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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 6 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 0161

Express decimal 480 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 480.
  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: skip it because remaining 480 is below 512; write bit 0; new remainder 480.
  5. Weight 256: take it because remaining 480 is at least 256; write bit 1; new remainder 224.
  6. Weight 128: take it because remaining 224 is at least 128; write bit 1; 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 0111100000.
  16. Confirm the bit weights sum back to 480.
  17. Answer: 480₁₀ = 0111100000₂.

Final answer: 0111100000

Foundations · Problem 0162

Express decimal 481 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 481.
  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: skip it because remaining 481 is below 512; write bit 0; new remainder 481.
  5. Weight 256: take it because remaining 481 is at least 256; write bit 1; new remainder 225.
  6. Weight 128: take it because remaining 225 is at least 128; write bit 1; 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 0111100001.
  16. Confirm the bit weights sum back to 481.
  17. Answer: 481₁₀ = 0111100001₂.

Final answer: 0111100001

Foundations · Problem 0163

Express decimal 482 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 482.
  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: skip it because remaining 482 is below 512; write bit 0; new remainder 482.
  5. Weight 256: take it because remaining 482 is at least 256; write bit 1; new remainder 226.
  6. Weight 128: take it because remaining 226 is at least 128; write bit 1; 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 0111100010.
  16. Confirm the bit weights sum back to 482.
  17. Answer: 482₁₀ = 0111100010₂.

Final answer: 0111100010

Foundations · Problem 0164

Express decimal 483 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 483.
  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: skip it because remaining 483 is below 512; write bit 0; new remainder 483.
  5. Weight 256: take it because remaining 483 is at least 256; write bit 1; new remainder 227.
  6. Weight 128: take it because remaining 227 is at least 128; write bit 1; 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 0111100011.
  16. Confirm the bit weights sum back to 483.
  17. Answer: 483₁₀ = 0111100011₂.

Final answer: 0111100011

Foundations · Problem 0165

Express decimal 484 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 484.
  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: skip it because remaining 484 is below 512; write bit 0; new remainder 484.
  5. Weight 256: take it because remaining 484 is at least 256; write bit 1; new remainder 228.
  6. Weight 128: take it because remaining 228 is at least 128; write bit 1; 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 0111100100.
  16. Confirm the bit weights sum back to 484.
  17. Answer: 484₁₀ = 0111100100₂.

Final answer: 0111100100

Foundations · Problem 0166

Express decimal 485 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 485.
  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: skip it because remaining 485 is below 512; write bit 0; new remainder 485.
  5. Weight 256: take it because remaining 485 is at least 256; write bit 1; new remainder 229.
  6. Weight 128: take it because remaining 229 is at least 128; write bit 1; 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 0111100101.
  16. Confirm the bit weights sum back to 485.
  17. Answer: 485₁₀ = 0111100101₂.

Final answer: 0111100101

Foundations · Problem 0167

Express decimal 486 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 486.
  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: skip it because remaining 486 is below 512; write bit 0; new remainder 486.
  5. Weight 256: take it because remaining 486 is at least 256; write bit 1; new remainder 230.
  6. Weight 128: take it because remaining 230 is at least 128; write bit 1; 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 0111100110.
  16. Confirm the bit weights sum back to 486.
  17. Answer: 486₁₀ = 0111100110₂.

Final answer: 0111100110

Foundations · Problem 0168

Express decimal 487 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 487.
  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: skip it because remaining 487 is below 512; write bit 0; new remainder 487.
  5. Weight 256: take it because remaining 487 is at least 256; write bit 1; new remainder 231.
  6. Weight 128: take it because remaining 231 is at least 128; write bit 1; 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 0111100111.
  16. Confirm the bit weights sum back to 487.
  17. Answer: 487₁₀ = 0111100111₂.

Final answer: 0111100111

Foundations · Problem 0169

Express decimal 488 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 488.
  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: skip it because remaining 488 is below 512; write bit 0; new remainder 488.
  5. Weight 256: take it because remaining 488 is at least 256; write bit 1; new remainder 232.
  6. Weight 128: take it because remaining 232 is at least 128; write bit 1; 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 0111101000.
  16. Confirm the bit weights sum back to 488.
  17. Answer: 488₁₀ = 0111101000₂.

Final answer: 0111101000

Foundations · Problem 0170

Express decimal 489 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 489.
  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: skip it because remaining 489 is below 512; write bit 0; new remainder 489.
  5. Weight 256: take it because remaining 489 is at least 256; write bit 1; new remainder 233.
  6. Weight 128: take it because remaining 233 is at least 128; write bit 1; 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 0111101001.
  16. Confirm the bit weights sum back to 489.
  17. Answer: 489₁₀ = 0111101001₂.

Final answer: 0111101001

Foundations · Problem 0171

Express decimal 490 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 490.
  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: skip it because remaining 490 is below 512; write bit 0; new remainder 490.
  5. Weight 256: take it because remaining 490 is at least 256; write bit 1; new remainder 234.
  6. Weight 128: take it because remaining 234 is at least 128; write bit 1; 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 0111101010.
  16. Confirm the bit weights sum back to 490.
  17. Answer: 490₁₀ = 0111101010₂.

Final answer: 0111101010

Foundations · Problem 0172

Express decimal 491 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 491.
  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: skip it because remaining 491 is below 512; write bit 0; new remainder 491.
  5. Weight 256: take it because remaining 491 is at least 256; write bit 1; new remainder 235.
  6. Weight 128: take it because remaining 235 is at least 128; write bit 1; 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 0111101011.
  16. Confirm the bit weights sum back to 491.
  17. Answer: 491₁₀ = 0111101011₂.

Final answer: 0111101011

Foundations · Problem 0173

Express decimal 492 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 492.
  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: skip it because remaining 492 is below 512; write bit 0; new remainder 492.
  5. Weight 256: take it because remaining 492 is at least 256; write bit 1; new remainder 236.
  6. Weight 128: take it because remaining 236 is at least 128; write bit 1; 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 0111101100.
  16. Confirm the bit weights sum back to 492.
  17. Answer: 492₁₀ = 0111101100₂.

Final answer: 0111101100

Foundations · Problem 0174

Express decimal 493 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 493.
  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: skip it because remaining 493 is below 512; write bit 0; new remainder 493.
  5. Weight 256: take it because remaining 493 is at least 256; write bit 1; new remainder 237.
  6. Weight 128: take it because remaining 237 is at least 128; write bit 1; 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 0111101101.
  16. Confirm the bit weights sum back to 493.
  17. Answer: 493₁₀ = 0111101101₂.

Final answer: 0111101101

Foundations · Problem 0175

Express decimal 494 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 494.
  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: skip it because remaining 494 is below 512; write bit 0; new remainder 494.
  5. Weight 256: take it because remaining 494 is at least 256; write bit 1; new remainder 238.
  6. Weight 128: take it because remaining 238 is at least 128; write bit 1; 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 0111101110.
  16. Confirm the bit weights sum back to 494.
  17. Answer: 494₁₀ = 0111101110₂.

Final answer: 0111101110

Foundations · Problem 0176

Express decimal 495 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 495.
  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: skip it because remaining 495 is below 512; write bit 0; new remainder 495.
  5. Weight 256: take it because remaining 495 is at least 256; write bit 1; new remainder 239.
  6. Weight 128: take it because remaining 239 is at least 128; write bit 1; 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 0111101111.
  16. Confirm the bit weights sum back to 495.
  17. Answer: 495₁₀ = 0111101111₂.

Final answer: 0111101111

Foundations · Problem 0177

Express decimal 496 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 496.
  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: skip it because remaining 496 is below 512; write bit 0; new remainder 496.
  5. Weight 256: take it because remaining 496 is at least 256; write bit 1; new remainder 240.
  6. Weight 128: take it because remaining 240 is at least 128; write bit 1; 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 0111110000.
  16. Confirm the bit weights sum back to 496.
  17. Answer: 496₁₀ = 0111110000₂.

Final answer: 0111110000

Foundations · Problem 0178

Express decimal 497 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 497.
  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: skip it because remaining 497 is below 512; write bit 0; new remainder 497.
  5. Weight 256: take it because remaining 497 is at least 256; write bit 1; new remainder 241.
  6. Weight 128: take it because remaining 241 is at least 128; write bit 1; 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 0111110001.
  16. Confirm the bit weights sum back to 497.
  17. Answer: 497₁₀ = 0111110001₂.

Final answer: 0111110001

Foundations · Problem 0179

Express decimal 498 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 498.
  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: skip it because remaining 498 is below 512; write bit 0; new remainder 498.
  5. Weight 256: take it because remaining 498 is at least 256; write bit 1; new remainder 242.
  6. Weight 128: take it because remaining 242 is at least 128; write bit 1; 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 0111110010.
  16. Confirm the bit weights sum back to 498.
  17. Answer: 498₁₀ = 0111110010₂.

Final answer: 0111110010

Foundations · Problem 0180

Express decimal 499 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 499.
  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: skip it because remaining 499 is below 512; write bit 0; new remainder 499.
  5. Weight 256: take it because remaining 499 is at least 256; write bit 1; new remainder 243.
  6. Weight 128: take it because remaining 243 is at least 128; write bit 1; 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 0111110011.
  16. Confirm the bit weights sum back to 499.
  17. Answer: 499₁₀ = 0111110011₂.

Final answer: 0111110011

Foundations · Problem 0181

Express decimal 500 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 500.
  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: skip it because remaining 500 is below 512; write bit 0; new remainder 500.
  5. Weight 256: take it because remaining 500 is at least 256; write bit 1; new remainder 244.
  6. Weight 128: take it because remaining 244 is at least 128; write bit 1; 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 0111110100.
  16. Confirm the bit weights sum back to 500.
  17. Answer: 500₁₀ = 0111110100₂.

Final answer: 0111110100

Foundations · Problem 0182

Express decimal 501 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 501.
  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: skip it because remaining 501 is below 512; write bit 0; new remainder 501.
  5. Weight 256: take it because remaining 501 is at least 256; write bit 1; new remainder 245.
  6. Weight 128: take it because remaining 245 is at least 128; write bit 1; 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 0111110101.
  16. Confirm the bit weights sum back to 501.
  17. Answer: 501₁₀ = 0111110101₂.

Final answer: 0111110101

Foundations · Problem 0183

Express decimal 502 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 502.
  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: skip it because remaining 502 is below 512; write bit 0; new remainder 502.
  5. Weight 256: take it because remaining 502 is at least 256; write bit 1; new remainder 246.
  6. Weight 128: take it because remaining 246 is at least 128; write bit 1; 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 0111110110.
  16. Confirm the bit weights sum back to 502.
  17. Answer: 502₁₀ = 0111110110₂.

Final answer: 0111110110

Foundations · Problem 0184

Express decimal 503 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 503.
  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: skip it because remaining 503 is below 512; write bit 0; new remainder 503.
  5. Weight 256: take it because remaining 503 is at least 256; write bit 1; new remainder 247.
  6. Weight 128: take it because remaining 247 is at least 128; write bit 1; 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 0111110111.
  16. Confirm the bit weights sum back to 503.
  17. Answer: 503₁₀ = 0111110111₂.

Final answer: 0111110111

Foundations · Problem 0185

Express decimal 504 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 504.
  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: skip it because remaining 504 is below 512; write bit 0; new remainder 504.
  5. Weight 256: take it because remaining 504 is at least 256; write bit 1; new remainder 248.
  6. Weight 128: take it because remaining 248 is at least 128; write bit 1; 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 0111111000.
  16. Confirm the bit weights sum back to 504.
  17. Answer: 504₁₀ = 0111111000₂.

Final answer: 0111111000

Foundations · Problem 0186

Express decimal 505 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 505.
  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: skip it because remaining 505 is below 512; write bit 0; new remainder 505.
  5. Weight 256: take it because remaining 505 is at least 256; write bit 1; new remainder 249.
  6. Weight 128: take it because remaining 249 is at least 128; write bit 1; 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 0111111001.
  16. Confirm the bit weights sum back to 505.
  17. Answer: 505₁₀ = 0111111001₂.

Final answer: 0111111001

Foundations · Problem 0187

Express decimal 506 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 506.
  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: skip it because remaining 506 is below 512; write bit 0; new remainder 506.
  5. Weight 256: take it because remaining 506 is at least 256; write bit 1; new remainder 250.
  6. Weight 128: take it because remaining 250 is at least 128; write bit 1; 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 0111111010.
  16. Confirm the bit weights sum back to 506.
  17. Answer: 506₁₀ = 0111111010₂.

Final answer: 0111111010

Foundations · Problem 0188

Express decimal 507 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 507.
  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: skip it because remaining 507 is below 512; write bit 0; new remainder 507.
  5. Weight 256: take it because remaining 507 is at least 256; write bit 1; new remainder 251.
  6. Weight 128: take it because remaining 251 is at least 128; write bit 1; 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 0111111011.
  16. Confirm the bit weights sum back to 507.
  17. Answer: 507₁₀ = 0111111011₂.

Final answer: 0111111011

Foundations · Problem 0189

Express decimal 508 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 508.
  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: skip it because remaining 508 is below 512; write bit 0; new remainder 508.
  5. Weight 256: take it because remaining 508 is at least 256; write bit 1; new remainder 252.
  6. Weight 128: take it because remaining 252 is at least 128; write bit 1; 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 0111111100.
  16. Confirm the bit weights sum back to 508.
  17. Answer: 508₁₀ = 0111111100₂.

Final answer: 0111111100

Foundations · Problem 0190

Express decimal 509 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 509.
  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: skip it because remaining 509 is below 512; write bit 0; new remainder 509.
  5. Weight 256: take it because remaining 509 is at least 256; write bit 1; new remainder 253.
  6. Weight 128: take it because remaining 253 is at least 128; write bit 1; 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 0111111101.
  16. Confirm the bit weights sum back to 509.
  17. Answer: 509₁₀ = 0111111101₂.

Final answer: 0111111101

Foundations · Problem 0191

Express decimal 510 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 510.
  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: skip it because remaining 510 is below 512; write bit 0; new remainder 510.
  5. Weight 256: take it because remaining 510 is at least 256; write bit 1; new remainder 254.
  6. Weight 128: take it because remaining 254 is at least 128; write bit 1; 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 0111111110.
  16. Confirm the bit weights sum back to 510.
  17. Answer: 510₁₀ = 0111111110₂.

Final answer: 0111111110

Foundations · Problem 0192

Express decimal 511 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 511.
  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: skip it because remaining 511 is below 512; write bit 0; new remainder 511.
  5. Weight 256: take it because remaining 511 is at least 256; write bit 1; new remainder 255.
  6. Weight 128: take it because remaining 255 is at least 128; write bit 1; 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 0111111111.
  16. Confirm the bit weights sum back to 511.
  17. Answer: 511₁₀ = 0111111111₂.

Final answer: 0111111111