Express decimal 593 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 593.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 593 is at least 512; write bit 1; new remainder 81.
  5. Weight 256: skip it because remaining 81 is below 256; write bit 0; new remainder 81.
  6. Weight 128: skip it because remaining 81 is below 128; write bit 0; new remainder 81.
  7. Weight 64: take it because remaining 81 is at least 64; write bit 1; new remainder 17.
  8. Weight 32: skip it because remaining 17 is below 32; write bit 0; new remainder 17.
  9. Weight 16: take it because remaining 17 is at least 16; write bit 1; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010001.
  16. Confirm the bit weights sum back to 593.
  17. Answer: 593₁₀ = 1001010001₂.

Final answer: 1001010001

Foundations · Problem 0275

Express decimal 594 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 594.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 594 is at least 512; write bit 1; new remainder 82.
  5. Weight 256: skip it because remaining 82 is below 256; write bit 0; new remainder 82.
  6. Weight 128: skip it because remaining 82 is below 128; write bit 0; new remainder 82.
  7. Weight 64: take it because remaining 82 is at least 64; write bit 1; new remainder 18.
  8. Weight 32: skip it because remaining 18 is below 32; write bit 0; new remainder 18.
  9. Weight 16: take it because remaining 18 is at least 16; write bit 1; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010010.
  16. Confirm the bit weights sum back to 594.
  17. Answer: 594₁₀ = 1001010010₂.

Final answer: 1001010010

Foundations · Problem 0276

Express decimal 595 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 595.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 595 is at least 512; write bit 1; new remainder 83.
  5. Weight 256: skip it because remaining 83 is below 256; write bit 0; new remainder 83.
  6. Weight 128: skip it because remaining 83 is below 128; write bit 0; new remainder 83.
  7. Weight 64: take it because remaining 83 is at least 64; write bit 1; new remainder 19.
  8. Weight 32: skip it because remaining 19 is below 32; write bit 0; new remainder 19.
  9. Weight 16: take it because remaining 19 is at least 16; write bit 1; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010011.
  16. Confirm the bit weights sum back to 595.
  17. Answer: 595₁₀ = 1001010011₂.

Final answer: 1001010011

Foundations · Problem 0277

Express decimal 596 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 596.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 596 is at least 512; write bit 1; new remainder 84.
  5. Weight 256: skip it because remaining 84 is below 256; write bit 0; new remainder 84.
  6. Weight 128: skip it because remaining 84 is below 128; write bit 0; new remainder 84.
  7. Weight 64: take it because remaining 84 is at least 64; write bit 1; new remainder 20.
  8. Weight 32: skip it because remaining 20 is below 32; write bit 0; new remainder 20.
  9. Weight 16: take it because remaining 20 is at least 16; write bit 1; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010100.
  16. Confirm the bit weights sum back to 596.
  17. Answer: 596₁₀ = 1001010100₂.

Final answer: 1001010100

Foundations · Problem 0278

Express decimal 597 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 597.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 597 is at least 512; write bit 1; new remainder 85.
  5. Weight 256: skip it because remaining 85 is below 256; write bit 0; new remainder 85.
  6. Weight 128: skip it because remaining 85 is below 128; write bit 0; new remainder 85.
  7. Weight 64: take it because remaining 85 is at least 64; write bit 1; new remainder 21.
  8. Weight 32: skip it because remaining 21 is below 32; write bit 0; new remainder 21.
  9. Weight 16: take it because remaining 21 is at least 16; write bit 1; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010101.
  16. Confirm the bit weights sum back to 597.
  17. Answer: 597₁₀ = 1001010101₂.

Final answer: 1001010101

Foundations · Problem 0279

Express decimal 598 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 598.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 598 is at least 512; write bit 1; new remainder 86.
  5. Weight 256: skip it because remaining 86 is below 256; write bit 0; new remainder 86.
  6. Weight 128: skip it because remaining 86 is below 128; write bit 0; new remainder 86.
  7. Weight 64: take it because remaining 86 is at least 64; write bit 1; new remainder 22.
  8. Weight 32: skip it because remaining 22 is below 32; write bit 0; new remainder 22.
  9. Weight 16: take it because remaining 22 is at least 16; write bit 1; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010110.
  16. Confirm the bit weights sum back to 598.
  17. Answer: 598₁₀ = 1001010110₂.

Final answer: 1001010110

Foundations · Problem 0280

Express decimal 599 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 599.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 599 is at least 512; write bit 1; new remainder 87.
  5. Weight 256: skip it because remaining 87 is below 256; write bit 0; new remainder 87.
  6. Weight 128: skip it because remaining 87 is below 128; write bit 0; new remainder 87.
  7. Weight 64: take it because remaining 87 is at least 64; write bit 1; new remainder 23.
  8. Weight 32: skip it because remaining 23 is below 32; write bit 0; new remainder 23.
  9. Weight 16: take it because remaining 23 is at least 16; write bit 1; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001010111.
  16. Confirm the bit weights sum back to 599.
  17. Answer: 599₁₀ = 1001010111₂.

Final answer: 1001010111

Foundations · Problem 0281

Express decimal 600 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 600.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 600 is at least 512; write bit 1; new remainder 88.
  5. Weight 256: skip it because remaining 88 is below 256; write bit 0; new remainder 88.
  6. Weight 128: skip it because remaining 88 is below 128; write bit 0; new remainder 88.
  7. Weight 64: take it because remaining 88 is at least 64; write bit 1; new remainder 24.
  8. Weight 32: skip it because remaining 24 is below 32; write bit 0; new remainder 24.
  9. Weight 16: take it because remaining 24 is at least 16; write bit 1; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011000.
  16. Confirm the bit weights sum back to 600.
  17. Answer: 600₁₀ = 1001011000₂.

Final answer: 1001011000

Foundations · Problem 0282

Express decimal 601 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 601.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 601 is at least 512; write bit 1; new remainder 89.
  5. Weight 256: skip it because remaining 89 is below 256; write bit 0; new remainder 89.
  6. Weight 128: skip it because remaining 89 is below 128; write bit 0; new remainder 89.
  7. Weight 64: take it because remaining 89 is at least 64; write bit 1; new remainder 25.
  8. Weight 32: skip it because remaining 25 is below 32; write bit 0; new remainder 25.
  9. Weight 16: take it because remaining 25 is at least 16; write bit 1; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011001.
  16. Confirm the bit weights sum back to 601.
  17. Answer: 601₁₀ = 1001011001₂.

Final answer: 1001011001

Foundations · Problem 0283

Express decimal 602 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 602.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 602 is at least 512; write bit 1; new remainder 90.
  5. Weight 256: skip it because remaining 90 is below 256; write bit 0; new remainder 90.
  6. Weight 128: skip it because remaining 90 is below 128; write bit 0; new remainder 90.
  7. Weight 64: take it because remaining 90 is at least 64; write bit 1; new remainder 26.
  8. Weight 32: skip it because remaining 26 is below 32; write bit 0; new remainder 26.
  9. Weight 16: take it because remaining 26 is at least 16; write bit 1; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011010.
  16. Confirm the bit weights sum back to 602.
  17. Answer: 602₁₀ = 1001011010₂.

Final answer: 1001011010

Foundations · Problem 0284

Express decimal 603 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 603.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 603 is at least 512; write bit 1; new remainder 91.
  5. Weight 256: skip it because remaining 91 is below 256; write bit 0; new remainder 91.
  6. Weight 128: skip it because remaining 91 is below 128; write bit 0; new remainder 91.
  7. Weight 64: take it because remaining 91 is at least 64; write bit 1; new remainder 27.
  8. Weight 32: skip it because remaining 27 is below 32; write bit 0; new remainder 27.
  9. Weight 16: take it because remaining 27 is at least 16; write bit 1; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011011.
  16. Confirm the bit weights sum back to 603.
  17. Answer: 603₁₀ = 1001011011₂.

Final answer: 1001011011

Foundations · Problem 0285

Express decimal 604 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 604.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 604 is at least 512; write bit 1; new remainder 92.
  5. Weight 256: skip it because remaining 92 is below 256; write bit 0; new remainder 92.
  6. Weight 128: skip it because remaining 92 is below 128; write bit 0; new remainder 92.
  7. Weight 64: take it because remaining 92 is at least 64; write bit 1; new remainder 28.
  8. Weight 32: skip it because remaining 28 is below 32; write bit 0; new remainder 28.
  9. Weight 16: take it because remaining 28 is at least 16; write bit 1; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011100.
  16. Confirm the bit weights sum back to 604.
  17. Answer: 604₁₀ = 1001011100₂.

Final answer: 1001011100

Foundations · Problem 0286

Express decimal 605 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 605.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 605 is at least 512; write bit 1; new remainder 93.
  5. Weight 256: skip it because remaining 93 is below 256; write bit 0; new remainder 93.
  6. Weight 128: skip it because remaining 93 is below 128; write bit 0; new remainder 93.
  7. Weight 64: take it because remaining 93 is at least 64; write bit 1; new remainder 29.
  8. Weight 32: skip it because remaining 29 is below 32; write bit 0; new remainder 29.
  9. Weight 16: take it because remaining 29 is at least 16; write bit 1; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011101.
  16. Confirm the bit weights sum back to 605.
  17. Answer: 605₁₀ = 1001011101₂.

Final answer: 1001011101

Foundations · Problem 0287

Express decimal 606 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 606.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 606 is at least 512; write bit 1; new remainder 94.
  5. Weight 256: skip it because remaining 94 is below 256; write bit 0; new remainder 94.
  6. Weight 128: skip it because remaining 94 is below 128; write bit 0; new remainder 94.
  7. Weight 64: take it because remaining 94 is at least 64; write bit 1; new remainder 30.
  8. Weight 32: skip it because remaining 30 is below 32; write bit 0; new remainder 30.
  9. Weight 16: take it because remaining 30 is at least 16; write bit 1; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011110.
  16. Confirm the bit weights sum back to 606.
  17. Answer: 606₁₀ = 1001011110₂.

Final answer: 1001011110

Foundations · Problem 0288

Express decimal 607 in 10-bit unsigned binary.

Try it on paper first. The worked answer is hidden below.

Reveal detailed solution →
  1. The target is the ordinary decimal value 607.
  2. Choose the smallest practical fixed width: 10 bits can hold the target.
  3. The highest available weight at that width is 512.
  4. Weight 512: take it because remaining 607 is at least 512; write bit 1; new remainder 95.
  5. Weight 256: skip it because remaining 95 is below 256; write bit 0; new remainder 95.
  6. Weight 128: skip it because remaining 95 is below 128; write bit 0; new remainder 95.
  7. Weight 64: take it because remaining 95 is at least 64; write bit 1; new remainder 31.
  8. Weight 32: skip it because remaining 31 is below 32; write bit 0; new remainder 31.
  9. Weight 16: take it because remaining 31 is at least 16; write bit 1; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 1001011111.
  16. Confirm the bit weights sum back to 607.
  17. Answer: 607₁₀ = 1001011111₂.

Final answer: 1001011111