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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 4 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 0097

Express decimal 416 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 416.
  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 416 is below 512; write bit 0; new remainder 416.
  5. Weight 256: take it because remaining 416 is at least 256; write bit 1; new remainder 160.
  6. Weight 128: take it because remaining 160 is at least 128; write bit 1; new remainder 32.
  7. Weight 64: skip it because remaining 32 is below 64; write bit 0; new remainder 32.
  8. Weight 32: take it because remaining 32 is at least 32; write bit 1; new remainder 0.
  9. Weight 16: skip it because remaining 0 is below 16; write bit 0; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100000.
  16. Confirm the bit weights sum back to 416.
  17. Answer: 416₁₀ = 0110100000₂.

Final answer: 0110100000

Foundations · Problem 0098

Express decimal 417 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 417.
  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 417 is below 512; write bit 0; new remainder 417.
  5. Weight 256: take it because remaining 417 is at least 256; write bit 1; new remainder 161.
  6. Weight 128: take it because remaining 161 is at least 128; write bit 1; new remainder 33.
  7. Weight 64: skip it because remaining 33 is below 64; write bit 0; new remainder 33.
  8. Weight 32: take it because remaining 33 is at least 32; write bit 1; new remainder 1.
  9. Weight 16: skip it because remaining 1 is below 16; write bit 0; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100001.
  16. Confirm the bit weights sum back to 417.
  17. Answer: 417₁₀ = 0110100001₂.

Final answer: 0110100001

Foundations · Problem 0099

Express decimal 418 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 418.
  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 418 is below 512; write bit 0; new remainder 418.
  5. Weight 256: take it because remaining 418 is at least 256; write bit 1; new remainder 162.
  6. Weight 128: take it because remaining 162 is at least 128; write bit 1; new remainder 34.
  7. Weight 64: skip it because remaining 34 is below 64; write bit 0; new remainder 34.
  8. Weight 32: take it because remaining 34 is at least 32; write bit 1; new remainder 2.
  9. Weight 16: skip it because remaining 2 is below 16; write bit 0; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100010.
  16. Confirm the bit weights sum back to 418.
  17. Answer: 418₁₀ = 0110100010₂.

Final answer: 0110100010

Foundations · Problem 0100

Express decimal 419 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 419.
  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 419 is below 512; write bit 0; new remainder 419.
  5. Weight 256: take it because remaining 419 is at least 256; write bit 1; new remainder 163.
  6. Weight 128: take it because remaining 163 is at least 128; write bit 1; new remainder 35.
  7. Weight 64: skip it because remaining 35 is below 64; write bit 0; new remainder 35.
  8. Weight 32: take it because remaining 35 is at least 32; write bit 1; new remainder 3.
  9. Weight 16: skip it because remaining 3 is below 16; write bit 0; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100011.
  16. Confirm the bit weights sum back to 419.
  17. Answer: 419₁₀ = 0110100011₂.

Final answer: 0110100011

Foundations · Problem 0101

Express decimal 420 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 420.
  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 420 is below 512; write bit 0; new remainder 420.
  5. Weight 256: take it because remaining 420 is at least 256; write bit 1; new remainder 164.
  6. Weight 128: take it because remaining 164 is at least 128; write bit 1; new remainder 36.
  7. Weight 64: skip it because remaining 36 is below 64; write bit 0; new remainder 36.
  8. Weight 32: take it because remaining 36 is at least 32; write bit 1; new remainder 4.
  9. Weight 16: skip it because remaining 4 is below 16; write bit 0; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100100.
  16. Confirm the bit weights sum back to 420.
  17. Answer: 420₁₀ = 0110100100₂.

Final answer: 0110100100

Foundations · Problem 0102

Express decimal 421 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 421.
  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 421 is below 512; write bit 0; new remainder 421.
  5. Weight 256: take it because remaining 421 is at least 256; write bit 1; new remainder 165.
  6. Weight 128: take it because remaining 165 is at least 128; write bit 1; new remainder 37.
  7. Weight 64: skip it because remaining 37 is below 64; write bit 0; new remainder 37.
  8. Weight 32: take it because remaining 37 is at least 32; write bit 1; new remainder 5.
  9. Weight 16: skip it because remaining 5 is below 16; write bit 0; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100101.
  16. Confirm the bit weights sum back to 421.
  17. Answer: 421₁₀ = 0110100101₂.

Final answer: 0110100101

Foundations · Problem 0103

Express decimal 422 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 422.
  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 422 is below 512; write bit 0; new remainder 422.
  5. Weight 256: take it because remaining 422 is at least 256; write bit 1; new remainder 166.
  6. Weight 128: take it because remaining 166 is at least 128; write bit 1; new remainder 38.
  7. Weight 64: skip it because remaining 38 is below 64; write bit 0; new remainder 38.
  8. Weight 32: take it because remaining 38 is at least 32; write bit 1; new remainder 6.
  9. Weight 16: skip it because remaining 6 is below 16; write bit 0; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100110.
  16. Confirm the bit weights sum back to 422.
  17. Answer: 422₁₀ = 0110100110₂.

Final answer: 0110100110

Foundations · Problem 0104

Express decimal 423 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 423.
  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 423 is below 512; write bit 0; new remainder 423.
  5. Weight 256: take it because remaining 423 is at least 256; write bit 1; new remainder 167.
  6. Weight 128: take it because remaining 167 is at least 128; write bit 1; new remainder 39.
  7. Weight 64: skip it because remaining 39 is below 64; write bit 0; new remainder 39.
  8. Weight 32: take it because remaining 39 is at least 32; write bit 1; new remainder 7.
  9. Weight 16: skip it because remaining 7 is below 16; write bit 0; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110100111.
  16. Confirm the bit weights sum back to 423.
  17. Answer: 423₁₀ = 0110100111₂.

Final answer: 0110100111

Foundations · Problem 0105

Express decimal 424 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 424.
  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 424 is below 512; write bit 0; new remainder 424.
  5. Weight 256: take it because remaining 424 is at least 256; write bit 1; new remainder 168.
  6. Weight 128: take it because remaining 168 is at least 128; write bit 1; new remainder 40.
  7. Weight 64: skip it because remaining 40 is below 64; write bit 0; new remainder 40.
  8. Weight 32: take it because remaining 40 is at least 32; write bit 1; new remainder 8.
  9. Weight 16: skip it because remaining 8 is below 16; write bit 0; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101000.
  16. Confirm the bit weights sum back to 424.
  17. Answer: 424₁₀ = 0110101000₂.

Final answer: 0110101000

Foundations · Problem 0106

Express decimal 425 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 425.
  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 425 is below 512; write bit 0; new remainder 425.
  5. Weight 256: take it because remaining 425 is at least 256; write bit 1; new remainder 169.
  6. Weight 128: take it because remaining 169 is at least 128; write bit 1; new remainder 41.
  7. Weight 64: skip it because remaining 41 is below 64; write bit 0; new remainder 41.
  8. Weight 32: take it because remaining 41 is at least 32; write bit 1; new remainder 9.
  9. Weight 16: skip it because remaining 9 is below 16; write bit 0; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101001.
  16. Confirm the bit weights sum back to 425.
  17. Answer: 425₁₀ = 0110101001₂.

Final answer: 0110101001

Foundations · Problem 0107

Express decimal 426 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 426.
  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 426 is below 512; write bit 0; new remainder 426.
  5. Weight 256: take it because remaining 426 is at least 256; write bit 1; new remainder 170.
  6. Weight 128: take it because remaining 170 is at least 128; write bit 1; new remainder 42.
  7. Weight 64: skip it because remaining 42 is below 64; write bit 0; new remainder 42.
  8. Weight 32: take it because remaining 42 is at least 32; write bit 1; new remainder 10.
  9. Weight 16: skip it because remaining 10 is below 16; write bit 0; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101010.
  16. Confirm the bit weights sum back to 426.
  17. Answer: 426₁₀ = 0110101010₂.

Final answer: 0110101010

Foundations · Problem 0108

Express decimal 427 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 427.
  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 427 is below 512; write bit 0; new remainder 427.
  5. Weight 256: take it because remaining 427 is at least 256; write bit 1; new remainder 171.
  6. Weight 128: take it because remaining 171 is at least 128; write bit 1; new remainder 43.
  7. Weight 64: skip it because remaining 43 is below 64; write bit 0; new remainder 43.
  8. Weight 32: take it because remaining 43 is at least 32; write bit 1; new remainder 11.
  9. Weight 16: skip it because remaining 11 is below 16; write bit 0; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101011.
  16. Confirm the bit weights sum back to 427.
  17. Answer: 427₁₀ = 0110101011₂.

Final answer: 0110101011

Foundations · Problem 0109

Express decimal 428 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 428.
  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 428 is below 512; write bit 0; new remainder 428.
  5. Weight 256: take it because remaining 428 is at least 256; write bit 1; new remainder 172.
  6. Weight 128: take it because remaining 172 is at least 128; write bit 1; new remainder 44.
  7. Weight 64: skip it because remaining 44 is below 64; write bit 0; new remainder 44.
  8. Weight 32: take it because remaining 44 is at least 32; write bit 1; new remainder 12.
  9. Weight 16: skip it because remaining 12 is below 16; write bit 0; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101100.
  16. Confirm the bit weights sum back to 428.
  17. Answer: 428₁₀ = 0110101100₂.

Final answer: 0110101100

Foundations · Problem 0110

Express decimal 429 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 429.
  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 429 is below 512; write bit 0; new remainder 429.
  5. Weight 256: take it because remaining 429 is at least 256; write bit 1; new remainder 173.
  6. Weight 128: take it because remaining 173 is at least 128; write bit 1; new remainder 45.
  7. Weight 64: skip it because remaining 45 is below 64; write bit 0; new remainder 45.
  8. Weight 32: take it because remaining 45 is at least 32; write bit 1; new remainder 13.
  9. Weight 16: skip it because remaining 13 is below 16; write bit 0; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101101.
  16. Confirm the bit weights sum back to 429.
  17. Answer: 429₁₀ = 0110101101₂.

Final answer: 0110101101

Foundations · Problem 0111

Express decimal 430 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 430.
  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 430 is below 512; write bit 0; new remainder 430.
  5. Weight 256: take it because remaining 430 is at least 256; write bit 1; new remainder 174.
  6. Weight 128: take it because remaining 174 is at least 128; write bit 1; new remainder 46.
  7. Weight 64: skip it because remaining 46 is below 64; write bit 0; new remainder 46.
  8. Weight 32: take it because remaining 46 is at least 32; write bit 1; new remainder 14.
  9. Weight 16: skip it because remaining 14 is below 16; write bit 0; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101110.
  16. Confirm the bit weights sum back to 430.
  17. Answer: 430₁₀ = 0110101110₂.

Final answer: 0110101110

Foundations · Problem 0112

Express decimal 431 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 431.
  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 431 is below 512; write bit 0; new remainder 431.
  5. Weight 256: take it because remaining 431 is at least 256; write bit 1; new remainder 175.
  6. Weight 128: take it because remaining 175 is at least 128; write bit 1; new remainder 47.
  7. Weight 64: skip it because remaining 47 is below 64; write bit 0; new remainder 47.
  8. Weight 32: take it because remaining 47 is at least 32; write bit 1; new remainder 15.
  9. Weight 16: skip it because remaining 15 is below 16; write bit 0; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110101111.
  16. Confirm the bit weights sum back to 431.
  17. Answer: 431₁₀ = 0110101111₂.

Final answer: 0110101111

Foundations · Problem 0113

Express decimal 432 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 432.
  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 432 is below 512; write bit 0; new remainder 432.
  5. Weight 256: take it because remaining 432 is at least 256; write bit 1; new remainder 176.
  6. Weight 128: take it because remaining 176 is at least 128; write bit 1; new remainder 48.
  7. Weight 64: skip it because remaining 48 is below 64; write bit 0; new remainder 48.
  8. Weight 32: take it because remaining 48 is at least 32; write bit 1; new remainder 16.
  9. Weight 16: take it because remaining 16 is at least 16; write bit 1; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110000.
  16. Confirm the bit weights sum back to 432.
  17. Answer: 432₁₀ = 0110110000₂.

Final answer: 0110110000

Foundations · Problem 0114

Express decimal 433 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 433.
  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 433 is below 512; write bit 0; new remainder 433.
  5. Weight 256: take it because remaining 433 is at least 256; write bit 1; new remainder 177.
  6. Weight 128: take it because remaining 177 is at least 128; write bit 1; new remainder 49.
  7. Weight 64: skip it because remaining 49 is below 64; write bit 0; new remainder 49.
  8. Weight 32: take it because remaining 49 is at least 32; write bit 1; new remainder 17.
  9. Weight 16: take it because remaining 17 is at least 16; write bit 1; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110001.
  16. Confirm the bit weights sum back to 433.
  17. Answer: 433₁₀ = 0110110001₂.

Final answer: 0110110001

Foundations · Problem 0115

Express decimal 434 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 434.
  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 434 is below 512; write bit 0; new remainder 434.
  5. Weight 256: take it because remaining 434 is at least 256; write bit 1; new remainder 178.
  6. Weight 128: take it because remaining 178 is at least 128; write bit 1; new remainder 50.
  7. Weight 64: skip it because remaining 50 is below 64; write bit 0; new remainder 50.
  8. Weight 32: take it because remaining 50 is at least 32; write bit 1; new remainder 18.
  9. Weight 16: take it because remaining 18 is at least 16; write bit 1; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110010.
  16. Confirm the bit weights sum back to 434.
  17. Answer: 434₁₀ = 0110110010₂.

Final answer: 0110110010

Foundations · Problem 0116

Express decimal 435 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 435.
  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 435 is below 512; write bit 0; new remainder 435.
  5. Weight 256: take it because remaining 435 is at least 256; write bit 1; new remainder 179.
  6. Weight 128: take it because remaining 179 is at least 128; write bit 1; new remainder 51.
  7. Weight 64: skip it because remaining 51 is below 64; write bit 0; new remainder 51.
  8. Weight 32: take it because remaining 51 is at least 32; write bit 1; new remainder 19.
  9. Weight 16: take it because remaining 19 is at least 16; write bit 1; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110011.
  16. Confirm the bit weights sum back to 435.
  17. Answer: 435₁₀ = 0110110011₂.

Final answer: 0110110011

Foundations · Problem 0117

Express decimal 436 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 436.
  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 436 is below 512; write bit 0; new remainder 436.
  5. Weight 256: take it because remaining 436 is at least 256; write bit 1; new remainder 180.
  6. Weight 128: take it because remaining 180 is at least 128; write bit 1; new remainder 52.
  7. Weight 64: skip it because remaining 52 is below 64; write bit 0; new remainder 52.
  8. Weight 32: take it because remaining 52 is at least 32; write bit 1; new remainder 20.
  9. Weight 16: take it because remaining 20 is at least 16; write bit 1; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110100.
  16. Confirm the bit weights sum back to 436.
  17. Answer: 436₁₀ = 0110110100₂.

Final answer: 0110110100

Foundations · Problem 0118

Express decimal 437 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 437.
  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 437 is below 512; write bit 0; new remainder 437.
  5. Weight 256: take it because remaining 437 is at least 256; write bit 1; new remainder 181.
  6. Weight 128: take it because remaining 181 is at least 128; write bit 1; new remainder 53.
  7. Weight 64: skip it because remaining 53 is below 64; write bit 0; new remainder 53.
  8. Weight 32: take it because remaining 53 is at least 32; write bit 1; new remainder 21.
  9. Weight 16: take it because remaining 21 is at least 16; write bit 1; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110101.
  16. Confirm the bit weights sum back to 437.
  17. Answer: 437₁₀ = 0110110101₂.

Final answer: 0110110101

Foundations · Problem 0119

Express decimal 438 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 438.
  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 438 is below 512; write bit 0; new remainder 438.
  5. Weight 256: take it because remaining 438 is at least 256; write bit 1; new remainder 182.
  6. Weight 128: take it because remaining 182 is at least 128; write bit 1; new remainder 54.
  7. Weight 64: skip it because remaining 54 is below 64; write bit 0; new remainder 54.
  8. Weight 32: take it because remaining 54 is at least 32; write bit 1; new remainder 22.
  9. Weight 16: take it because remaining 22 is at least 16; write bit 1; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110110.
  16. Confirm the bit weights sum back to 438.
  17. Answer: 438₁₀ = 0110110110₂.

Final answer: 0110110110

Foundations · Problem 0120

Express decimal 439 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 439.
  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 439 is below 512; write bit 0; new remainder 439.
  5. Weight 256: take it because remaining 439 is at least 256; write bit 1; new remainder 183.
  6. Weight 128: take it because remaining 183 is at least 128; write bit 1; new remainder 55.
  7. Weight 64: skip it because remaining 55 is below 64; write bit 0; new remainder 55.
  8. Weight 32: take it because remaining 55 is at least 32; write bit 1; new remainder 23.
  9. Weight 16: take it because remaining 23 is at least 16; write bit 1; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110110111.
  16. Confirm the bit weights sum back to 439.
  17. Answer: 439₁₀ = 0110110111₂.

Final answer: 0110110111

Foundations · Problem 0121

Express decimal 440 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 440.
  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 440 is below 512; write bit 0; new remainder 440.
  5. Weight 256: take it because remaining 440 is at least 256; write bit 1; new remainder 184.
  6. Weight 128: take it because remaining 184 is at least 128; write bit 1; new remainder 56.
  7. Weight 64: skip it because remaining 56 is below 64; write bit 0; new remainder 56.
  8. Weight 32: take it because remaining 56 is at least 32; write bit 1; new remainder 24.
  9. Weight 16: take it because remaining 24 is at least 16; write bit 1; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111000.
  16. Confirm the bit weights sum back to 440.
  17. Answer: 440₁₀ = 0110111000₂.

Final answer: 0110111000

Foundations · Problem 0122

Express decimal 441 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 441.
  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 441 is below 512; write bit 0; new remainder 441.
  5. Weight 256: take it because remaining 441 is at least 256; write bit 1; new remainder 185.
  6. Weight 128: take it because remaining 185 is at least 128; write bit 1; new remainder 57.
  7. Weight 64: skip it because remaining 57 is below 64; write bit 0; new remainder 57.
  8. Weight 32: take it because remaining 57 is at least 32; write bit 1; new remainder 25.
  9. Weight 16: take it because remaining 25 is at least 16; write bit 1; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111001.
  16. Confirm the bit weights sum back to 441.
  17. Answer: 441₁₀ = 0110111001₂.

Final answer: 0110111001

Foundations · Problem 0123

Express decimal 442 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 442.
  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 442 is below 512; write bit 0; new remainder 442.
  5. Weight 256: take it because remaining 442 is at least 256; write bit 1; new remainder 186.
  6. Weight 128: take it because remaining 186 is at least 128; write bit 1; new remainder 58.
  7. Weight 64: skip it because remaining 58 is below 64; write bit 0; new remainder 58.
  8. Weight 32: take it because remaining 58 is at least 32; write bit 1; new remainder 26.
  9. Weight 16: take it because remaining 26 is at least 16; write bit 1; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111010.
  16. Confirm the bit weights sum back to 442.
  17. Answer: 442₁₀ = 0110111010₂.

Final answer: 0110111010

Foundations · Problem 0124

Express decimal 443 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 443.
  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 443 is below 512; write bit 0; new remainder 443.
  5. Weight 256: take it because remaining 443 is at least 256; write bit 1; new remainder 187.
  6. Weight 128: take it because remaining 187 is at least 128; write bit 1; new remainder 59.
  7. Weight 64: skip it because remaining 59 is below 64; write bit 0; new remainder 59.
  8. Weight 32: take it because remaining 59 is at least 32; write bit 1; new remainder 27.
  9. Weight 16: take it because remaining 27 is at least 16; write bit 1; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111011.
  16. Confirm the bit weights sum back to 443.
  17. Answer: 443₁₀ = 0110111011₂.

Final answer: 0110111011

Foundations · Problem 0125

Express decimal 444 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 444.
  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 444 is below 512; write bit 0; new remainder 444.
  5. Weight 256: take it because remaining 444 is at least 256; write bit 1; new remainder 188.
  6. Weight 128: take it because remaining 188 is at least 128; write bit 1; new remainder 60.
  7. Weight 64: skip it because remaining 60 is below 64; write bit 0; new remainder 60.
  8. Weight 32: take it because remaining 60 is at least 32; write bit 1; new remainder 28.
  9. Weight 16: take it because remaining 28 is at least 16; write bit 1; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111100.
  16. Confirm the bit weights sum back to 444.
  17. Answer: 444₁₀ = 0110111100₂.

Final answer: 0110111100

Foundations · Problem 0126

Express decimal 445 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 445.
  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 445 is below 512; write bit 0; new remainder 445.
  5. Weight 256: take it because remaining 445 is at least 256; write bit 1; new remainder 189.
  6. Weight 128: take it because remaining 189 is at least 128; write bit 1; new remainder 61.
  7. Weight 64: skip it because remaining 61 is below 64; write bit 0; new remainder 61.
  8. Weight 32: take it because remaining 61 is at least 32; write bit 1; new remainder 29.
  9. Weight 16: take it because remaining 29 is at least 16; write bit 1; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111101.
  16. Confirm the bit weights sum back to 445.
  17. Answer: 445₁₀ = 0110111101₂.

Final answer: 0110111101

Foundations · Problem 0127

Express decimal 446 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 446.
  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 446 is below 512; write bit 0; new remainder 446.
  5. Weight 256: take it because remaining 446 is at least 256; write bit 1; new remainder 190.
  6. Weight 128: take it because remaining 190 is at least 128; write bit 1; new remainder 62.
  7. Weight 64: skip it because remaining 62 is below 64; write bit 0; new remainder 62.
  8. Weight 32: take it because remaining 62 is at least 32; write bit 1; new remainder 30.
  9. Weight 16: take it because remaining 30 is at least 16; write bit 1; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111110.
  16. Confirm the bit weights sum back to 446.
  17. Answer: 446₁₀ = 0110111110₂.

Final answer: 0110111110

Foundations · Problem 0128

Express decimal 447 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 447.
  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 447 is below 512; write bit 0; new remainder 447.
  5. Weight 256: take it because remaining 447 is at least 256; write bit 1; new remainder 191.
  6. Weight 128: take it because remaining 191 is at least 128; write bit 1; new remainder 63.
  7. Weight 64: skip it because remaining 63 is below 64; write bit 0; new remainder 63.
  8. Weight 32: take it because remaining 63 is at least 32; write bit 1; new remainder 31.
  9. Weight 16: take it because remaining 31 is at least 16; write bit 1; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0110111111.
  16. Confirm the bit weights sum back to 447.
  17. Answer: 447₁₀ = 0110111111₂.

Final answer: 0110111111