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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 5 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 0129

Express decimal 448 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 448.
  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 448 is below 512; write bit 0; new remainder 448.
  5. Weight 256: take it because remaining 448 is at least 256; write bit 1; new remainder 192.
  6. Weight 128: take it because remaining 192 is at least 128; write bit 1; new remainder 64.
  7. Weight 64: take it because remaining 64 is at least 64; write bit 1; new remainder 0.
  8. Weight 32: skip it because remaining 0 is below 32; write bit 0; new remainder 0.
  9. Weight 16: skip it because remaining 0 is below 16; write bit 0; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000000.
  16. Confirm the bit weights sum back to 448.
  17. Answer: 448₁₀ = 0111000000₂.

Final answer: 0111000000

Foundations · Problem 0130

Express decimal 449 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 449.
  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 449 is below 512; write bit 0; new remainder 449.
  5. Weight 256: take it because remaining 449 is at least 256; write bit 1; new remainder 193.
  6. Weight 128: take it because remaining 193 is at least 128; write bit 1; new remainder 65.
  7. Weight 64: take it because remaining 65 is at least 64; write bit 1; new remainder 1.
  8. Weight 32: skip it because remaining 1 is below 32; write bit 0; new remainder 1.
  9. Weight 16: skip it because remaining 1 is below 16; write bit 0; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000001.
  16. Confirm the bit weights sum back to 449.
  17. Answer: 449₁₀ = 0111000001₂.

Final answer: 0111000001

Foundations · Problem 0131

Express decimal 450 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 450.
  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 450 is below 512; write bit 0; new remainder 450.
  5. Weight 256: take it because remaining 450 is at least 256; write bit 1; new remainder 194.
  6. Weight 128: take it because remaining 194 is at least 128; write bit 1; new remainder 66.
  7. Weight 64: take it because remaining 66 is at least 64; write bit 1; new remainder 2.
  8. Weight 32: skip it because remaining 2 is below 32; write bit 0; new remainder 2.
  9. Weight 16: skip it because remaining 2 is below 16; write bit 0; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000010.
  16. Confirm the bit weights sum back to 450.
  17. Answer: 450₁₀ = 0111000010₂.

Final answer: 0111000010

Foundations · Problem 0132

Express decimal 451 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 451.
  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 451 is below 512; write bit 0; new remainder 451.
  5. Weight 256: take it because remaining 451 is at least 256; write bit 1; new remainder 195.
  6. Weight 128: take it because remaining 195 is at least 128; write bit 1; new remainder 67.
  7. Weight 64: take it because remaining 67 is at least 64; write bit 1; new remainder 3.
  8. Weight 32: skip it because remaining 3 is below 32; write bit 0; new remainder 3.
  9. Weight 16: skip it because remaining 3 is below 16; write bit 0; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000011.
  16. Confirm the bit weights sum back to 451.
  17. Answer: 451₁₀ = 0111000011₂.

Final answer: 0111000011

Foundations · Problem 0133

Express decimal 452 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 452.
  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 452 is below 512; write bit 0; new remainder 452.
  5. Weight 256: take it because remaining 452 is at least 256; write bit 1; new remainder 196.
  6. Weight 128: take it because remaining 196 is at least 128; write bit 1; new remainder 68.
  7. Weight 64: take it because remaining 68 is at least 64; write bit 1; new remainder 4.
  8. Weight 32: skip it because remaining 4 is below 32; write bit 0; new remainder 4.
  9. Weight 16: skip it because remaining 4 is below 16; write bit 0; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000100.
  16. Confirm the bit weights sum back to 452.
  17. Answer: 452₁₀ = 0111000100₂.

Final answer: 0111000100

Foundations · Problem 0134

Express decimal 453 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 453.
  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 453 is below 512; write bit 0; new remainder 453.
  5. Weight 256: take it because remaining 453 is at least 256; write bit 1; new remainder 197.
  6. Weight 128: take it because remaining 197 is at least 128; write bit 1; new remainder 69.
  7. Weight 64: take it because remaining 69 is at least 64; write bit 1; new remainder 5.
  8. Weight 32: skip it because remaining 5 is below 32; write bit 0; new remainder 5.
  9. Weight 16: skip it because remaining 5 is below 16; write bit 0; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000101.
  16. Confirm the bit weights sum back to 453.
  17. Answer: 453₁₀ = 0111000101₂.

Final answer: 0111000101

Foundations · Problem 0135

Express decimal 454 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 454.
  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 454 is below 512; write bit 0; new remainder 454.
  5. Weight 256: take it because remaining 454 is at least 256; write bit 1; new remainder 198.
  6. Weight 128: take it because remaining 198 is at least 128; write bit 1; new remainder 70.
  7. Weight 64: take it because remaining 70 is at least 64; write bit 1; new remainder 6.
  8. Weight 32: skip it because remaining 6 is below 32; write bit 0; new remainder 6.
  9. Weight 16: skip it because remaining 6 is below 16; write bit 0; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000110.
  16. Confirm the bit weights sum back to 454.
  17. Answer: 454₁₀ = 0111000110₂.

Final answer: 0111000110

Foundations · Problem 0136

Express decimal 455 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 455.
  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 455 is below 512; write bit 0; new remainder 455.
  5. Weight 256: take it because remaining 455 is at least 256; write bit 1; new remainder 199.
  6. Weight 128: take it because remaining 199 is at least 128; write bit 1; new remainder 71.
  7. Weight 64: take it because remaining 71 is at least 64; write bit 1; new remainder 7.
  8. Weight 32: skip it because remaining 7 is below 32; write bit 0; new remainder 7.
  9. Weight 16: skip it because remaining 7 is below 16; write bit 0; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111000111.
  16. Confirm the bit weights sum back to 455.
  17. Answer: 455₁₀ = 0111000111₂.

Final answer: 0111000111

Foundations · Problem 0137

Express decimal 456 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 456.
  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 456 is below 512; write bit 0; new remainder 456.
  5. Weight 256: take it because remaining 456 is at least 256; write bit 1; new remainder 200.
  6. Weight 128: take it because remaining 200 is at least 128; write bit 1; new remainder 72.
  7. Weight 64: take it because remaining 72 is at least 64; write bit 1; new remainder 8.
  8. Weight 32: skip it because remaining 8 is below 32; write bit 0; new remainder 8.
  9. Weight 16: skip it because remaining 8 is below 16; write bit 0; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001000.
  16. Confirm the bit weights sum back to 456.
  17. Answer: 456₁₀ = 0111001000₂.

Final answer: 0111001000

Foundations · Problem 0138

Express decimal 457 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 457.
  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 457 is below 512; write bit 0; new remainder 457.
  5. Weight 256: take it because remaining 457 is at least 256; write bit 1; new remainder 201.
  6. Weight 128: take it because remaining 201 is at least 128; write bit 1; new remainder 73.
  7. Weight 64: take it because remaining 73 is at least 64; write bit 1; new remainder 9.
  8. Weight 32: skip it because remaining 9 is below 32; write bit 0; new remainder 9.
  9. Weight 16: skip it because remaining 9 is below 16; write bit 0; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001001.
  16. Confirm the bit weights sum back to 457.
  17. Answer: 457₁₀ = 0111001001₂.

Final answer: 0111001001

Foundations · Problem 0139

Express decimal 458 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 458.
  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 458 is below 512; write bit 0; new remainder 458.
  5. Weight 256: take it because remaining 458 is at least 256; write bit 1; new remainder 202.
  6. Weight 128: take it because remaining 202 is at least 128; write bit 1; new remainder 74.
  7. Weight 64: take it because remaining 74 is at least 64; write bit 1; new remainder 10.
  8. Weight 32: skip it because remaining 10 is below 32; write bit 0; new remainder 10.
  9. Weight 16: skip it because remaining 10 is below 16; write bit 0; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001010.
  16. Confirm the bit weights sum back to 458.
  17. Answer: 458₁₀ = 0111001010₂.

Final answer: 0111001010

Foundations · Problem 0140

Express decimal 459 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 459.
  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 459 is below 512; write bit 0; new remainder 459.
  5. Weight 256: take it because remaining 459 is at least 256; write bit 1; new remainder 203.
  6. Weight 128: take it because remaining 203 is at least 128; write bit 1; new remainder 75.
  7. Weight 64: take it because remaining 75 is at least 64; write bit 1; new remainder 11.
  8. Weight 32: skip it because remaining 11 is below 32; write bit 0; new remainder 11.
  9. Weight 16: skip it because remaining 11 is below 16; write bit 0; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001011.
  16. Confirm the bit weights sum back to 459.
  17. Answer: 459₁₀ = 0111001011₂.

Final answer: 0111001011

Foundations · Problem 0141

Express decimal 460 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 460.
  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 460 is below 512; write bit 0; new remainder 460.
  5. Weight 256: take it because remaining 460 is at least 256; write bit 1; new remainder 204.
  6. Weight 128: take it because remaining 204 is at least 128; write bit 1; new remainder 76.
  7. Weight 64: take it because remaining 76 is at least 64; write bit 1; new remainder 12.
  8. Weight 32: skip it because remaining 12 is below 32; write bit 0; new remainder 12.
  9. Weight 16: skip it because remaining 12 is below 16; write bit 0; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001100.
  16. Confirm the bit weights sum back to 460.
  17. Answer: 460₁₀ = 0111001100₂.

Final answer: 0111001100

Foundations · Problem 0142

Express decimal 461 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 461.
  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 461 is below 512; write bit 0; new remainder 461.
  5. Weight 256: take it because remaining 461 is at least 256; write bit 1; new remainder 205.
  6. Weight 128: take it because remaining 205 is at least 128; write bit 1; new remainder 77.
  7. Weight 64: take it because remaining 77 is at least 64; write bit 1; new remainder 13.
  8. Weight 32: skip it because remaining 13 is below 32; write bit 0; new remainder 13.
  9. Weight 16: skip it because remaining 13 is below 16; write bit 0; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001101.
  16. Confirm the bit weights sum back to 461.
  17. Answer: 461₁₀ = 0111001101₂.

Final answer: 0111001101

Foundations · Problem 0143

Express decimal 462 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 462.
  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 462 is below 512; write bit 0; new remainder 462.
  5. Weight 256: take it because remaining 462 is at least 256; write bit 1; new remainder 206.
  6. Weight 128: take it because remaining 206 is at least 128; write bit 1; new remainder 78.
  7. Weight 64: take it because remaining 78 is at least 64; write bit 1; new remainder 14.
  8. Weight 32: skip it because remaining 14 is below 32; write bit 0; new remainder 14.
  9. Weight 16: skip it because remaining 14 is below 16; write bit 0; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001110.
  16. Confirm the bit weights sum back to 462.
  17. Answer: 462₁₀ = 0111001110₂.

Final answer: 0111001110

Foundations · Problem 0144

Express decimal 463 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 463.
  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 463 is below 512; write bit 0; new remainder 463.
  5. Weight 256: take it because remaining 463 is at least 256; write bit 1; new remainder 207.
  6. Weight 128: take it because remaining 207 is at least 128; write bit 1; new remainder 79.
  7. Weight 64: take it because remaining 79 is at least 64; write bit 1; new remainder 15.
  8. Weight 32: skip it because remaining 15 is below 32; write bit 0; new remainder 15.
  9. Weight 16: skip it because remaining 15 is below 16; write bit 0; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111001111.
  16. Confirm the bit weights sum back to 463.
  17. Answer: 463₁₀ = 0111001111₂.

Final answer: 0111001111

Foundations · Problem 0145

Express decimal 464 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 464.
  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 464 is below 512; write bit 0; new remainder 464.
  5. Weight 256: take it because remaining 464 is at least 256; write bit 1; new remainder 208.
  6. Weight 128: take it because remaining 208 is at least 128; write bit 1; new remainder 80.
  7. Weight 64: take it because remaining 80 is at least 64; write bit 1; new remainder 16.
  8. Weight 32: skip it because remaining 16 is below 32; write bit 0; new remainder 16.
  9. Weight 16: take it because remaining 16 is at least 16; write bit 1; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010000.
  16. Confirm the bit weights sum back to 464.
  17. Answer: 464₁₀ = 0111010000₂.

Final answer: 0111010000

Foundations · Problem 0146

Express decimal 465 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 465.
  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 465 is below 512; write bit 0; new remainder 465.
  5. Weight 256: take it because remaining 465 is at least 256; write bit 1; new remainder 209.
  6. Weight 128: take it because remaining 209 is at least 128; write bit 1; new remainder 81.
  7. Weight 64: take it because remaining 81 is at least 64; write bit 1; new remainder 17.
  8. Weight 32: skip it because remaining 17 is below 32; write bit 0; new remainder 17.
  9. Weight 16: take it because remaining 17 is at least 16; write bit 1; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010001.
  16. Confirm the bit weights sum back to 465.
  17. Answer: 465₁₀ = 0111010001₂.

Final answer: 0111010001

Foundations · Problem 0147

Express decimal 466 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 466.
  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 466 is below 512; write bit 0; new remainder 466.
  5. Weight 256: take it because remaining 466 is at least 256; write bit 1; new remainder 210.
  6. Weight 128: take it because remaining 210 is at least 128; write bit 1; new remainder 82.
  7. Weight 64: take it because remaining 82 is at least 64; write bit 1; new remainder 18.
  8. Weight 32: skip it because remaining 18 is below 32; write bit 0; new remainder 18.
  9. Weight 16: take it because remaining 18 is at least 16; write bit 1; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010010.
  16. Confirm the bit weights sum back to 466.
  17. Answer: 466₁₀ = 0111010010₂.

Final answer: 0111010010

Foundations · Problem 0148

Express decimal 467 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 467.
  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 467 is below 512; write bit 0; new remainder 467.
  5. Weight 256: take it because remaining 467 is at least 256; write bit 1; new remainder 211.
  6. Weight 128: take it because remaining 211 is at least 128; write bit 1; new remainder 83.
  7. Weight 64: take it because remaining 83 is at least 64; write bit 1; new remainder 19.
  8. Weight 32: skip it because remaining 19 is below 32; write bit 0; new remainder 19.
  9. Weight 16: take it because remaining 19 is at least 16; write bit 1; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010011.
  16. Confirm the bit weights sum back to 467.
  17. Answer: 467₁₀ = 0111010011₂.

Final answer: 0111010011

Foundations · Problem 0149

Express decimal 468 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 468.
  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 468 is below 512; write bit 0; new remainder 468.
  5. Weight 256: take it because remaining 468 is at least 256; write bit 1; new remainder 212.
  6. Weight 128: take it because remaining 212 is at least 128; write bit 1; new remainder 84.
  7. Weight 64: take it because remaining 84 is at least 64; write bit 1; new remainder 20.
  8. Weight 32: skip it because remaining 20 is below 32; write bit 0; new remainder 20.
  9. Weight 16: take it because remaining 20 is at least 16; write bit 1; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010100.
  16. Confirm the bit weights sum back to 468.
  17. Answer: 468₁₀ = 0111010100₂.

Final answer: 0111010100

Foundations · Problem 0150

Express decimal 469 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 469.
  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 469 is below 512; write bit 0; new remainder 469.
  5. Weight 256: take it because remaining 469 is at least 256; write bit 1; new remainder 213.
  6. Weight 128: take it because remaining 213 is at least 128; write bit 1; new remainder 85.
  7. Weight 64: take it because remaining 85 is at least 64; write bit 1; new remainder 21.
  8. Weight 32: skip it because remaining 21 is below 32; write bit 0; new remainder 21.
  9. Weight 16: take it because remaining 21 is at least 16; write bit 1; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010101.
  16. Confirm the bit weights sum back to 469.
  17. Answer: 469₁₀ = 0111010101₂.

Final answer: 0111010101

Foundations · Problem 0151

Express decimal 470 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 470.
  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 470 is below 512; write bit 0; new remainder 470.
  5. Weight 256: take it because remaining 470 is at least 256; write bit 1; new remainder 214.
  6. Weight 128: take it because remaining 214 is at least 128; write bit 1; new remainder 86.
  7. Weight 64: take it because remaining 86 is at least 64; write bit 1; new remainder 22.
  8. Weight 32: skip it because remaining 22 is below 32; write bit 0; new remainder 22.
  9. Weight 16: take it because remaining 22 is at least 16; write bit 1; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010110.
  16. Confirm the bit weights sum back to 470.
  17. Answer: 470₁₀ = 0111010110₂.

Final answer: 0111010110

Foundations · Problem 0152

Express decimal 471 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 471.
  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 471 is below 512; write bit 0; new remainder 471.
  5. Weight 256: take it because remaining 471 is at least 256; write bit 1; new remainder 215.
  6. Weight 128: take it because remaining 215 is at least 128; write bit 1; new remainder 87.
  7. Weight 64: take it because remaining 87 is at least 64; write bit 1; new remainder 23.
  8. Weight 32: skip it because remaining 23 is below 32; write bit 0; new remainder 23.
  9. Weight 16: take it because remaining 23 is at least 16; write bit 1; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111010111.
  16. Confirm the bit weights sum back to 471.
  17. Answer: 471₁₀ = 0111010111₂.

Final answer: 0111010111

Foundations · Problem 0153

Express decimal 472 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 472.
  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 472 is below 512; write bit 0; new remainder 472.
  5. Weight 256: take it because remaining 472 is at least 256; write bit 1; new remainder 216.
  6. Weight 128: take it because remaining 216 is at least 128; write bit 1; new remainder 88.
  7. Weight 64: take it because remaining 88 is at least 64; write bit 1; new remainder 24.
  8. Weight 32: skip it because remaining 24 is below 32; write bit 0; new remainder 24.
  9. Weight 16: take it because remaining 24 is at least 16; write bit 1; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011000.
  16. Confirm the bit weights sum back to 472.
  17. Answer: 472₁₀ = 0111011000₂.

Final answer: 0111011000

Foundations · Problem 0154

Express decimal 473 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 473.
  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 473 is below 512; write bit 0; new remainder 473.
  5. Weight 256: take it because remaining 473 is at least 256; write bit 1; new remainder 217.
  6. Weight 128: take it because remaining 217 is at least 128; write bit 1; new remainder 89.
  7. Weight 64: take it because remaining 89 is at least 64; write bit 1; new remainder 25.
  8. Weight 32: skip it because remaining 25 is below 32; write bit 0; new remainder 25.
  9. Weight 16: take it because remaining 25 is at least 16; write bit 1; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011001.
  16. Confirm the bit weights sum back to 473.
  17. Answer: 473₁₀ = 0111011001₂.

Final answer: 0111011001

Foundations · Problem 0155

Express decimal 474 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 474.
  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 474 is below 512; write bit 0; new remainder 474.
  5. Weight 256: take it because remaining 474 is at least 256; write bit 1; new remainder 218.
  6. Weight 128: take it because remaining 218 is at least 128; write bit 1; new remainder 90.
  7. Weight 64: take it because remaining 90 is at least 64; write bit 1; new remainder 26.
  8. Weight 32: skip it because remaining 26 is below 32; write bit 0; new remainder 26.
  9. Weight 16: take it because remaining 26 is at least 16; write bit 1; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011010.
  16. Confirm the bit weights sum back to 474.
  17. Answer: 474₁₀ = 0111011010₂.

Final answer: 0111011010

Foundations · Problem 0156

Express decimal 475 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 475.
  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 475 is below 512; write bit 0; new remainder 475.
  5. Weight 256: take it because remaining 475 is at least 256; write bit 1; new remainder 219.
  6. Weight 128: take it because remaining 219 is at least 128; write bit 1; new remainder 91.
  7. Weight 64: take it because remaining 91 is at least 64; write bit 1; new remainder 27.
  8. Weight 32: skip it because remaining 27 is below 32; write bit 0; new remainder 27.
  9. Weight 16: take it because remaining 27 is at least 16; write bit 1; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011011.
  16. Confirm the bit weights sum back to 475.
  17. Answer: 475₁₀ = 0111011011₂.

Final answer: 0111011011

Foundations · Problem 0157

Express decimal 476 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 476.
  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 476 is below 512; write bit 0; new remainder 476.
  5. Weight 256: take it because remaining 476 is at least 256; write bit 1; new remainder 220.
  6. Weight 128: take it because remaining 220 is at least 128; write bit 1; new remainder 92.
  7. Weight 64: take it because remaining 92 is at least 64; write bit 1; new remainder 28.
  8. Weight 32: skip it because remaining 28 is below 32; write bit 0; new remainder 28.
  9. Weight 16: take it because remaining 28 is at least 16; write bit 1; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011100.
  16. Confirm the bit weights sum back to 476.
  17. Answer: 476₁₀ = 0111011100₂.

Final answer: 0111011100

Foundations · Problem 0158

Express decimal 477 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 477.
  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 477 is below 512; write bit 0; new remainder 477.
  5. Weight 256: take it because remaining 477 is at least 256; write bit 1; new remainder 221.
  6. Weight 128: take it because remaining 221 is at least 128; write bit 1; new remainder 93.
  7. Weight 64: take it because remaining 93 is at least 64; write bit 1; new remainder 29.
  8. Weight 32: skip it because remaining 29 is below 32; write bit 0; new remainder 29.
  9. Weight 16: take it because remaining 29 is at least 16; write bit 1; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011101.
  16. Confirm the bit weights sum back to 477.
  17. Answer: 477₁₀ = 0111011101₂.

Final answer: 0111011101

Foundations · Problem 0159

Express decimal 478 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 478.
  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 478 is below 512; write bit 0; new remainder 478.
  5. Weight 256: take it because remaining 478 is at least 256; write bit 1; new remainder 222.
  6. Weight 128: take it because remaining 222 is at least 128; write bit 1; new remainder 94.
  7. Weight 64: take it because remaining 94 is at least 64; write bit 1; new remainder 30.
  8. Weight 32: skip it because remaining 30 is below 32; write bit 0; new remainder 30.
  9. Weight 16: take it because remaining 30 is at least 16; write bit 1; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011110.
  16. Confirm the bit weights sum back to 478.
  17. Answer: 478₁₀ = 0111011110₂.

Final answer: 0111011110

Foundations · Problem 0160

Express decimal 479 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 479.
  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 479 is below 512; write bit 0; new remainder 479.
  5. Weight 256: take it because remaining 479 is at least 256; write bit 1; new remainder 223.
  6. Weight 128: take it because remaining 223 is at least 128; write bit 1; new remainder 95.
  7. Weight 64: take it because remaining 95 is at least 64; write bit 1; new remainder 31.
  8. Weight 32: skip it because remaining 31 is below 32; write bit 0; new remainder 31.
  9. Weight 16: take it because remaining 31 is at least 16; write bit 1; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0111011111.
  16. Confirm the bit weights sum back to 479.
  17. Answer: 479₁₀ = 0111011111₂.

Final answer: 0111011111