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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 10 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 0289

Express decimal 608 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 608.
  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 608 is at least 512; write bit 1; new remainder 96.
  5. Weight 256: skip it because remaining 96 is below 256; write bit 0; new remainder 96.
  6. Weight 128: skip it because remaining 96 is below 128; write bit 0; new remainder 96.
  7. Weight 64: take it because remaining 96 is at least 64; write bit 1; 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 1001100000.
  16. Confirm the bit weights sum back to 608.
  17. Answer: 608₁₀ = 1001100000₂.

Final answer: 1001100000

Foundations · Problem 0290

Express decimal 609 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 609.
  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 609 is at least 512; write bit 1; new remainder 97.
  5. Weight 256: skip it because remaining 97 is below 256; write bit 0; new remainder 97.
  6. Weight 128: skip it because remaining 97 is below 128; write bit 0; new remainder 97.
  7. Weight 64: take it because remaining 97 is at least 64; write bit 1; 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 1001100001.
  16. Confirm the bit weights sum back to 609.
  17. Answer: 609₁₀ = 1001100001₂.

Final answer: 1001100001

Foundations · Problem 0291

Express decimal 610 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 610.
  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 610 is at least 512; write bit 1; new remainder 98.
  5. Weight 256: skip it because remaining 98 is below 256; write bit 0; new remainder 98.
  6. Weight 128: skip it because remaining 98 is below 128; write bit 0; new remainder 98.
  7. Weight 64: take it because remaining 98 is at least 64; write bit 1; 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 1001100010.
  16. Confirm the bit weights sum back to 610.
  17. Answer: 610₁₀ = 1001100010₂.

Final answer: 1001100010

Foundations · Problem 0292

Express decimal 611 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 611.
  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 611 is at least 512; write bit 1; new remainder 99.
  5. Weight 256: skip it because remaining 99 is below 256; write bit 0; new remainder 99.
  6. Weight 128: skip it because remaining 99 is below 128; write bit 0; new remainder 99.
  7. Weight 64: take it because remaining 99 is at least 64; write bit 1; 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 1001100011.
  16. Confirm the bit weights sum back to 611.
  17. Answer: 611₁₀ = 1001100011₂.

Final answer: 1001100011

Foundations · Problem 0293

Express decimal 612 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 612.
  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 612 is at least 512; write bit 1; new remainder 100.
  5. Weight 256: skip it because remaining 100 is below 256; write bit 0; new remainder 100.
  6. Weight 128: skip it because remaining 100 is below 128; write bit 0; new remainder 100.
  7. Weight 64: take it because remaining 100 is at least 64; write bit 1; 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 1001100100.
  16. Confirm the bit weights sum back to 612.
  17. Answer: 612₁₀ = 1001100100₂.

Final answer: 1001100100

Foundations · Problem 0294

Express decimal 613 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 613.
  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 613 is at least 512; write bit 1; new remainder 101.
  5. Weight 256: skip it because remaining 101 is below 256; write bit 0; new remainder 101.
  6. Weight 128: skip it because remaining 101 is below 128; write bit 0; new remainder 101.
  7. Weight 64: take it because remaining 101 is at least 64; write bit 1; 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 1001100101.
  16. Confirm the bit weights sum back to 613.
  17. Answer: 613₁₀ = 1001100101₂.

Final answer: 1001100101

Foundations · Problem 0295

Express decimal 614 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 614.
  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 614 is at least 512; write bit 1; new remainder 102.
  5. Weight 256: skip it because remaining 102 is below 256; write bit 0; new remainder 102.
  6. Weight 128: skip it because remaining 102 is below 128; write bit 0; new remainder 102.
  7. Weight 64: take it because remaining 102 is at least 64; write bit 1; 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 1001100110.
  16. Confirm the bit weights sum back to 614.
  17. Answer: 614₁₀ = 1001100110₂.

Final answer: 1001100110

Foundations · Problem 0296

Express decimal 615 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 615.
  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 615 is at least 512; write bit 1; new remainder 103.
  5. Weight 256: skip it because remaining 103 is below 256; write bit 0; new remainder 103.
  6. Weight 128: skip it because remaining 103 is below 128; write bit 0; new remainder 103.
  7. Weight 64: take it because remaining 103 is at least 64; write bit 1; 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 1001100111.
  16. Confirm the bit weights sum back to 615.
  17. Answer: 615₁₀ = 1001100111₂.

Final answer: 1001100111

Foundations · Problem 0297

Express decimal 616 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 616.
  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 616 is at least 512; write bit 1; new remainder 104.
  5. Weight 256: skip it because remaining 104 is below 256; write bit 0; new remainder 104.
  6. Weight 128: skip it because remaining 104 is below 128; write bit 0; new remainder 104.
  7. Weight 64: take it because remaining 104 is at least 64; write bit 1; 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 1001101000.
  16. Confirm the bit weights sum back to 616.
  17. Answer: 616₁₀ = 1001101000₂.

Final answer: 1001101000

Foundations · Problem 0298

Express decimal 617 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 617.
  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 617 is at least 512; write bit 1; new remainder 105.
  5. Weight 256: skip it because remaining 105 is below 256; write bit 0; new remainder 105.
  6. Weight 128: skip it because remaining 105 is below 128; write bit 0; new remainder 105.
  7. Weight 64: take it because remaining 105 is at least 64; write bit 1; 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 1001101001.
  16. Confirm the bit weights sum back to 617.
  17. Answer: 617₁₀ = 1001101001₂.

Final answer: 1001101001

Foundations · Problem 0299

Express decimal 618 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 618.
  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 618 is at least 512; write bit 1; new remainder 106.
  5. Weight 256: skip it because remaining 106 is below 256; write bit 0; new remainder 106.
  6. Weight 128: skip it because remaining 106 is below 128; write bit 0; new remainder 106.
  7. Weight 64: take it because remaining 106 is at least 64; write bit 1; 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 1001101010.
  16. Confirm the bit weights sum back to 618.
  17. Answer: 618₁₀ = 1001101010₂.

Final answer: 1001101010

Foundations · Problem 0300

Express decimal 619 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 619.
  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 619 is at least 512; write bit 1; new remainder 107.
  5. Weight 256: skip it because remaining 107 is below 256; write bit 0; new remainder 107.
  6. Weight 128: skip it because remaining 107 is below 128; write bit 0; new remainder 107.
  7. Weight 64: take it because remaining 107 is at least 64; write bit 1; 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 1001101011.
  16. Confirm the bit weights sum back to 619.
  17. Answer: 619₁₀ = 1001101011₂.

Final answer: 1001101011

Foundations · Problem 0301

Express decimal 620 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 620.
  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 620 is at least 512; write bit 1; new remainder 108.
  5. Weight 256: skip it because remaining 108 is below 256; write bit 0; new remainder 108.
  6. Weight 128: skip it because remaining 108 is below 128; write bit 0; new remainder 108.
  7. Weight 64: take it because remaining 108 is at least 64; write bit 1; 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 1001101100.
  16. Confirm the bit weights sum back to 620.
  17. Answer: 620₁₀ = 1001101100₂.

Final answer: 1001101100

Foundations · Problem 0302

Express decimal 621 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 621.
  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 621 is at least 512; write bit 1; new remainder 109.
  5. Weight 256: skip it because remaining 109 is below 256; write bit 0; new remainder 109.
  6. Weight 128: skip it because remaining 109 is below 128; write bit 0; new remainder 109.
  7. Weight 64: take it because remaining 109 is at least 64; write bit 1; 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 1001101101.
  16. Confirm the bit weights sum back to 621.
  17. Answer: 621₁₀ = 1001101101₂.

Final answer: 1001101101

Foundations · Problem 0303

Express decimal 622 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 622.
  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 622 is at least 512; write bit 1; new remainder 110.
  5. Weight 256: skip it because remaining 110 is below 256; write bit 0; new remainder 110.
  6. Weight 128: skip it because remaining 110 is below 128; write bit 0; new remainder 110.
  7. Weight 64: take it because remaining 110 is at least 64; write bit 1; 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 1001101110.
  16. Confirm the bit weights sum back to 622.
  17. Answer: 622₁₀ = 1001101110₂.

Final answer: 1001101110

Foundations · Problem 0304

Express decimal 623 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 623.
  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 623 is at least 512; write bit 1; new remainder 111.
  5. Weight 256: skip it because remaining 111 is below 256; write bit 0; new remainder 111.
  6. Weight 128: skip it because remaining 111 is below 128; write bit 0; new remainder 111.
  7. Weight 64: take it because remaining 111 is at least 64; write bit 1; 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 1001101111.
  16. Confirm the bit weights sum back to 623.
  17. Answer: 623₁₀ = 1001101111₂.

Final answer: 1001101111

Foundations · Problem 0305

Express decimal 624 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 624.
  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 624 is at least 512; write bit 1; new remainder 112.
  5. Weight 256: skip it because remaining 112 is below 256; write bit 0; new remainder 112.
  6. Weight 128: skip it because remaining 112 is below 128; write bit 0; new remainder 112.
  7. Weight 64: take it because remaining 112 is at least 64; write bit 1; 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 1001110000.
  16. Confirm the bit weights sum back to 624.
  17. Answer: 624₁₀ = 1001110000₂.

Final answer: 1001110000

Foundations · Problem 0306

Express decimal 625 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 625.
  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 625 is at least 512; write bit 1; new remainder 113.
  5. Weight 256: skip it because remaining 113 is below 256; write bit 0; new remainder 113.
  6. Weight 128: skip it because remaining 113 is below 128; write bit 0; new remainder 113.
  7. Weight 64: take it because remaining 113 is at least 64; write bit 1; 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 1001110001.
  16. Confirm the bit weights sum back to 625.
  17. Answer: 625₁₀ = 1001110001₂.

Final answer: 1001110001

Foundations · Problem 0307

Express decimal 626 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 626.
  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 626 is at least 512; write bit 1; new remainder 114.
  5. Weight 256: skip it because remaining 114 is below 256; write bit 0; new remainder 114.
  6. Weight 128: skip it because remaining 114 is below 128; write bit 0; new remainder 114.
  7. Weight 64: take it because remaining 114 is at least 64; write bit 1; 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 1001110010.
  16. Confirm the bit weights sum back to 626.
  17. Answer: 626₁₀ = 1001110010₂.

Final answer: 1001110010

Foundations · Problem 0308

Express decimal 627 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 627.
  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 627 is at least 512; write bit 1; new remainder 115.
  5. Weight 256: skip it because remaining 115 is below 256; write bit 0; new remainder 115.
  6. Weight 128: skip it because remaining 115 is below 128; write bit 0; new remainder 115.
  7. Weight 64: take it because remaining 115 is at least 64; write bit 1; 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 1001110011.
  16. Confirm the bit weights sum back to 627.
  17. Answer: 627₁₀ = 1001110011₂.

Final answer: 1001110011

Foundations · Problem 0309

Express decimal 628 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 628.
  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 628 is at least 512; write bit 1; new remainder 116.
  5. Weight 256: skip it because remaining 116 is below 256; write bit 0; new remainder 116.
  6. Weight 128: skip it because remaining 116 is below 128; write bit 0; new remainder 116.
  7. Weight 64: take it because remaining 116 is at least 64; write bit 1; 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 1001110100.
  16. Confirm the bit weights sum back to 628.
  17. Answer: 628₁₀ = 1001110100₂.

Final answer: 1001110100

Foundations · Problem 0310

Express decimal 629 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 629.
  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 629 is at least 512; write bit 1; new remainder 117.
  5. Weight 256: skip it because remaining 117 is below 256; write bit 0; new remainder 117.
  6. Weight 128: skip it because remaining 117 is below 128; write bit 0; new remainder 117.
  7. Weight 64: take it because remaining 117 is at least 64; write bit 1; 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 1001110101.
  16. Confirm the bit weights sum back to 629.
  17. Answer: 629₁₀ = 1001110101₂.

Final answer: 1001110101

Foundations · Problem 0311

Express decimal 630 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 630.
  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 630 is at least 512; write bit 1; new remainder 118.
  5. Weight 256: skip it because remaining 118 is below 256; write bit 0; new remainder 118.
  6. Weight 128: skip it because remaining 118 is below 128; write bit 0; new remainder 118.
  7. Weight 64: take it because remaining 118 is at least 64; write bit 1; 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 1001110110.
  16. Confirm the bit weights sum back to 630.
  17. Answer: 630₁₀ = 1001110110₂.

Final answer: 1001110110

Foundations · Problem 0312

Express decimal 631 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 631.
  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 631 is at least 512; write bit 1; new remainder 119.
  5. Weight 256: skip it because remaining 119 is below 256; write bit 0; new remainder 119.
  6. Weight 128: skip it because remaining 119 is below 128; write bit 0; new remainder 119.
  7. Weight 64: take it because remaining 119 is at least 64; write bit 1; 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 1001110111.
  16. Confirm the bit weights sum back to 631.
  17. Answer: 631₁₀ = 1001110111₂.

Final answer: 1001110111

Foundations · Problem 0313

Express decimal 632 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 632.
  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 632 is at least 512; write bit 1; new remainder 120.
  5. Weight 256: skip it because remaining 120 is below 256; write bit 0; new remainder 120.
  6. Weight 128: skip it because remaining 120 is below 128; write bit 0; new remainder 120.
  7. Weight 64: take it because remaining 120 is at least 64; write bit 1; 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 1001111000.
  16. Confirm the bit weights sum back to 632.
  17. Answer: 632₁₀ = 1001111000₂.

Final answer: 1001111000

Foundations · Problem 0314

Express decimal 633 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 633.
  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 633 is at least 512; write bit 1; new remainder 121.
  5. Weight 256: skip it because remaining 121 is below 256; write bit 0; new remainder 121.
  6. Weight 128: skip it because remaining 121 is below 128; write bit 0; new remainder 121.
  7. Weight 64: take it because remaining 121 is at least 64; write bit 1; 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 1001111001.
  16. Confirm the bit weights sum back to 633.
  17. Answer: 633₁₀ = 1001111001₂.

Final answer: 1001111001

Foundations · Problem 0315

Express decimal 634 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 634.
  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 634 is at least 512; write bit 1; new remainder 122.
  5. Weight 256: skip it because remaining 122 is below 256; write bit 0; new remainder 122.
  6. Weight 128: skip it because remaining 122 is below 128; write bit 0; new remainder 122.
  7. Weight 64: take it because remaining 122 is at least 64; write bit 1; 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 1001111010.
  16. Confirm the bit weights sum back to 634.
  17. Answer: 634₁₀ = 1001111010₂.

Final answer: 1001111010

Foundations · Problem 0316

Express decimal 635 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 635.
  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 635 is at least 512; write bit 1; new remainder 123.
  5. Weight 256: skip it because remaining 123 is below 256; write bit 0; new remainder 123.
  6. Weight 128: skip it because remaining 123 is below 128; write bit 0; new remainder 123.
  7. Weight 64: take it because remaining 123 is at least 64; write bit 1; 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 1001111011.
  16. Confirm the bit weights sum back to 635.
  17. Answer: 635₁₀ = 1001111011₂.

Final answer: 1001111011

Foundations · Problem 0317

Express decimal 636 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 636.
  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 636 is at least 512; write bit 1; new remainder 124.
  5. Weight 256: skip it because remaining 124 is below 256; write bit 0; new remainder 124.
  6. Weight 128: skip it because remaining 124 is below 128; write bit 0; new remainder 124.
  7. Weight 64: take it because remaining 124 is at least 64; write bit 1; 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 1001111100.
  16. Confirm the bit weights sum back to 636.
  17. Answer: 636₁₀ = 1001111100₂.

Final answer: 1001111100

Foundations · Problem 0318

Express decimal 637 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 637.
  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 637 is at least 512; write bit 1; new remainder 125.
  5. Weight 256: skip it because remaining 125 is below 256; write bit 0; new remainder 125.
  6. Weight 128: skip it because remaining 125 is below 128; write bit 0; new remainder 125.
  7. Weight 64: take it because remaining 125 is at least 64; write bit 1; 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 1001111101.
  16. Confirm the bit weights sum back to 637.
  17. Answer: 637₁₀ = 1001111101₂.

Final answer: 1001111101

Foundations · Problem 0319

Express decimal 638 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 638.
  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 638 is at least 512; write bit 1; new remainder 126.
  5. Weight 256: skip it because remaining 126 is below 256; write bit 0; new remainder 126.
  6. Weight 128: skip it because remaining 126 is below 128; write bit 0; new remainder 126.
  7. Weight 64: take it because remaining 126 is at least 64; write bit 1; 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 1001111110.
  16. Confirm the bit weights sum back to 638.
  17. Answer: 638₁₀ = 1001111110₂.

Final answer: 1001111110

Foundations · Problem 0320

Express decimal 639 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 639.
  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 639 is at least 512; write bit 1; new remainder 127.
  5. Weight 256: skip it because remaining 127 is below 256; write bit 0; new remainder 127.
  6. Weight 128: skip it because remaining 127 is below 128; write bit 0; new remainder 127.
  7. Weight 64: take it because remaining 127 is at least 64; write bit 1; 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 1001111111.
  16. Confirm the bit weights sum back to 639.
  17. Answer: 639₁₀ = 1001111111₂.

Final answer: 1001111111