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Practice library / Writing binary

Fully solved problem book

Writing binary

Set 2 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 0033

Express decimal 352 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 352.
  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 352 is below 512; write bit 0; new remainder 352.
  5. Weight 256: take it because remaining 352 is at least 256; write bit 1; new remainder 96.
  6. Weight 128: skip it because remaining 96 is below 128; write bit 0; new remainder 96.
  7. Weight 64: take it because remaining 96 is at least 64; write bit 1; new remainder 32.
  8. Weight 32: take it because remaining 32 is at least 32; write bit 1; new remainder 0.
  9. Weight 16: skip it because remaining 0 is below 16; write bit 0; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100000.
  16. Confirm the bit weights sum back to 352.
  17. Answer: 352₁₀ = 0101100000₂.

Final answer: 0101100000

Foundations · Problem 0034

Express decimal 353 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 353.
  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 353 is below 512; write bit 0; new remainder 353.
  5. Weight 256: take it because remaining 353 is at least 256; write bit 1; new remainder 97.
  6. Weight 128: skip it because remaining 97 is below 128; write bit 0; new remainder 97.
  7. Weight 64: take it because remaining 97 is at least 64; write bit 1; new remainder 33.
  8. Weight 32: take it because remaining 33 is at least 32; write bit 1; new remainder 1.
  9. Weight 16: skip it because remaining 1 is below 16; write bit 0; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100001.
  16. Confirm the bit weights sum back to 353.
  17. Answer: 353₁₀ = 0101100001₂.

Final answer: 0101100001

Foundations · Problem 0035

Express decimal 354 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 354.
  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 354 is below 512; write bit 0; new remainder 354.
  5. Weight 256: take it because remaining 354 is at least 256; write bit 1; new remainder 98.
  6. Weight 128: skip it because remaining 98 is below 128; write bit 0; new remainder 98.
  7. Weight 64: take it because remaining 98 is at least 64; write bit 1; new remainder 34.
  8. Weight 32: take it because remaining 34 is at least 32; write bit 1; new remainder 2.
  9. Weight 16: skip it because remaining 2 is below 16; write bit 0; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100010.
  16. Confirm the bit weights sum back to 354.
  17. Answer: 354₁₀ = 0101100010₂.

Final answer: 0101100010

Foundations · Problem 0036

Express decimal 355 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 355.
  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 355 is below 512; write bit 0; new remainder 355.
  5. Weight 256: take it because remaining 355 is at least 256; write bit 1; new remainder 99.
  6. Weight 128: skip it because remaining 99 is below 128; write bit 0; new remainder 99.
  7. Weight 64: take it because remaining 99 is at least 64; write bit 1; new remainder 35.
  8. Weight 32: take it because remaining 35 is at least 32; write bit 1; new remainder 3.
  9. Weight 16: skip it because remaining 3 is below 16; write bit 0; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100011.
  16. Confirm the bit weights sum back to 355.
  17. Answer: 355₁₀ = 0101100011₂.

Final answer: 0101100011

Foundations · Problem 0037

Express decimal 356 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 356.
  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 356 is below 512; write bit 0; new remainder 356.
  5. Weight 256: take it because remaining 356 is at least 256; write bit 1; new remainder 100.
  6. Weight 128: skip it because remaining 100 is below 128; write bit 0; new remainder 100.
  7. Weight 64: take it because remaining 100 is at least 64; write bit 1; new remainder 36.
  8. Weight 32: take it because remaining 36 is at least 32; write bit 1; new remainder 4.
  9. Weight 16: skip it because remaining 4 is below 16; write bit 0; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100100.
  16. Confirm the bit weights sum back to 356.
  17. Answer: 356₁₀ = 0101100100₂.

Final answer: 0101100100

Foundations · Problem 0038

Express decimal 357 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 357.
  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 357 is below 512; write bit 0; new remainder 357.
  5. Weight 256: take it because remaining 357 is at least 256; write bit 1; new remainder 101.
  6. Weight 128: skip it because remaining 101 is below 128; write bit 0; new remainder 101.
  7. Weight 64: take it because remaining 101 is at least 64; write bit 1; new remainder 37.
  8. Weight 32: take it because remaining 37 is at least 32; write bit 1; new remainder 5.
  9. Weight 16: skip it because remaining 5 is below 16; write bit 0; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100101.
  16. Confirm the bit weights sum back to 357.
  17. Answer: 357₁₀ = 0101100101₂.

Final answer: 0101100101

Foundations · Problem 0039

Express decimal 358 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 358.
  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 358 is below 512; write bit 0; new remainder 358.
  5. Weight 256: take it because remaining 358 is at least 256; write bit 1; new remainder 102.
  6. Weight 128: skip it because remaining 102 is below 128; write bit 0; new remainder 102.
  7. Weight 64: take it because remaining 102 is at least 64; write bit 1; new remainder 38.
  8. Weight 32: take it because remaining 38 is at least 32; write bit 1; new remainder 6.
  9. Weight 16: skip it because remaining 6 is below 16; write bit 0; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100110.
  16. Confirm the bit weights sum back to 358.
  17. Answer: 358₁₀ = 0101100110₂.

Final answer: 0101100110

Foundations · Problem 0040

Express decimal 359 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 359.
  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 359 is below 512; write bit 0; new remainder 359.
  5. Weight 256: take it because remaining 359 is at least 256; write bit 1; new remainder 103.
  6. Weight 128: skip it because remaining 103 is below 128; write bit 0; new remainder 103.
  7. Weight 64: take it because remaining 103 is at least 64; write bit 1; new remainder 39.
  8. Weight 32: take it because remaining 39 is at least 32; write bit 1; new remainder 7.
  9. Weight 16: skip it because remaining 7 is below 16; write bit 0; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101100111.
  16. Confirm the bit weights sum back to 359.
  17. Answer: 359₁₀ = 0101100111₂.

Final answer: 0101100111

Foundations · Problem 0041

Express decimal 360 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 360.
  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 360 is below 512; write bit 0; new remainder 360.
  5. Weight 256: take it because remaining 360 is at least 256; write bit 1; new remainder 104.
  6. Weight 128: skip it because remaining 104 is below 128; write bit 0; new remainder 104.
  7. Weight 64: take it because remaining 104 is at least 64; write bit 1; new remainder 40.
  8. Weight 32: take it because remaining 40 is at least 32; write bit 1; new remainder 8.
  9. Weight 16: skip it because remaining 8 is below 16; write bit 0; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101000.
  16. Confirm the bit weights sum back to 360.
  17. Answer: 360₁₀ = 0101101000₂.

Final answer: 0101101000

Foundations · Problem 0042

Express decimal 361 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 361.
  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 361 is below 512; write bit 0; new remainder 361.
  5. Weight 256: take it because remaining 361 is at least 256; write bit 1; new remainder 105.
  6. Weight 128: skip it because remaining 105 is below 128; write bit 0; new remainder 105.
  7. Weight 64: take it because remaining 105 is at least 64; write bit 1; new remainder 41.
  8. Weight 32: take it because remaining 41 is at least 32; write bit 1; new remainder 9.
  9. Weight 16: skip it because remaining 9 is below 16; write bit 0; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101001.
  16. Confirm the bit weights sum back to 361.
  17. Answer: 361₁₀ = 0101101001₂.

Final answer: 0101101001

Foundations · Problem 0043

Express decimal 362 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 362.
  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 362 is below 512; write bit 0; new remainder 362.
  5. Weight 256: take it because remaining 362 is at least 256; write bit 1; new remainder 106.
  6. Weight 128: skip it because remaining 106 is below 128; write bit 0; new remainder 106.
  7. Weight 64: take it because remaining 106 is at least 64; write bit 1; new remainder 42.
  8. Weight 32: take it because remaining 42 is at least 32; write bit 1; new remainder 10.
  9. Weight 16: skip it because remaining 10 is below 16; write bit 0; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101010.
  16. Confirm the bit weights sum back to 362.
  17. Answer: 362₁₀ = 0101101010₂.

Final answer: 0101101010

Foundations · Problem 0044

Express decimal 363 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 363.
  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 363 is below 512; write bit 0; new remainder 363.
  5. Weight 256: take it because remaining 363 is at least 256; write bit 1; new remainder 107.
  6. Weight 128: skip it because remaining 107 is below 128; write bit 0; new remainder 107.
  7. Weight 64: take it because remaining 107 is at least 64; write bit 1; new remainder 43.
  8. Weight 32: take it because remaining 43 is at least 32; write bit 1; new remainder 11.
  9. Weight 16: skip it because remaining 11 is below 16; write bit 0; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101011.
  16. Confirm the bit weights sum back to 363.
  17. Answer: 363₁₀ = 0101101011₂.

Final answer: 0101101011

Foundations · Problem 0045

Express decimal 364 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 364.
  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 364 is below 512; write bit 0; new remainder 364.
  5. Weight 256: take it because remaining 364 is at least 256; write bit 1; new remainder 108.
  6. Weight 128: skip it because remaining 108 is below 128; write bit 0; new remainder 108.
  7. Weight 64: take it because remaining 108 is at least 64; write bit 1; new remainder 44.
  8. Weight 32: take it because remaining 44 is at least 32; write bit 1; new remainder 12.
  9. Weight 16: skip it because remaining 12 is below 16; write bit 0; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101100.
  16. Confirm the bit weights sum back to 364.
  17. Answer: 364₁₀ = 0101101100₂.

Final answer: 0101101100

Foundations · Problem 0046

Express decimal 365 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 365.
  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 365 is below 512; write bit 0; new remainder 365.
  5. Weight 256: take it because remaining 365 is at least 256; write bit 1; new remainder 109.
  6. Weight 128: skip it because remaining 109 is below 128; write bit 0; new remainder 109.
  7. Weight 64: take it because remaining 109 is at least 64; write bit 1; new remainder 45.
  8. Weight 32: take it because remaining 45 is at least 32; write bit 1; new remainder 13.
  9. Weight 16: skip it because remaining 13 is below 16; write bit 0; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101101.
  16. Confirm the bit weights sum back to 365.
  17. Answer: 365₁₀ = 0101101101₂.

Final answer: 0101101101

Foundations · Problem 0047

Express decimal 366 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 366.
  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 366 is below 512; write bit 0; new remainder 366.
  5. Weight 256: take it because remaining 366 is at least 256; write bit 1; new remainder 110.
  6. Weight 128: skip it because remaining 110 is below 128; write bit 0; new remainder 110.
  7. Weight 64: take it because remaining 110 is at least 64; write bit 1; new remainder 46.
  8. Weight 32: take it because remaining 46 is at least 32; write bit 1; new remainder 14.
  9. Weight 16: skip it because remaining 14 is below 16; write bit 0; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101110.
  16. Confirm the bit weights sum back to 366.
  17. Answer: 366₁₀ = 0101101110₂.

Final answer: 0101101110

Foundations · Problem 0048

Express decimal 367 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 367.
  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 367 is below 512; write bit 0; new remainder 367.
  5. Weight 256: take it because remaining 367 is at least 256; write bit 1; new remainder 111.
  6. Weight 128: skip it because remaining 111 is below 128; write bit 0; new remainder 111.
  7. Weight 64: take it because remaining 111 is at least 64; write bit 1; new remainder 47.
  8. Weight 32: take it because remaining 47 is at least 32; write bit 1; new remainder 15.
  9. Weight 16: skip it because remaining 15 is below 16; write bit 0; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101101111.
  16. Confirm the bit weights sum back to 367.
  17. Answer: 367₁₀ = 0101101111₂.

Final answer: 0101101111

Foundations · Problem 0049

Express decimal 368 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 368.
  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 368 is below 512; write bit 0; new remainder 368.
  5. Weight 256: take it because remaining 368 is at least 256; write bit 1; new remainder 112.
  6. Weight 128: skip it because remaining 112 is below 128; write bit 0; new remainder 112.
  7. Weight 64: take it because remaining 112 is at least 64; write bit 1; new remainder 48.
  8. Weight 32: take it because remaining 48 is at least 32; write bit 1; new remainder 16.
  9. Weight 16: take it because remaining 16 is at least 16; write bit 1; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110000.
  16. Confirm the bit weights sum back to 368.
  17. Answer: 368₁₀ = 0101110000₂.

Final answer: 0101110000

Foundations · Problem 0050

Express decimal 369 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 369.
  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 369 is below 512; write bit 0; new remainder 369.
  5. Weight 256: take it because remaining 369 is at least 256; write bit 1; new remainder 113.
  6. Weight 128: skip it because remaining 113 is below 128; write bit 0; new remainder 113.
  7. Weight 64: take it because remaining 113 is at least 64; write bit 1; new remainder 49.
  8. Weight 32: take it because remaining 49 is at least 32; write bit 1; new remainder 17.
  9. Weight 16: take it because remaining 17 is at least 16; write bit 1; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110001.
  16. Confirm the bit weights sum back to 369.
  17. Answer: 369₁₀ = 0101110001₂.

Final answer: 0101110001

Foundations · Problem 0051

Express decimal 370 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 370.
  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 370 is below 512; write bit 0; new remainder 370.
  5. Weight 256: take it because remaining 370 is at least 256; write bit 1; new remainder 114.
  6. Weight 128: skip it because remaining 114 is below 128; write bit 0; new remainder 114.
  7. Weight 64: take it because remaining 114 is at least 64; write bit 1; new remainder 50.
  8. Weight 32: take it because remaining 50 is at least 32; write bit 1; new remainder 18.
  9. Weight 16: take it because remaining 18 is at least 16; write bit 1; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110010.
  16. Confirm the bit weights sum back to 370.
  17. Answer: 370₁₀ = 0101110010₂.

Final answer: 0101110010

Foundations · Problem 0052

Express decimal 371 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 371.
  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 371 is below 512; write bit 0; new remainder 371.
  5. Weight 256: take it because remaining 371 is at least 256; write bit 1; new remainder 115.
  6. Weight 128: skip it because remaining 115 is below 128; write bit 0; new remainder 115.
  7. Weight 64: take it because remaining 115 is at least 64; write bit 1; new remainder 51.
  8. Weight 32: take it because remaining 51 is at least 32; write bit 1; new remainder 19.
  9. Weight 16: take it because remaining 19 is at least 16; write bit 1; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110011.
  16. Confirm the bit weights sum back to 371.
  17. Answer: 371₁₀ = 0101110011₂.

Final answer: 0101110011

Foundations · Problem 0053

Express decimal 372 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 372.
  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 372 is below 512; write bit 0; new remainder 372.
  5. Weight 256: take it because remaining 372 is at least 256; write bit 1; new remainder 116.
  6. Weight 128: skip it because remaining 116 is below 128; write bit 0; new remainder 116.
  7. Weight 64: take it because remaining 116 is at least 64; write bit 1; new remainder 52.
  8. Weight 32: take it because remaining 52 is at least 32; write bit 1; new remainder 20.
  9. Weight 16: take it because remaining 20 is at least 16; write bit 1; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110100.
  16. Confirm the bit weights sum back to 372.
  17. Answer: 372₁₀ = 0101110100₂.

Final answer: 0101110100

Foundations · Problem 0054

Express decimal 373 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 373.
  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 373 is below 512; write bit 0; new remainder 373.
  5. Weight 256: take it because remaining 373 is at least 256; write bit 1; new remainder 117.
  6. Weight 128: skip it because remaining 117 is below 128; write bit 0; new remainder 117.
  7. Weight 64: take it because remaining 117 is at least 64; write bit 1; new remainder 53.
  8. Weight 32: take it because remaining 53 is at least 32; write bit 1; new remainder 21.
  9. Weight 16: take it because remaining 21 is at least 16; write bit 1; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110101.
  16. Confirm the bit weights sum back to 373.
  17. Answer: 373₁₀ = 0101110101₂.

Final answer: 0101110101

Foundations · Problem 0055

Express decimal 374 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 374.
  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 374 is below 512; write bit 0; new remainder 374.
  5. Weight 256: take it because remaining 374 is at least 256; write bit 1; new remainder 118.
  6. Weight 128: skip it because remaining 118 is below 128; write bit 0; new remainder 118.
  7. Weight 64: take it because remaining 118 is at least 64; write bit 1; new remainder 54.
  8. Weight 32: take it because remaining 54 is at least 32; write bit 1; new remainder 22.
  9. Weight 16: take it because remaining 22 is at least 16; write bit 1; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110110.
  16. Confirm the bit weights sum back to 374.
  17. Answer: 374₁₀ = 0101110110₂.

Final answer: 0101110110

Foundations · Problem 0056

Express decimal 375 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 375.
  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 375 is below 512; write bit 0; new remainder 375.
  5. Weight 256: take it because remaining 375 is at least 256; write bit 1; new remainder 119.
  6. Weight 128: skip it because remaining 119 is below 128; write bit 0; new remainder 119.
  7. Weight 64: take it because remaining 119 is at least 64; write bit 1; new remainder 55.
  8. Weight 32: take it because remaining 55 is at least 32; write bit 1; new remainder 23.
  9. Weight 16: take it because remaining 23 is at least 16; write bit 1; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101110111.
  16. Confirm the bit weights sum back to 375.
  17. Answer: 375₁₀ = 0101110111₂.

Final answer: 0101110111

Foundations · Problem 0057

Express decimal 376 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 376.
  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 376 is below 512; write bit 0; new remainder 376.
  5. Weight 256: take it because remaining 376 is at least 256; write bit 1; new remainder 120.
  6. Weight 128: skip it because remaining 120 is below 128; write bit 0; new remainder 120.
  7. Weight 64: take it because remaining 120 is at least 64; write bit 1; new remainder 56.
  8. Weight 32: take it because remaining 56 is at least 32; write bit 1; new remainder 24.
  9. Weight 16: take it because remaining 24 is at least 16; write bit 1; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111000.
  16. Confirm the bit weights sum back to 376.
  17. Answer: 376₁₀ = 0101111000₂.

Final answer: 0101111000

Foundations · Problem 0058

Express decimal 377 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 377.
  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 377 is below 512; write bit 0; new remainder 377.
  5. Weight 256: take it because remaining 377 is at least 256; write bit 1; new remainder 121.
  6. Weight 128: skip it because remaining 121 is below 128; write bit 0; new remainder 121.
  7. Weight 64: take it because remaining 121 is at least 64; write bit 1; new remainder 57.
  8. Weight 32: take it because remaining 57 is at least 32; write bit 1; new remainder 25.
  9. Weight 16: take it because remaining 25 is at least 16; write bit 1; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111001.
  16. Confirm the bit weights sum back to 377.
  17. Answer: 377₁₀ = 0101111001₂.

Final answer: 0101111001

Foundations · Problem 0059

Express decimal 378 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 378.
  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 378 is below 512; write bit 0; new remainder 378.
  5. Weight 256: take it because remaining 378 is at least 256; write bit 1; new remainder 122.
  6. Weight 128: skip it because remaining 122 is below 128; write bit 0; new remainder 122.
  7. Weight 64: take it because remaining 122 is at least 64; write bit 1; new remainder 58.
  8. Weight 32: take it because remaining 58 is at least 32; write bit 1; new remainder 26.
  9. Weight 16: take it because remaining 26 is at least 16; write bit 1; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111010.
  16. Confirm the bit weights sum back to 378.
  17. Answer: 378₁₀ = 0101111010₂.

Final answer: 0101111010

Foundations · Problem 0060

Express decimal 379 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 379.
  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 379 is below 512; write bit 0; new remainder 379.
  5. Weight 256: take it because remaining 379 is at least 256; write bit 1; new remainder 123.
  6. Weight 128: skip it because remaining 123 is below 128; write bit 0; new remainder 123.
  7. Weight 64: take it because remaining 123 is at least 64; write bit 1; new remainder 59.
  8. Weight 32: take it because remaining 59 is at least 32; write bit 1; new remainder 27.
  9. Weight 16: take it because remaining 27 is at least 16; write bit 1; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111011.
  16. Confirm the bit weights sum back to 379.
  17. Answer: 379₁₀ = 0101111011₂.

Final answer: 0101111011

Foundations · Problem 0061

Express decimal 380 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 380.
  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 380 is below 512; write bit 0; new remainder 380.
  5. Weight 256: take it because remaining 380 is at least 256; write bit 1; new remainder 124.
  6. Weight 128: skip it because remaining 124 is below 128; write bit 0; new remainder 124.
  7. Weight 64: take it because remaining 124 is at least 64; write bit 1; new remainder 60.
  8. Weight 32: take it because remaining 60 is at least 32; write bit 1; new remainder 28.
  9. Weight 16: take it because remaining 28 is at least 16; write bit 1; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111100.
  16. Confirm the bit weights sum back to 380.
  17. Answer: 380₁₀ = 0101111100₂.

Final answer: 0101111100

Foundations · Problem 0062

Express decimal 381 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 381.
  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 381 is below 512; write bit 0; new remainder 381.
  5. Weight 256: take it because remaining 381 is at least 256; write bit 1; new remainder 125.
  6. Weight 128: skip it because remaining 125 is below 128; write bit 0; new remainder 125.
  7. Weight 64: take it because remaining 125 is at least 64; write bit 1; new remainder 61.
  8. Weight 32: take it because remaining 61 is at least 32; write bit 1; new remainder 29.
  9. Weight 16: take it because remaining 29 is at least 16; write bit 1; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111101.
  16. Confirm the bit weights sum back to 381.
  17. Answer: 381₁₀ = 0101111101₂.

Final answer: 0101111101

Foundations · Problem 0063

Express decimal 382 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 382.
  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 382 is below 512; write bit 0; new remainder 382.
  5. Weight 256: take it because remaining 382 is at least 256; write bit 1; new remainder 126.
  6. Weight 128: skip it because remaining 126 is below 128; write bit 0; new remainder 126.
  7. Weight 64: take it because remaining 126 is at least 64; write bit 1; new remainder 62.
  8. Weight 32: take it because remaining 62 is at least 32; write bit 1; new remainder 30.
  9. Weight 16: take it because remaining 30 is at least 16; write bit 1; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111110.
  16. Confirm the bit weights sum back to 382.
  17. Answer: 382₁₀ = 0101111110₂.

Final answer: 0101111110

Foundations · Problem 0064

Express decimal 383 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 383.
  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 383 is below 512; write bit 0; new remainder 383.
  5. Weight 256: take it because remaining 383 is at least 256; write bit 1; new remainder 127.
  6. Weight 128: skip it because remaining 127 is below 128; write bit 0; new remainder 127.
  7. Weight 64: take it because remaining 127 is at least 64; write bit 1; new remainder 63.
  8. Weight 32: take it because remaining 63 is at least 32; write bit 1; new remainder 31.
  9. Weight 16: take it because remaining 31 is at least 16; write bit 1; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101111111.
  16. Confirm the bit weights sum back to 383.
  17. Answer: 383₁₀ = 0101111111₂.

Final answer: 0101111111