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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 8 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 0225

Express decimal 544 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 544.
  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 544 is at least 512; write bit 1; new remainder 32.
  5. Weight 256: skip it because remaining 32 is below 256; write bit 0; new remainder 32.
  6. Weight 128: skip it because remaining 32 is below 128; write bit 0; new remainder 32.
  7. Weight 64: skip it because remaining 32 is below 64; write bit 0; new remainder 32.
  8. Weight 32: take it because remaining 32 is at least 32; write bit 1; 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 1000100000.
  16. Confirm the bit weights sum back to 544.
  17. Answer: 544₁₀ = 1000100000₂.

Final answer: 1000100000

Foundations · Problem 0226

Express decimal 545 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 545.
  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 545 is at least 512; write bit 1; new remainder 33.
  5. Weight 256: skip it because remaining 33 is below 256; write bit 0; new remainder 33.
  6. Weight 128: skip it because remaining 33 is below 128; write bit 0; new remainder 33.
  7. Weight 64: skip it because remaining 33 is below 64; write bit 0; new remainder 33.
  8. Weight 32: take it because remaining 33 is at least 32; write bit 1; 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 1000100001.
  16. Confirm the bit weights sum back to 545.
  17. Answer: 545₁₀ = 1000100001₂.

Final answer: 1000100001

Foundations · Problem 0227

Express decimal 546 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 546.
  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 546 is at least 512; write bit 1; new remainder 34.
  5. Weight 256: skip it because remaining 34 is below 256; write bit 0; new remainder 34.
  6. Weight 128: skip it because remaining 34 is below 128; write bit 0; new remainder 34.
  7. Weight 64: skip it because remaining 34 is below 64; write bit 0; new remainder 34.
  8. Weight 32: take it because remaining 34 is at least 32; write bit 1; 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 1000100010.
  16. Confirm the bit weights sum back to 546.
  17. Answer: 546₁₀ = 1000100010₂.

Final answer: 1000100010

Foundations · Problem 0228

Express decimal 547 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 547.
  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 547 is at least 512; write bit 1; new remainder 35.
  5. Weight 256: skip it because remaining 35 is below 256; write bit 0; new remainder 35.
  6. Weight 128: skip it because remaining 35 is below 128; write bit 0; new remainder 35.
  7. Weight 64: skip it because remaining 35 is below 64; write bit 0; new remainder 35.
  8. Weight 32: take it because remaining 35 is at least 32; write bit 1; 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 1000100011.
  16. Confirm the bit weights sum back to 547.
  17. Answer: 547₁₀ = 1000100011₂.

Final answer: 1000100011

Foundations · Problem 0229

Express decimal 548 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 548.
  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 548 is at least 512; write bit 1; new remainder 36.
  5. Weight 256: skip it because remaining 36 is below 256; write bit 0; new remainder 36.
  6. Weight 128: skip it because remaining 36 is below 128; write bit 0; new remainder 36.
  7. Weight 64: skip it because remaining 36 is below 64; write bit 0; new remainder 36.
  8. Weight 32: take it because remaining 36 is at least 32; write bit 1; 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 1000100100.
  16. Confirm the bit weights sum back to 548.
  17. Answer: 548₁₀ = 1000100100₂.

Final answer: 1000100100

Foundations · Problem 0230

Express decimal 549 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 549.
  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 549 is at least 512; write bit 1; new remainder 37.
  5. Weight 256: skip it because remaining 37 is below 256; write bit 0; new remainder 37.
  6. Weight 128: skip it because remaining 37 is below 128; write bit 0; new remainder 37.
  7. Weight 64: skip it because remaining 37 is below 64; write bit 0; new remainder 37.
  8. Weight 32: take it because remaining 37 is at least 32; write bit 1; 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 1000100101.
  16. Confirm the bit weights sum back to 549.
  17. Answer: 549₁₀ = 1000100101₂.

Final answer: 1000100101

Foundations · Problem 0231

Express decimal 550 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 550.
  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 550 is at least 512; write bit 1; new remainder 38.
  5. Weight 256: skip it because remaining 38 is below 256; write bit 0; new remainder 38.
  6. Weight 128: skip it because remaining 38 is below 128; write bit 0; new remainder 38.
  7. Weight 64: skip it because remaining 38 is below 64; write bit 0; new remainder 38.
  8. Weight 32: take it because remaining 38 is at least 32; write bit 1; 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 1000100110.
  16. Confirm the bit weights sum back to 550.
  17. Answer: 550₁₀ = 1000100110₂.

Final answer: 1000100110

Foundations · Problem 0232

Express decimal 551 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 551.
  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 551 is at least 512; write bit 1; new remainder 39.
  5. Weight 256: skip it because remaining 39 is below 256; write bit 0; new remainder 39.
  6. Weight 128: skip it because remaining 39 is below 128; write bit 0; new remainder 39.
  7. Weight 64: skip it because remaining 39 is below 64; write bit 0; new remainder 39.
  8. Weight 32: take it because remaining 39 is at least 32; write bit 1; 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 1000100111.
  16. Confirm the bit weights sum back to 551.
  17. Answer: 551₁₀ = 1000100111₂.

Final answer: 1000100111

Foundations · Problem 0233

Express decimal 552 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 552.
  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 552 is at least 512; write bit 1; new remainder 40.
  5. Weight 256: skip it because remaining 40 is below 256; write bit 0; new remainder 40.
  6. Weight 128: skip it because remaining 40 is below 128; write bit 0; new remainder 40.
  7. Weight 64: skip it because remaining 40 is below 64; write bit 0; new remainder 40.
  8. Weight 32: take it because remaining 40 is at least 32; write bit 1; 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 1000101000.
  16. Confirm the bit weights sum back to 552.
  17. Answer: 552₁₀ = 1000101000₂.

Final answer: 1000101000

Foundations · Problem 0234

Express decimal 553 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 553.
  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 553 is at least 512; write bit 1; new remainder 41.
  5. Weight 256: skip it because remaining 41 is below 256; write bit 0; new remainder 41.
  6. Weight 128: skip it because remaining 41 is below 128; write bit 0; new remainder 41.
  7. Weight 64: skip it because remaining 41 is below 64; write bit 0; new remainder 41.
  8. Weight 32: take it because remaining 41 is at least 32; write bit 1; 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 1000101001.
  16. Confirm the bit weights sum back to 553.
  17. Answer: 553₁₀ = 1000101001₂.

Final answer: 1000101001

Foundations · Problem 0235

Express decimal 554 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 554.
  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 554 is at least 512; write bit 1; new remainder 42.
  5. Weight 256: skip it because remaining 42 is below 256; write bit 0; new remainder 42.
  6. Weight 128: skip it because remaining 42 is below 128; write bit 0; new remainder 42.
  7. Weight 64: skip it because remaining 42 is below 64; write bit 0; new remainder 42.
  8. Weight 32: take it because remaining 42 is at least 32; write bit 1; 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 1000101010.
  16. Confirm the bit weights sum back to 554.
  17. Answer: 554₁₀ = 1000101010₂.

Final answer: 1000101010

Foundations · Problem 0236

Express decimal 555 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 555.
  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 555 is at least 512; write bit 1; new remainder 43.
  5. Weight 256: skip it because remaining 43 is below 256; write bit 0; new remainder 43.
  6. Weight 128: skip it because remaining 43 is below 128; write bit 0; new remainder 43.
  7. Weight 64: skip it because remaining 43 is below 64; write bit 0; new remainder 43.
  8. Weight 32: take it because remaining 43 is at least 32; write bit 1; 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 1000101011.
  16. Confirm the bit weights sum back to 555.
  17. Answer: 555₁₀ = 1000101011₂.

Final answer: 1000101011

Foundations · Problem 0237

Express decimal 556 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 556.
  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 556 is at least 512; write bit 1; new remainder 44.
  5. Weight 256: skip it because remaining 44 is below 256; write bit 0; new remainder 44.
  6. Weight 128: skip it because remaining 44 is below 128; write bit 0; new remainder 44.
  7. Weight 64: skip it because remaining 44 is below 64; write bit 0; new remainder 44.
  8. Weight 32: take it because remaining 44 is at least 32; write bit 1; 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 1000101100.
  16. Confirm the bit weights sum back to 556.
  17. Answer: 556₁₀ = 1000101100₂.

Final answer: 1000101100

Foundations · Problem 0238

Express decimal 557 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 557.
  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 557 is at least 512; write bit 1; new remainder 45.
  5. Weight 256: skip it because remaining 45 is below 256; write bit 0; new remainder 45.
  6. Weight 128: skip it because remaining 45 is below 128; write bit 0; new remainder 45.
  7. Weight 64: skip it because remaining 45 is below 64; write bit 0; new remainder 45.
  8. Weight 32: take it because remaining 45 is at least 32; write bit 1; 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 1000101101.
  16. Confirm the bit weights sum back to 557.
  17. Answer: 557₁₀ = 1000101101₂.

Final answer: 1000101101

Foundations · Problem 0239

Express decimal 558 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 558.
  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 558 is at least 512; write bit 1; new remainder 46.
  5. Weight 256: skip it because remaining 46 is below 256; write bit 0; new remainder 46.
  6. Weight 128: skip it because remaining 46 is below 128; write bit 0; new remainder 46.
  7. Weight 64: skip it because remaining 46 is below 64; write bit 0; new remainder 46.
  8. Weight 32: take it because remaining 46 is at least 32; write bit 1; 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 1000101110.
  16. Confirm the bit weights sum back to 558.
  17. Answer: 558₁₀ = 1000101110₂.

Final answer: 1000101110

Foundations · Problem 0240

Express decimal 559 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 559.
  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 559 is at least 512; write bit 1; new remainder 47.
  5. Weight 256: skip it because remaining 47 is below 256; write bit 0; new remainder 47.
  6. Weight 128: skip it because remaining 47 is below 128; write bit 0; new remainder 47.
  7. Weight 64: skip it because remaining 47 is below 64; write bit 0; new remainder 47.
  8. Weight 32: take it because remaining 47 is at least 32; write bit 1; 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 1000101111.
  16. Confirm the bit weights sum back to 559.
  17. Answer: 559₁₀ = 1000101111₂.

Final answer: 1000101111

Foundations · Problem 0241

Express decimal 560 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 560.
  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 560 is at least 512; write bit 1; new remainder 48.
  5. Weight 256: skip it because remaining 48 is below 256; write bit 0; new remainder 48.
  6. Weight 128: skip it because remaining 48 is below 128; write bit 0; new remainder 48.
  7. Weight 64: skip it because remaining 48 is below 64; write bit 0; new remainder 48.
  8. Weight 32: take it because remaining 48 is at least 32; write bit 1; 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 1000110000.
  16. Confirm the bit weights sum back to 560.
  17. Answer: 560₁₀ = 1000110000₂.

Final answer: 1000110000

Foundations · Problem 0242

Express decimal 561 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 561.
  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 561 is at least 512; write bit 1; new remainder 49.
  5. Weight 256: skip it because remaining 49 is below 256; write bit 0; new remainder 49.
  6. Weight 128: skip it because remaining 49 is below 128; write bit 0; new remainder 49.
  7. Weight 64: skip it because remaining 49 is below 64; write bit 0; new remainder 49.
  8. Weight 32: take it because remaining 49 is at least 32; write bit 1; 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 1000110001.
  16. Confirm the bit weights sum back to 561.
  17. Answer: 561₁₀ = 1000110001₂.

Final answer: 1000110001

Foundations · Problem 0243

Express decimal 562 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 562.
  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 562 is at least 512; write bit 1; new remainder 50.
  5. Weight 256: skip it because remaining 50 is below 256; write bit 0; new remainder 50.
  6. Weight 128: skip it because remaining 50 is below 128; write bit 0; new remainder 50.
  7. Weight 64: skip it because remaining 50 is below 64; write bit 0; new remainder 50.
  8. Weight 32: take it because remaining 50 is at least 32; write bit 1; 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 1000110010.
  16. Confirm the bit weights sum back to 562.
  17. Answer: 562₁₀ = 1000110010₂.

Final answer: 1000110010

Foundations · Problem 0244

Express decimal 563 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 563.
  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 563 is at least 512; write bit 1; new remainder 51.
  5. Weight 256: skip it because remaining 51 is below 256; write bit 0; new remainder 51.
  6. Weight 128: skip it because remaining 51 is below 128; write bit 0; new remainder 51.
  7. Weight 64: skip it because remaining 51 is below 64; write bit 0; new remainder 51.
  8. Weight 32: take it because remaining 51 is at least 32; write bit 1; 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 1000110011.
  16. Confirm the bit weights sum back to 563.
  17. Answer: 563₁₀ = 1000110011₂.

Final answer: 1000110011

Foundations · Problem 0245

Express decimal 564 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 564.
  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 564 is at least 512; write bit 1; new remainder 52.
  5. Weight 256: skip it because remaining 52 is below 256; write bit 0; new remainder 52.
  6. Weight 128: skip it because remaining 52 is below 128; write bit 0; new remainder 52.
  7. Weight 64: skip it because remaining 52 is below 64; write bit 0; new remainder 52.
  8. Weight 32: take it because remaining 52 is at least 32; write bit 1; 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 1000110100.
  16. Confirm the bit weights sum back to 564.
  17. Answer: 564₁₀ = 1000110100₂.

Final answer: 1000110100

Foundations · Problem 0246

Express decimal 565 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 565.
  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 565 is at least 512; write bit 1; new remainder 53.
  5. Weight 256: skip it because remaining 53 is below 256; write bit 0; new remainder 53.
  6. Weight 128: skip it because remaining 53 is below 128; write bit 0; new remainder 53.
  7. Weight 64: skip it because remaining 53 is below 64; write bit 0; new remainder 53.
  8. Weight 32: take it because remaining 53 is at least 32; write bit 1; 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 1000110101.
  16. Confirm the bit weights sum back to 565.
  17. Answer: 565₁₀ = 1000110101₂.

Final answer: 1000110101

Foundations · Problem 0247

Express decimal 566 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 566.
  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 566 is at least 512; write bit 1; new remainder 54.
  5. Weight 256: skip it because remaining 54 is below 256; write bit 0; new remainder 54.
  6. Weight 128: skip it because remaining 54 is below 128; write bit 0; new remainder 54.
  7. Weight 64: skip it because remaining 54 is below 64; write bit 0; new remainder 54.
  8. Weight 32: take it because remaining 54 is at least 32; write bit 1; 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 1000110110.
  16. Confirm the bit weights sum back to 566.
  17. Answer: 566₁₀ = 1000110110₂.

Final answer: 1000110110

Foundations · Problem 0248

Express decimal 567 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 567.
  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 567 is at least 512; write bit 1; new remainder 55.
  5. Weight 256: skip it because remaining 55 is below 256; write bit 0; new remainder 55.
  6. Weight 128: skip it because remaining 55 is below 128; write bit 0; new remainder 55.
  7. Weight 64: skip it because remaining 55 is below 64; write bit 0; new remainder 55.
  8. Weight 32: take it because remaining 55 is at least 32; write bit 1; 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 1000110111.
  16. Confirm the bit weights sum back to 567.
  17. Answer: 567₁₀ = 1000110111₂.

Final answer: 1000110111

Foundations · Problem 0249

Express decimal 568 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 568.
  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 568 is at least 512; write bit 1; new remainder 56.
  5. Weight 256: skip it because remaining 56 is below 256; write bit 0; new remainder 56.
  6. Weight 128: skip it because remaining 56 is below 128; write bit 0; new remainder 56.
  7. Weight 64: skip it because remaining 56 is below 64; write bit 0; new remainder 56.
  8. Weight 32: take it because remaining 56 is at least 32; write bit 1; 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 1000111000.
  16. Confirm the bit weights sum back to 568.
  17. Answer: 568₁₀ = 1000111000₂.

Final answer: 1000111000

Foundations · Problem 0250

Express decimal 569 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 569.
  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 569 is at least 512; write bit 1; new remainder 57.
  5. Weight 256: skip it because remaining 57 is below 256; write bit 0; new remainder 57.
  6. Weight 128: skip it because remaining 57 is below 128; write bit 0; new remainder 57.
  7. Weight 64: skip it because remaining 57 is below 64; write bit 0; new remainder 57.
  8. Weight 32: take it because remaining 57 is at least 32; write bit 1; 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 1000111001.
  16. Confirm the bit weights sum back to 569.
  17. Answer: 569₁₀ = 1000111001₂.

Final answer: 1000111001

Foundations · Problem 0251

Express decimal 570 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 570.
  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 570 is at least 512; write bit 1; new remainder 58.
  5. Weight 256: skip it because remaining 58 is below 256; write bit 0; new remainder 58.
  6. Weight 128: skip it because remaining 58 is below 128; write bit 0; new remainder 58.
  7. Weight 64: skip it because remaining 58 is below 64; write bit 0; new remainder 58.
  8. Weight 32: take it because remaining 58 is at least 32; write bit 1; 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 1000111010.
  16. Confirm the bit weights sum back to 570.
  17. Answer: 570₁₀ = 1000111010₂.

Final answer: 1000111010

Foundations · Problem 0252

Express decimal 571 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 571.
  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 571 is at least 512; write bit 1; new remainder 59.
  5. Weight 256: skip it because remaining 59 is below 256; write bit 0; new remainder 59.
  6. Weight 128: skip it because remaining 59 is below 128; write bit 0; new remainder 59.
  7. Weight 64: skip it because remaining 59 is below 64; write bit 0; new remainder 59.
  8. Weight 32: take it because remaining 59 is at least 32; write bit 1; 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 1000111011.
  16. Confirm the bit weights sum back to 571.
  17. Answer: 571₁₀ = 1000111011₂.

Final answer: 1000111011

Foundations · Problem 0253

Express decimal 572 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 572.
  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 572 is at least 512; write bit 1; new remainder 60.
  5. Weight 256: skip it because remaining 60 is below 256; write bit 0; new remainder 60.
  6. Weight 128: skip it because remaining 60 is below 128; write bit 0; new remainder 60.
  7. Weight 64: skip it because remaining 60 is below 64; write bit 0; new remainder 60.
  8. Weight 32: take it because remaining 60 is at least 32; write bit 1; 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 1000111100.
  16. Confirm the bit weights sum back to 572.
  17. Answer: 572₁₀ = 1000111100₂.

Final answer: 1000111100

Foundations · Problem 0254

Express decimal 573 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 573.
  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 573 is at least 512; write bit 1; new remainder 61.
  5. Weight 256: skip it because remaining 61 is below 256; write bit 0; new remainder 61.
  6. Weight 128: skip it because remaining 61 is below 128; write bit 0; new remainder 61.
  7. Weight 64: skip it because remaining 61 is below 64; write bit 0; new remainder 61.
  8. Weight 32: take it because remaining 61 is at least 32; write bit 1; 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 1000111101.
  16. Confirm the bit weights sum back to 573.
  17. Answer: 573₁₀ = 1000111101₂.

Final answer: 1000111101

Foundations · Problem 0255

Express decimal 574 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 574.
  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 574 is at least 512; write bit 1; new remainder 62.
  5. Weight 256: skip it because remaining 62 is below 256; write bit 0; new remainder 62.
  6. Weight 128: skip it because remaining 62 is below 128; write bit 0; new remainder 62.
  7. Weight 64: skip it because remaining 62 is below 64; write bit 0; new remainder 62.
  8. Weight 32: take it because remaining 62 is at least 32; write bit 1; 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 1000111110.
  16. Confirm the bit weights sum back to 574.
  17. Answer: 574₁₀ = 1000111110₂.

Final answer: 1000111110

Foundations · Problem 0256

Express decimal 575 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 575.
  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 575 is at least 512; write bit 1; new remainder 63.
  5. Weight 256: skip it because remaining 63 is below 256; write bit 0; new remainder 63.
  6. Weight 128: skip it because remaining 63 is below 128; write bit 0; new remainder 63.
  7. Weight 64: skip it because remaining 63 is below 64; write bit 0; new remainder 63.
  8. Weight 32: take it because remaining 63 is at least 32; write bit 1; 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 1000111111.
  16. Confirm the bit weights sum back to 575.
  17. Answer: 575₁₀ = 1000111111₂.

Final answer: 1000111111