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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 3 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 0065

Express decimal 384 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 384.
  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 384 is below 512; write bit 0; new remainder 384.
  5. Weight 256: take it because remaining 384 is at least 256; write bit 1; new remainder 128.
  6. Weight 128: take it because remaining 128 is at least 128; write bit 1; new remainder 0.
  7. Weight 64: skip it because remaining 0 is below 64; write bit 0; 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 0110000000.
  16. Confirm the bit weights sum back to 384.
  17. Answer: 384₁₀ = 0110000000₂.

Final answer: 0110000000

Foundations · Problem 0066

Express decimal 385 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 385.
  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 385 is below 512; write bit 0; new remainder 385.
  5. Weight 256: take it because remaining 385 is at least 256; write bit 1; new remainder 129.
  6. Weight 128: take it because remaining 129 is at least 128; write bit 1; new remainder 1.
  7. Weight 64: skip it because remaining 1 is below 64; write bit 0; 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 0110000001.
  16. Confirm the bit weights sum back to 385.
  17. Answer: 385₁₀ = 0110000001₂.

Final answer: 0110000001

Foundations · Problem 0067

Express decimal 386 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 386.
  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 386 is below 512; write bit 0; new remainder 386.
  5. Weight 256: take it because remaining 386 is at least 256; write bit 1; new remainder 130.
  6. Weight 128: take it because remaining 130 is at least 128; write bit 1; new remainder 2.
  7. Weight 64: skip it because remaining 2 is below 64; write bit 0; 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 0110000010.
  16. Confirm the bit weights sum back to 386.
  17. Answer: 386₁₀ = 0110000010₂.

Final answer: 0110000010

Foundations · Problem 0068

Express decimal 387 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 387.
  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 387 is below 512; write bit 0; new remainder 387.
  5. Weight 256: take it because remaining 387 is at least 256; write bit 1; new remainder 131.
  6. Weight 128: take it because remaining 131 is at least 128; write bit 1; new remainder 3.
  7. Weight 64: skip it because remaining 3 is below 64; write bit 0; 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 0110000011.
  16. Confirm the bit weights sum back to 387.
  17. Answer: 387₁₀ = 0110000011₂.

Final answer: 0110000011

Foundations · Problem 0069

Express decimal 388 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 388.
  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 388 is below 512; write bit 0; new remainder 388.
  5. Weight 256: take it because remaining 388 is at least 256; write bit 1; new remainder 132.
  6. Weight 128: take it because remaining 132 is at least 128; write bit 1; new remainder 4.
  7. Weight 64: skip it because remaining 4 is below 64; write bit 0; 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 0110000100.
  16. Confirm the bit weights sum back to 388.
  17. Answer: 388₁₀ = 0110000100₂.

Final answer: 0110000100

Foundations · Problem 0070

Express decimal 389 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 389.
  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 389 is below 512; write bit 0; new remainder 389.
  5. Weight 256: take it because remaining 389 is at least 256; write bit 1; new remainder 133.
  6. Weight 128: take it because remaining 133 is at least 128; write bit 1; new remainder 5.
  7. Weight 64: skip it because remaining 5 is below 64; write bit 0; 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 0110000101.
  16. Confirm the bit weights sum back to 389.
  17. Answer: 389₁₀ = 0110000101₂.

Final answer: 0110000101

Foundations · Problem 0071

Express decimal 390 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 390.
  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 390 is below 512; write bit 0; new remainder 390.
  5. Weight 256: take it because remaining 390 is at least 256; write bit 1; new remainder 134.
  6. Weight 128: take it because remaining 134 is at least 128; write bit 1; new remainder 6.
  7. Weight 64: skip it because remaining 6 is below 64; write bit 0; 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 0110000110.
  16. Confirm the bit weights sum back to 390.
  17. Answer: 390₁₀ = 0110000110₂.

Final answer: 0110000110

Foundations · Problem 0072

Express decimal 391 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 391.
  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 391 is below 512; write bit 0; new remainder 391.
  5. Weight 256: take it because remaining 391 is at least 256; write bit 1; new remainder 135.
  6. Weight 128: take it because remaining 135 is at least 128; write bit 1; new remainder 7.
  7. Weight 64: skip it because remaining 7 is below 64; write bit 0; 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 0110000111.
  16. Confirm the bit weights sum back to 391.
  17. Answer: 391₁₀ = 0110000111₂.

Final answer: 0110000111

Foundations · Problem 0073

Express decimal 392 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 392.
  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 392 is below 512; write bit 0; new remainder 392.
  5. Weight 256: take it because remaining 392 is at least 256; write bit 1; new remainder 136.
  6. Weight 128: take it because remaining 136 is at least 128; write bit 1; new remainder 8.
  7. Weight 64: skip it because remaining 8 is below 64; write bit 0; 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 0110001000.
  16. Confirm the bit weights sum back to 392.
  17. Answer: 392₁₀ = 0110001000₂.

Final answer: 0110001000

Foundations · Problem 0074

Express decimal 393 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 393.
  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 393 is below 512; write bit 0; new remainder 393.
  5. Weight 256: take it because remaining 393 is at least 256; write bit 1; new remainder 137.
  6. Weight 128: take it because remaining 137 is at least 128; write bit 1; new remainder 9.
  7. Weight 64: skip it because remaining 9 is below 64; write bit 0; 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 0110001001.
  16. Confirm the bit weights sum back to 393.
  17. Answer: 393₁₀ = 0110001001₂.

Final answer: 0110001001

Foundations · Problem 0075

Express decimal 394 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 394.
  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 394 is below 512; write bit 0; new remainder 394.
  5. Weight 256: take it because remaining 394 is at least 256; write bit 1; new remainder 138.
  6. Weight 128: take it because remaining 138 is at least 128; write bit 1; new remainder 10.
  7. Weight 64: skip it because remaining 10 is below 64; write bit 0; 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 0110001010.
  16. Confirm the bit weights sum back to 394.
  17. Answer: 394₁₀ = 0110001010₂.

Final answer: 0110001010

Foundations · Problem 0076

Express decimal 395 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 395.
  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 395 is below 512; write bit 0; new remainder 395.
  5. Weight 256: take it because remaining 395 is at least 256; write bit 1; new remainder 139.
  6. Weight 128: take it because remaining 139 is at least 128; write bit 1; new remainder 11.
  7. Weight 64: skip it because remaining 11 is below 64; write bit 0; 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 0110001011.
  16. Confirm the bit weights sum back to 395.
  17. Answer: 395₁₀ = 0110001011₂.

Final answer: 0110001011

Foundations · Problem 0077

Express decimal 396 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 396.
  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 396 is below 512; write bit 0; new remainder 396.
  5. Weight 256: take it because remaining 396 is at least 256; write bit 1; new remainder 140.
  6. Weight 128: take it because remaining 140 is at least 128; write bit 1; new remainder 12.
  7. Weight 64: skip it because remaining 12 is below 64; write bit 0; 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 0110001100.
  16. Confirm the bit weights sum back to 396.
  17. Answer: 396₁₀ = 0110001100₂.

Final answer: 0110001100

Foundations · Problem 0078

Express decimal 397 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 397.
  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 397 is below 512; write bit 0; new remainder 397.
  5. Weight 256: take it because remaining 397 is at least 256; write bit 1; new remainder 141.
  6. Weight 128: take it because remaining 141 is at least 128; write bit 1; new remainder 13.
  7. Weight 64: skip it because remaining 13 is below 64; write bit 0; 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 0110001101.
  16. Confirm the bit weights sum back to 397.
  17. Answer: 397₁₀ = 0110001101₂.

Final answer: 0110001101

Foundations · Problem 0079

Express decimal 398 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 398.
  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 398 is below 512; write bit 0; new remainder 398.
  5. Weight 256: take it because remaining 398 is at least 256; write bit 1; new remainder 142.
  6. Weight 128: take it because remaining 142 is at least 128; write bit 1; new remainder 14.
  7. Weight 64: skip it because remaining 14 is below 64; write bit 0; 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 0110001110.
  16. Confirm the bit weights sum back to 398.
  17. Answer: 398₁₀ = 0110001110₂.

Final answer: 0110001110

Foundations · Problem 0080

Express decimal 399 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 399.
  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 399 is below 512; write bit 0; new remainder 399.
  5. Weight 256: take it because remaining 399 is at least 256; write bit 1; new remainder 143.
  6. Weight 128: take it because remaining 143 is at least 128; write bit 1; new remainder 15.
  7. Weight 64: skip it because remaining 15 is below 64; write bit 0; 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 0110001111.
  16. Confirm the bit weights sum back to 399.
  17. Answer: 399₁₀ = 0110001111₂.

Final answer: 0110001111

Foundations · Problem 0081

Express decimal 400 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 400.
  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 400 is below 512; write bit 0; new remainder 400.
  5. Weight 256: take it because remaining 400 is at least 256; write bit 1; new remainder 144.
  6. Weight 128: take it because remaining 144 is at least 128; write bit 1; new remainder 16.
  7. Weight 64: skip it because remaining 16 is below 64; write bit 0; 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 0110010000.
  16. Confirm the bit weights sum back to 400.
  17. Answer: 400₁₀ = 0110010000₂.

Final answer: 0110010000

Foundations · Problem 0082

Express decimal 401 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 401.
  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 401 is below 512; write bit 0; new remainder 401.
  5. Weight 256: take it because remaining 401 is at least 256; write bit 1; new remainder 145.
  6. Weight 128: take it because remaining 145 is at least 128; write bit 1; new remainder 17.
  7. Weight 64: skip it because remaining 17 is below 64; write bit 0; 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 0110010001.
  16. Confirm the bit weights sum back to 401.
  17. Answer: 401₁₀ = 0110010001₂.

Final answer: 0110010001

Foundations · Problem 0083

Express decimal 402 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 402.
  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 402 is below 512; write bit 0; new remainder 402.
  5. Weight 256: take it because remaining 402 is at least 256; write bit 1; new remainder 146.
  6. Weight 128: take it because remaining 146 is at least 128; write bit 1; new remainder 18.
  7. Weight 64: skip it because remaining 18 is below 64; write bit 0; 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 0110010010.
  16. Confirm the bit weights sum back to 402.
  17. Answer: 402₁₀ = 0110010010₂.

Final answer: 0110010010

Foundations · Problem 0084

Express decimal 403 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 403.
  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 403 is below 512; write bit 0; new remainder 403.
  5. Weight 256: take it because remaining 403 is at least 256; write bit 1; new remainder 147.
  6. Weight 128: take it because remaining 147 is at least 128; write bit 1; new remainder 19.
  7. Weight 64: skip it because remaining 19 is below 64; write bit 0; 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 0110010011.
  16. Confirm the bit weights sum back to 403.
  17. Answer: 403₁₀ = 0110010011₂.

Final answer: 0110010011

Foundations · Problem 0085

Express decimal 404 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 404.
  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 404 is below 512; write bit 0; new remainder 404.
  5. Weight 256: take it because remaining 404 is at least 256; write bit 1; new remainder 148.
  6. Weight 128: take it because remaining 148 is at least 128; write bit 1; new remainder 20.
  7. Weight 64: skip it because remaining 20 is below 64; write bit 0; 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 0110010100.
  16. Confirm the bit weights sum back to 404.
  17. Answer: 404₁₀ = 0110010100₂.

Final answer: 0110010100

Foundations · Problem 0086

Express decimal 405 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 405.
  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 405 is below 512; write bit 0; new remainder 405.
  5. Weight 256: take it because remaining 405 is at least 256; write bit 1; new remainder 149.
  6. Weight 128: take it because remaining 149 is at least 128; write bit 1; new remainder 21.
  7. Weight 64: skip it because remaining 21 is below 64; write bit 0; 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 0110010101.
  16. Confirm the bit weights sum back to 405.
  17. Answer: 405₁₀ = 0110010101₂.

Final answer: 0110010101

Foundations · Problem 0087

Express decimal 406 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 406.
  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 406 is below 512; write bit 0; new remainder 406.
  5. Weight 256: take it because remaining 406 is at least 256; write bit 1; new remainder 150.
  6. Weight 128: take it because remaining 150 is at least 128; write bit 1; new remainder 22.
  7. Weight 64: skip it because remaining 22 is below 64; write bit 0; 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 0110010110.
  16. Confirm the bit weights sum back to 406.
  17. Answer: 406₁₀ = 0110010110₂.

Final answer: 0110010110

Foundations · Problem 0088

Express decimal 407 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 407.
  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 407 is below 512; write bit 0; new remainder 407.
  5. Weight 256: take it because remaining 407 is at least 256; write bit 1; new remainder 151.
  6. Weight 128: take it because remaining 151 is at least 128; write bit 1; new remainder 23.
  7. Weight 64: skip it because remaining 23 is below 64; write bit 0; 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 0110010111.
  16. Confirm the bit weights sum back to 407.
  17. Answer: 407₁₀ = 0110010111₂.

Final answer: 0110010111

Foundations · Problem 0089

Express decimal 408 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 408.
  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 408 is below 512; write bit 0; new remainder 408.
  5. Weight 256: take it because remaining 408 is at least 256; write bit 1; new remainder 152.
  6. Weight 128: take it because remaining 152 is at least 128; write bit 1; new remainder 24.
  7. Weight 64: skip it because remaining 24 is below 64; write bit 0; 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 0110011000.
  16. Confirm the bit weights sum back to 408.
  17. Answer: 408₁₀ = 0110011000₂.

Final answer: 0110011000

Foundations · Problem 0090

Express decimal 409 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 409.
  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 409 is below 512; write bit 0; new remainder 409.
  5. Weight 256: take it because remaining 409 is at least 256; write bit 1; new remainder 153.
  6. Weight 128: take it because remaining 153 is at least 128; write bit 1; new remainder 25.
  7. Weight 64: skip it because remaining 25 is below 64; write bit 0; 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 0110011001.
  16. Confirm the bit weights sum back to 409.
  17. Answer: 409₁₀ = 0110011001₂.

Final answer: 0110011001

Foundations · Problem 0091

Express decimal 410 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 410.
  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 410 is below 512; write bit 0; new remainder 410.
  5. Weight 256: take it because remaining 410 is at least 256; write bit 1; new remainder 154.
  6. Weight 128: take it because remaining 154 is at least 128; write bit 1; new remainder 26.
  7. Weight 64: skip it because remaining 26 is below 64; write bit 0; 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 0110011010.
  16. Confirm the bit weights sum back to 410.
  17. Answer: 410₁₀ = 0110011010₂.

Final answer: 0110011010

Foundations · Problem 0092

Express decimal 411 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 411.
  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 411 is below 512; write bit 0; new remainder 411.
  5. Weight 256: take it because remaining 411 is at least 256; write bit 1; new remainder 155.
  6. Weight 128: take it because remaining 155 is at least 128; write bit 1; new remainder 27.
  7. Weight 64: skip it because remaining 27 is below 64; write bit 0; 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 0110011011.
  16. Confirm the bit weights sum back to 411.
  17. Answer: 411₁₀ = 0110011011₂.

Final answer: 0110011011

Foundations · Problem 0093

Express decimal 412 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 412.
  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 412 is below 512; write bit 0; new remainder 412.
  5. Weight 256: take it because remaining 412 is at least 256; write bit 1; new remainder 156.
  6. Weight 128: take it because remaining 156 is at least 128; write bit 1; new remainder 28.
  7. Weight 64: skip it because remaining 28 is below 64; write bit 0; 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 0110011100.
  16. Confirm the bit weights sum back to 412.
  17. Answer: 412₁₀ = 0110011100₂.

Final answer: 0110011100

Foundations · Problem 0094

Express decimal 413 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 413.
  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 413 is below 512; write bit 0; new remainder 413.
  5. Weight 256: take it because remaining 413 is at least 256; write bit 1; new remainder 157.
  6. Weight 128: take it because remaining 157 is at least 128; write bit 1; new remainder 29.
  7. Weight 64: skip it because remaining 29 is below 64; write bit 0; 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 0110011101.
  16. Confirm the bit weights sum back to 413.
  17. Answer: 413₁₀ = 0110011101₂.

Final answer: 0110011101

Foundations · Problem 0095

Express decimal 414 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 414.
  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 414 is below 512; write bit 0; new remainder 414.
  5. Weight 256: take it because remaining 414 is at least 256; write bit 1; new remainder 158.
  6. Weight 128: take it because remaining 158 is at least 128; write bit 1; new remainder 30.
  7. Weight 64: skip it because remaining 30 is below 64; write bit 0; 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 0110011110.
  16. Confirm the bit weights sum back to 414.
  17. Answer: 414₁₀ = 0110011110₂.

Final answer: 0110011110

Foundations · Problem 0096

Express decimal 415 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 415.
  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 415 is below 512; write bit 0; new remainder 415.
  5. Weight 256: take it because remaining 415 is at least 256; write bit 1; new remainder 159.
  6. Weight 128: take it because remaining 159 is at least 128; write bit 1; new remainder 31.
  7. Weight 64: skip it because remaining 31 is below 64; write bit 0; 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 0110011111.
  16. Confirm the bit weights sum back to 415.
  17. Answer: 415₁₀ = 0110011111₂.

Final answer: 0110011111