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Writing binary

Set 7 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 0193

Express decimal 512 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 512.
  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 512 is at least 512; write bit 1; new remainder 0.
  5. Weight 256: skip it because remaining 0 is below 256; write bit 0; new remainder 0.
  6. Weight 128: skip it because remaining 0 is below 128; write bit 0; new remainder 0.
  7. Weight 64: skip it because remaining 0 is below 64; write bit 0; 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 1000000000.
  16. Confirm the bit weights sum back to 512.
  17. Answer: 512₁₀ = 1000000000₂.

Final answer: 1000000000

Foundations · Problem 0194

Express decimal 513 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 513.
  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 513 is at least 512; write bit 1; new remainder 1.
  5. Weight 256: skip it because remaining 1 is below 256; write bit 0; new remainder 1.
  6. Weight 128: skip it because remaining 1 is below 128; write bit 0; new remainder 1.
  7. Weight 64: skip it because remaining 1 is below 64; write bit 0; 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 1000000001.
  16. Confirm the bit weights sum back to 513.
  17. Answer: 513₁₀ = 1000000001₂.

Final answer: 1000000001

Foundations · Problem 0195

Express decimal 514 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 514.
  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 514 is at least 512; write bit 1; new remainder 2.
  5. Weight 256: skip it because remaining 2 is below 256; write bit 0; new remainder 2.
  6. Weight 128: skip it because remaining 2 is below 128; write bit 0; new remainder 2.
  7. Weight 64: skip it because remaining 2 is below 64; write bit 0; 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 1000000010.
  16. Confirm the bit weights sum back to 514.
  17. Answer: 514₁₀ = 1000000010₂.

Final answer: 1000000010

Foundations · Problem 0196

Express decimal 515 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 515.
  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 515 is at least 512; write bit 1; new remainder 3.
  5. Weight 256: skip it because remaining 3 is below 256; write bit 0; new remainder 3.
  6. Weight 128: skip it because remaining 3 is below 128; write bit 0; new remainder 3.
  7. Weight 64: skip it because remaining 3 is below 64; write bit 0; 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 1000000011.
  16. Confirm the bit weights sum back to 515.
  17. Answer: 515₁₀ = 1000000011₂.

Final answer: 1000000011

Foundations · Problem 0197

Express decimal 516 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 516.
  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 516 is at least 512; write bit 1; new remainder 4.
  5. Weight 256: skip it because remaining 4 is below 256; write bit 0; new remainder 4.
  6. Weight 128: skip it because remaining 4 is below 128; write bit 0; new remainder 4.
  7. Weight 64: skip it because remaining 4 is below 64; write bit 0; 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 1000000100.
  16. Confirm the bit weights sum back to 516.
  17. Answer: 516₁₀ = 1000000100₂.

Final answer: 1000000100

Foundations · Problem 0198

Express decimal 517 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 517.
  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 517 is at least 512; write bit 1; new remainder 5.
  5. Weight 256: skip it because remaining 5 is below 256; write bit 0; new remainder 5.
  6. Weight 128: skip it because remaining 5 is below 128; write bit 0; new remainder 5.
  7. Weight 64: skip it because remaining 5 is below 64; write bit 0; 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 1000000101.
  16. Confirm the bit weights sum back to 517.
  17. Answer: 517₁₀ = 1000000101₂.

Final answer: 1000000101

Foundations · Problem 0199

Express decimal 518 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 518.
  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 518 is at least 512; write bit 1; new remainder 6.
  5. Weight 256: skip it because remaining 6 is below 256; write bit 0; new remainder 6.
  6. Weight 128: skip it because remaining 6 is below 128; write bit 0; new remainder 6.
  7. Weight 64: skip it because remaining 6 is below 64; write bit 0; 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 1000000110.
  16. Confirm the bit weights sum back to 518.
  17. Answer: 518₁₀ = 1000000110₂.

Final answer: 1000000110

Foundations · Problem 0200

Express decimal 519 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 519.
  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 519 is at least 512; write bit 1; new remainder 7.
  5. Weight 256: skip it because remaining 7 is below 256; write bit 0; new remainder 7.
  6. Weight 128: skip it because remaining 7 is below 128; write bit 0; new remainder 7.
  7. Weight 64: skip it because remaining 7 is below 64; write bit 0; 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 1000000111.
  16. Confirm the bit weights sum back to 519.
  17. Answer: 519₁₀ = 1000000111₂.

Final answer: 1000000111

Foundations · Problem 0201

Express decimal 520 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 520.
  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 520 is at least 512; write bit 1; new remainder 8.
  5. Weight 256: skip it because remaining 8 is below 256; write bit 0; new remainder 8.
  6. Weight 128: skip it because remaining 8 is below 128; write bit 0; new remainder 8.
  7. Weight 64: skip it because remaining 8 is below 64; write bit 0; 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 1000001000.
  16. Confirm the bit weights sum back to 520.
  17. Answer: 520₁₀ = 1000001000₂.

Final answer: 1000001000

Foundations · Problem 0202

Express decimal 521 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 521.
  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 521 is at least 512; write bit 1; new remainder 9.
  5. Weight 256: skip it because remaining 9 is below 256; write bit 0; new remainder 9.
  6. Weight 128: skip it because remaining 9 is below 128; write bit 0; new remainder 9.
  7. Weight 64: skip it because remaining 9 is below 64; write bit 0; 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 1000001001.
  16. Confirm the bit weights sum back to 521.
  17. Answer: 521₁₀ = 1000001001₂.

Final answer: 1000001001

Foundations · Problem 0203

Express decimal 522 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 522.
  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 522 is at least 512; write bit 1; new remainder 10.
  5. Weight 256: skip it because remaining 10 is below 256; write bit 0; new remainder 10.
  6. Weight 128: skip it because remaining 10 is below 128; write bit 0; new remainder 10.
  7. Weight 64: skip it because remaining 10 is below 64; write bit 0; 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 1000001010.
  16. Confirm the bit weights sum back to 522.
  17. Answer: 522₁₀ = 1000001010₂.

Final answer: 1000001010

Foundations · Problem 0204

Express decimal 523 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 523.
  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 523 is at least 512; write bit 1; new remainder 11.
  5. Weight 256: skip it because remaining 11 is below 256; write bit 0; new remainder 11.
  6. Weight 128: skip it because remaining 11 is below 128; write bit 0; new remainder 11.
  7. Weight 64: skip it because remaining 11 is below 64; write bit 0; 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 1000001011.
  16. Confirm the bit weights sum back to 523.
  17. Answer: 523₁₀ = 1000001011₂.

Final answer: 1000001011

Foundations · Problem 0205

Express decimal 524 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 524.
  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 524 is at least 512; write bit 1; new remainder 12.
  5. Weight 256: skip it because remaining 12 is below 256; write bit 0; new remainder 12.
  6. Weight 128: skip it because remaining 12 is below 128; write bit 0; new remainder 12.
  7. Weight 64: skip it because remaining 12 is below 64; write bit 0; 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 1000001100.
  16. Confirm the bit weights sum back to 524.
  17. Answer: 524₁₀ = 1000001100₂.

Final answer: 1000001100

Foundations · Problem 0206

Express decimal 525 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 525.
  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 525 is at least 512; write bit 1; new remainder 13.
  5. Weight 256: skip it because remaining 13 is below 256; write bit 0; new remainder 13.
  6. Weight 128: skip it because remaining 13 is below 128; write bit 0; new remainder 13.
  7. Weight 64: skip it because remaining 13 is below 64; write bit 0; 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 1000001101.
  16. Confirm the bit weights sum back to 525.
  17. Answer: 525₁₀ = 1000001101₂.

Final answer: 1000001101

Foundations · Problem 0207

Express decimal 526 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 526.
  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 526 is at least 512; write bit 1; new remainder 14.
  5. Weight 256: skip it because remaining 14 is below 256; write bit 0; new remainder 14.
  6. Weight 128: skip it because remaining 14 is below 128; write bit 0; new remainder 14.
  7. Weight 64: skip it because remaining 14 is below 64; write bit 0; 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 1000001110.
  16. Confirm the bit weights sum back to 526.
  17. Answer: 526₁₀ = 1000001110₂.

Final answer: 1000001110

Foundations · Problem 0208

Express decimal 527 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 527.
  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 527 is at least 512; write bit 1; new remainder 15.
  5. Weight 256: skip it because remaining 15 is below 256; write bit 0; new remainder 15.
  6. Weight 128: skip it because remaining 15 is below 128; write bit 0; new remainder 15.
  7. Weight 64: skip it because remaining 15 is below 64; write bit 0; 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 1000001111.
  16. Confirm the bit weights sum back to 527.
  17. Answer: 527₁₀ = 1000001111₂.

Final answer: 1000001111

Foundations · Problem 0209

Express decimal 528 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 528.
  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 528 is at least 512; write bit 1; new remainder 16.
  5. Weight 256: skip it because remaining 16 is below 256; write bit 0; new remainder 16.
  6. Weight 128: skip it because remaining 16 is below 128; write bit 0; new remainder 16.
  7. Weight 64: skip it because remaining 16 is below 64; write bit 0; 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 1000010000.
  16. Confirm the bit weights sum back to 528.
  17. Answer: 528₁₀ = 1000010000₂.

Final answer: 1000010000

Foundations · Problem 0210

Express decimal 529 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 529.
  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 529 is at least 512; write bit 1; new remainder 17.
  5. Weight 256: skip it because remaining 17 is below 256; write bit 0; new remainder 17.
  6. Weight 128: skip it because remaining 17 is below 128; write bit 0; new remainder 17.
  7. Weight 64: skip it because remaining 17 is below 64; write bit 0; 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 1000010001.
  16. Confirm the bit weights sum back to 529.
  17. Answer: 529₁₀ = 1000010001₂.

Final answer: 1000010001

Foundations · Problem 0211

Express decimal 530 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 530.
  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 530 is at least 512; write bit 1; new remainder 18.
  5. Weight 256: skip it because remaining 18 is below 256; write bit 0; new remainder 18.
  6. Weight 128: skip it because remaining 18 is below 128; write bit 0; new remainder 18.
  7. Weight 64: skip it because remaining 18 is below 64; write bit 0; 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 1000010010.
  16. Confirm the bit weights sum back to 530.
  17. Answer: 530₁₀ = 1000010010₂.

Final answer: 1000010010

Foundations · Problem 0212

Express decimal 531 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 531.
  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 531 is at least 512; write bit 1; new remainder 19.
  5. Weight 256: skip it because remaining 19 is below 256; write bit 0; new remainder 19.
  6. Weight 128: skip it because remaining 19 is below 128; write bit 0; new remainder 19.
  7. Weight 64: skip it because remaining 19 is below 64; write bit 0; 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 1000010011.
  16. Confirm the bit weights sum back to 531.
  17. Answer: 531₁₀ = 1000010011₂.

Final answer: 1000010011

Foundations · Problem 0213

Express decimal 532 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 532.
  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 532 is at least 512; write bit 1; new remainder 20.
  5. Weight 256: skip it because remaining 20 is below 256; write bit 0; new remainder 20.
  6. Weight 128: skip it because remaining 20 is below 128; write bit 0; new remainder 20.
  7. Weight 64: skip it because remaining 20 is below 64; write bit 0; 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 1000010100.
  16. Confirm the bit weights sum back to 532.
  17. Answer: 532₁₀ = 1000010100₂.

Final answer: 1000010100

Foundations · Problem 0214

Express decimal 533 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 533.
  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 533 is at least 512; write bit 1; new remainder 21.
  5. Weight 256: skip it because remaining 21 is below 256; write bit 0; new remainder 21.
  6. Weight 128: skip it because remaining 21 is below 128; write bit 0; new remainder 21.
  7. Weight 64: skip it because remaining 21 is below 64; write bit 0; 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 1000010101.
  16. Confirm the bit weights sum back to 533.
  17. Answer: 533₁₀ = 1000010101₂.

Final answer: 1000010101

Foundations · Problem 0215

Express decimal 534 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 534.
  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 534 is at least 512; write bit 1; new remainder 22.
  5. Weight 256: skip it because remaining 22 is below 256; write bit 0; new remainder 22.
  6. Weight 128: skip it because remaining 22 is below 128; write bit 0; new remainder 22.
  7. Weight 64: skip it because remaining 22 is below 64; write bit 0; 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 1000010110.
  16. Confirm the bit weights sum back to 534.
  17. Answer: 534₁₀ = 1000010110₂.

Final answer: 1000010110

Foundations · Problem 0216

Express decimal 535 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 535.
  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 535 is at least 512; write bit 1; new remainder 23.
  5. Weight 256: skip it because remaining 23 is below 256; write bit 0; new remainder 23.
  6. Weight 128: skip it because remaining 23 is below 128; write bit 0; new remainder 23.
  7. Weight 64: skip it because remaining 23 is below 64; write bit 0; 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 1000010111.
  16. Confirm the bit weights sum back to 535.
  17. Answer: 535₁₀ = 1000010111₂.

Final answer: 1000010111

Foundations · Problem 0217

Express decimal 536 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 536.
  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 536 is at least 512; write bit 1; new remainder 24.
  5. Weight 256: skip it because remaining 24 is below 256; write bit 0; new remainder 24.
  6. Weight 128: skip it because remaining 24 is below 128; write bit 0; new remainder 24.
  7. Weight 64: skip it because remaining 24 is below 64; write bit 0; 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 1000011000.
  16. Confirm the bit weights sum back to 536.
  17. Answer: 536₁₀ = 1000011000₂.

Final answer: 1000011000

Foundations · Problem 0218

Express decimal 537 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 537.
  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 537 is at least 512; write bit 1; new remainder 25.
  5. Weight 256: skip it because remaining 25 is below 256; write bit 0; new remainder 25.
  6. Weight 128: skip it because remaining 25 is below 128; write bit 0; new remainder 25.
  7. Weight 64: skip it because remaining 25 is below 64; write bit 0; 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 1000011001.
  16. Confirm the bit weights sum back to 537.
  17. Answer: 537₁₀ = 1000011001₂.

Final answer: 1000011001

Foundations · Problem 0219

Express decimal 538 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 538.
  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 538 is at least 512; write bit 1; new remainder 26.
  5. Weight 256: skip it because remaining 26 is below 256; write bit 0; new remainder 26.
  6. Weight 128: skip it because remaining 26 is below 128; write bit 0; new remainder 26.
  7. Weight 64: skip it because remaining 26 is below 64; write bit 0; 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 1000011010.
  16. Confirm the bit weights sum back to 538.
  17. Answer: 538₁₀ = 1000011010₂.

Final answer: 1000011010

Foundations · Problem 0220

Express decimal 539 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 539.
  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 539 is at least 512; write bit 1; new remainder 27.
  5. Weight 256: skip it because remaining 27 is below 256; write bit 0; new remainder 27.
  6. Weight 128: skip it because remaining 27 is below 128; write bit 0; new remainder 27.
  7. Weight 64: skip it because remaining 27 is below 64; write bit 0; 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 1000011011.
  16. Confirm the bit weights sum back to 539.
  17. Answer: 539₁₀ = 1000011011₂.

Final answer: 1000011011

Foundations · Problem 0221

Express decimal 540 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 540.
  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 540 is at least 512; write bit 1; new remainder 28.
  5. Weight 256: skip it because remaining 28 is below 256; write bit 0; new remainder 28.
  6. Weight 128: skip it because remaining 28 is below 128; write bit 0; new remainder 28.
  7. Weight 64: skip it because remaining 28 is below 64; write bit 0; 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 1000011100.
  16. Confirm the bit weights sum back to 540.
  17. Answer: 540₁₀ = 1000011100₂.

Final answer: 1000011100

Foundations · Problem 0222

Express decimal 541 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 541.
  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 541 is at least 512; write bit 1; new remainder 29.
  5. Weight 256: skip it because remaining 29 is below 256; write bit 0; new remainder 29.
  6. Weight 128: skip it because remaining 29 is below 128; write bit 0; new remainder 29.
  7. Weight 64: skip it because remaining 29 is below 64; write bit 0; 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 1000011101.
  16. Confirm the bit weights sum back to 541.
  17. Answer: 541₁₀ = 1000011101₂.

Final answer: 1000011101

Foundations · Problem 0223

Express decimal 542 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 542.
  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 542 is at least 512; write bit 1; new remainder 30.
  5. Weight 256: skip it because remaining 30 is below 256; write bit 0; new remainder 30.
  6. Weight 128: skip it because remaining 30 is below 128; write bit 0; new remainder 30.
  7. Weight 64: skip it because remaining 30 is below 64; write bit 0; 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 1000011110.
  16. Confirm the bit weights sum back to 542.
  17. Answer: 542₁₀ = 1000011110₂.

Final answer: 1000011110

Foundations · Problem 0224

Express decimal 543 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 543.
  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 543 is at least 512; write bit 1; new remainder 31.
  5. Weight 256: skip it because remaining 31 is below 256; write bit 0; new remainder 31.
  6. Weight 128: skip it because remaining 31 is below 128; write bit 0; new remainder 31.
  7. Weight 64: skip it because remaining 31 is below 64; write bit 0; 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 1000011111.
  16. Confirm the bit weights sum back to 543.
  17. Answer: 543₁₀ = 1000011111₂.

Final answer: 1000011111