01How Computers Work (online)
Practice library (online) / Writing binary

Fully solved problem book

Writing binary

Set 1 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 0001

Express decimal 320 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 320.
  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 320 is below 512; write bit 0; new remainder 320.
  5. Weight 256: take it because remaining 320 is at least 256; write bit 1; new remainder 64.
  6. Weight 128: skip it because remaining 64 is below 128; write bit 0; new remainder 64.
  7. Weight 64: take it because remaining 64 is at least 64; write bit 1; new remainder 0.
  8. Weight 32: skip it because remaining 0 is below 32; write bit 0; new remainder 0.
  9. Weight 16: skip it because remaining 0 is below 16; write bit 0; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000000.
  16. Confirm the bit weights sum back to 320.
  17. Answer: 320₁₀ = 0101000000₂.

Final answer: 0101000000

Foundations · Problem 0002

Express decimal 321 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 321.
  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 321 is below 512; write bit 0; new remainder 321.
  5. Weight 256: take it because remaining 321 is at least 256; write bit 1; new remainder 65.
  6. Weight 128: skip it because remaining 65 is below 128; write bit 0; new remainder 65.
  7. Weight 64: take it because remaining 65 is at least 64; write bit 1; new remainder 1.
  8. Weight 32: skip it because remaining 1 is below 32; write bit 0; new remainder 1.
  9. Weight 16: skip it because remaining 1 is below 16; write bit 0; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000001.
  16. Confirm the bit weights sum back to 321.
  17. Answer: 321₁₀ = 0101000001₂.

Final answer: 0101000001

Foundations · Problem 0003

Express decimal 322 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 322.
  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 322 is below 512; write bit 0; new remainder 322.
  5. Weight 256: take it because remaining 322 is at least 256; write bit 1; new remainder 66.
  6. Weight 128: skip it because remaining 66 is below 128; write bit 0; new remainder 66.
  7. Weight 64: take it because remaining 66 is at least 64; write bit 1; new remainder 2.
  8. Weight 32: skip it because remaining 2 is below 32; write bit 0; new remainder 2.
  9. Weight 16: skip it because remaining 2 is below 16; write bit 0; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000010.
  16. Confirm the bit weights sum back to 322.
  17. Answer: 322₁₀ = 0101000010₂.

Final answer: 0101000010

Foundations · Problem 0004

Express decimal 323 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 323.
  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 323 is below 512; write bit 0; new remainder 323.
  5. Weight 256: take it because remaining 323 is at least 256; write bit 1; new remainder 67.
  6. Weight 128: skip it because remaining 67 is below 128; write bit 0; new remainder 67.
  7. Weight 64: take it because remaining 67 is at least 64; write bit 1; new remainder 3.
  8. Weight 32: skip it because remaining 3 is below 32; write bit 0; new remainder 3.
  9. Weight 16: skip it because remaining 3 is below 16; write bit 0; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000011.
  16. Confirm the bit weights sum back to 323.
  17. Answer: 323₁₀ = 0101000011₂.

Final answer: 0101000011

Foundations · Problem 0005

Express decimal 324 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 324.
  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 324 is below 512; write bit 0; new remainder 324.
  5. Weight 256: take it because remaining 324 is at least 256; write bit 1; new remainder 68.
  6. Weight 128: skip it because remaining 68 is below 128; write bit 0; new remainder 68.
  7. Weight 64: take it because remaining 68 is at least 64; write bit 1; new remainder 4.
  8. Weight 32: skip it because remaining 4 is below 32; write bit 0; new remainder 4.
  9. Weight 16: skip it because remaining 4 is below 16; write bit 0; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000100.
  16. Confirm the bit weights sum back to 324.
  17. Answer: 324₁₀ = 0101000100₂.

Final answer: 0101000100

Foundations · Problem 0006

Express decimal 325 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 325.
  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 325 is below 512; write bit 0; new remainder 325.
  5. Weight 256: take it because remaining 325 is at least 256; write bit 1; new remainder 69.
  6. Weight 128: skip it because remaining 69 is below 128; write bit 0; new remainder 69.
  7. Weight 64: take it because remaining 69 is at least 64; write bit 1; new remainder 5.
  8. Weight 32: skip it because remaining 5 is below 32; write bit 0; new remainder 5.
  9. Weight 16: skip it because remaining 5 is below 16; write bit 0; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000101.
  16. Confirm the bit weights sum back to 325.
  17. Answer: 325₁₀ = 0101000101₂.

Final answer: 0101000101

Foundations · Problem 0007

Express decimal 326 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 326.
  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 326 is below 512; write bit 0; new remainder 326.
  5. Weight 256: take it because remaining 326 is at least 256; write bit 1; new remainder 70.
  6. Weight 128: skip it because remaining 70 is below 128; write bit 0; new remainder 70.
  7. Weight 64: take it because remaining 70 is at least 64; write bit 1; new remainder 6.
  8. Weight 32: skip it because remaining 6 is below 32; write bit 0; new remainder 6.
  9. Weight 16: skip it because remaining 6 is below 16; write bit 0; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000110.
  16. Confirm the bit weights sum back to 326.
  17. Answer: 326₁₀ = 0101000110₂.

Final answer: 0101000110

Foundations · Problem 0008

Express decimal 327 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 327.
  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 327 is below 512; write bit 0; new remainder 327.
  5. Weight 256: take it because remaining 327 is at least 256; write bit 1; new remainder 71.
  6. Weight 128: skip it because remaining 71 is below 128; write bit 0; new remainder 71.
  7. Weight 64: take it because remaining 71 is at least 64; write bit 1; new remainder 7.
  8. Weight 32: skip it because remaining 7 is below 32; write bit 0; new remainder 7.
  9. Weight 16: skip it because remaining 7 is below 16; write bit 0; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101000111.
  16. Confirm the bit weights sum back to 327.
  17. Answer: 327₁₀ = 0101000111₂.

Final answer: 0101000111

Foundations · Problem 0009

Express decimal 328 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 328.
  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 328 is below 512; write bit 0; new remainder 328.
  5. Weight 256: take it because remaining 328 is at least 256; write bit 1; new remainder 72.
  6. Weight 128: skip it because remaining 72 is below 128; write bit 0; new remainder 72.
  7. Weight 64: take it because remaining 72 is at least 64; write bit 1; new remainder 8.
  8. Weight 32: skip it because remaining 8 is below 32; write bit 0; new remainder 8.
  9. Weight 16: skip it because remaining 8 is below 16; write bit 0; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001000.
  16. Confirm the bit weights sum back to 328.
  17. Answer: 328₁₀ = 0101001000₂.

Final answer: 0101001000

Foundations · Problem 0010

Express decimal 329 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 329.
  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 329 is below 512; write bit 0; new remainder 329.
  5. Weight 256: take it because remaining 329 is at least 256; write bit 1; new remainder 73.
  6. Weight 128: skip it because remaining 73 is below 128; write bit 0; new remainder 73.
  7. Weight 64: take it because remaining 73 is at least 64; write bit 1; new remainder 9.
  8. Weight 32: skip it because remaining 9 is below 32; write bit 0; new remainder 9.
  9. Weight 16: skip it because remaining 9 is below 16; write bit 0; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001001.
  16. Confirm the bit weights sum back to 329.
  17. Answer: 329₁₀ = 0101001001₂.

Final answer: 0101001001

Foundations · Problem 0011

Express decimal 330 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 330.
  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 330 is below 512; write bit 0; new remainder 330.
  5. Weight 256: take it because remaining 330 is at least 256; write bit 1; new remainder 74.
  6. Weight 128: skip it because remaining 74 is below 128; write bit 0; new remainder 74.
  7. Weight 64: take it because remaining 74 is at least 64; write bit 1; new remainder 10.
  8. Weight 32: skip it because remaining 10 is below 32; write bit 0; new remainder 10.
  9. Weight 16: skip it because remaining 10 is below 16; write bit 0; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001010.
  16. Confirm the bit weights sum back to 330.
  17. Answer: 330₁₀ = 0101001010₂.

Final answer: 0101001010

Foundations · Problem 0012

Express decimal 331 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 331.
  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 331 is below 512; write bit 0; new remainder 331.
  5. Weight 256: take it because remaining 331 is at least 256; write bit 1; new remainder 75.
  6. Weight 128: skip it because remaining 75 is below 128; write bit 0; new remainder 75.
  7. Weight 64: take it because remaining 75 is at least 64; write bit 1; new remainder 11.
  8. Weight 32: skip it because remaining 11 is below 32; write bit 0; new remainder 11.
  9. Weight 16: skip it because remaining 11 is below 16; write bit 0; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001011.
  16. Confirm the bit weights sum back to 331.
  17. Answer: 331₁₀ = 0101001011₂.

Final answer: 0101001011

Foundations · Problem 0013

Express decimal 332 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 332.
  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 332 is below 512; write bit 0; new remainder 332.
  5. Weight 256: take it because remaining 332 is at least 256; write bit 1; new remainder 76.
  6. Weight 128: skip it because remaining 76 is below 128; write bit 0; new remainder 76.
  7. Weight 64: take it because remaining 76 is at least 64; write bit 1; new remainder 12.
  8. Weight 32: skip it because remaining 12 is below 32; write bit 0; new remainder 12.
  9. Weight 16: skip it because remaining 12 is below 16; write bit 0; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001100.
  16. Confirm the bit weights sum back to 332.
  17. Answer: 332₁₀ = 0101001100₂.

Final answer: 0101001100

Foundations · Problem 0014

Express decimal 333 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 333.
  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 333 is below 512; write bit 0; new remainder 333.
  5. Weight 256: take it because remaining 333 is at least 256; write bit 1; new remainder 77.
  6. Weight 128: skip it because remaining 77 is below 128; write bit 0; new remainder 77.
  7. Weight 64: take it because remaining 77 is at least 64; write bit 1; new remainder 13.
  8. Weight 32: skip it because remaining 13 is below 32; write bit 0; new remainder 13.
  9. Weight 16: skip it because remaining 13 is below 16; write bit 0; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001101.
  16. Confirm the bit weights sum back to 333.
  17. Answer: 333₁₀ = 0101001101₂.

Final answer: 0101001101

Foundations · Problem 0015

Express decimal 334 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 334.
  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 334 is below 512; write bit 0; new remainder 334.
  5. Weight 256: take it because remaining 334 is at least 256; write bit 1; new remainder 78.
  6. Weight 128: skip it because remaining 78 is below 128; write bit 0; new remainder 78.
  7. Weight 64: take it because remaining 78 is at least 64; write bit 1; new remainder 14.
  8. Weight 32: skip it because remaining 14 is below 32; write bit 0; new remainder 14.
  9. Weight 16: skip it because remaining 14 is below 16; write bit 0; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001110.
  16. Confirm the bit weights sum back to 334.
  17. Answer: 334₁₀ = 0101001110₂.

Final answer: 0101001110

Foundations · Problem 0016

Express decimal 335 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 335.
  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 335 is below 512; write bit 0; new remainder 335.
  5. Weight 256: take it because remaining 335 is at least 256; write bit 1; new remainder 79.
  6. Weight 128: skip it because remaining 79 is below 128; write bit 0; new remainder 79.
  7. Weight 64: take it because remaining 79 is at least 64; write bit 1; new remainder 15.
  8. Weight 32: skip it because remaining 15 is below 32; write bit 0; new remainder 15.
  9. Weight 16: skip it because remaining 15 is below 16; write bit 0; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101001111.
  16. Confirm the bit weights sum back to 335.
  17. Answer: 335₁₀ = 0101001111₂.

Final answer: 0101001111

Foundations · Problem 0017

Express decimal 336 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 336.
  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 336 is below 512; write bit 0; new remainder 336.
  5. Weight 256: take it because remaining 336 is at least 256; write bit 1; new remainder 80.
  6. Weight 128: skip it because remaining 80 is below 128; write bit 0; new remainder 80.
  7. Weight 64: take it because remaining 80 is at least 64; write bit 1; new remainder 16.
  8. Weight 32: skip it because remaining 16 is below 32; write bit 0; new remainder 16.
  9. Weight 16: take it because remaining 16 is at least 16; write bit 1; new remainder 0.
  10. Weight 8: skip it because remaining 0 is below 8; write bit 0; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010000.
  16. Confirm the bit weights sum back to 336.
  17. Answer: 336₁₀ = 0101010000₂.

Final answer: 0101010000

Foundations · Problem 0018

Express decimal 337 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 337.
  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 337 is below 512; write bit 0; new remainder 337.
  5. Weight 256: take it because remaining 337 is at least 256; write bit 1; new remainder 81.
  6. Weight 128: skip it because remaining 81 is below 128; write bit 0; new remainder 81.
  7. Weight 64: take it because remaining 81 is at least 64; write bit 1; new remainder 17.
  8. Weight 32: skip it because remaining 17 is below 32; write bit 0; new remainder 17.
  9. Weight 16: take it because remaining 17 is at least 16; write bit 1; new remainder 1.
  10. Weight 8: skip it because remaining 1 is below 8; write bit 0; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010001.
  16. Confirm the bit weights sum back to 337.
  17. Answer: 337₁₀ = 0101010001₂.

Final answer: 0101010001

Foundations · Problem 0019

Express decimal 338 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 338.
  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 338 is below 512; write bit 0; new remainder 338.
  5. Weight 256: take it because remaining 338 is at least 256; write bit 1; new remainder 82.
  6. Weight 128: skip it because remaining 82 is below 128; write bit 0; new remainder 82.
  7. Weight 64: take it because remaining 82 is at least 64; write bit 1; new remainder 18.
  8. Weight 32: skip it because remaining 18 is below 32; write bit 0; new remainder 18.
  9. Weight 16: take it because remaining 18 is at least 16; write bit 1; new remainder 2.
  10. Weight 8: skip it because remaining 2 is below 8; write bit 0; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010010.
  16. Confirm the bit weights sum back to 338.
  17. Answer: 338₁₀ = 0101010010₂.

Final answer: 0101010010

Foundations · Problem 0020

Express decimal 339 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 339.
  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 339 is below 512; write bit 0; new remainder 339.
  5. Weight 256: take it because remaining 339 is at least 256; write bit 1; new remainder 83.
  6. Weight 128: skip it because remaining 83 is below 128; write bit 0; new remainder 83.
  7. Weight 64: take it because remaining 83 is at least 64; write bit 1; new remainder 19.
  8. Weight 32: skip it because remaining 19 is below 32; write bit 0; new remainder 19.
  9. Weight 16: take it because remaining 19 is at least 16; write bit 1; new remainder 3.
  10. Weight 8: skip it because remaining 3 is below 8; write bit 0; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010011.
  16. Confirm the bit weights sum back to 339.
  17. Answer: 339₁₀ = 0101010011₂.

Final answer: 0101010011

Foundations · Problem 0021

Express decimal 340 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 340.
  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 340 is below 512; write bit 0; new remainder 340.
  5. Weight 256: take it because remaining 340 is at least 256; write bit 1; new remainder 84.
  6. Weight 128: skip it because remaining 84 is below 128; write bit 0; new remainder 84.
  7. Weight 64: take it because remaining 84 is at least 64; write bit 1; new remainder 20.
  8. Weight 32: skip it because remaining 20 is below 32; write bit 0; new remainder 20.
  9. Weight 16: take it because remaining 20 is at least 16; write bit 1; new remainder 4.
  10. Weight 8: skip it because remaining 4 is below 8; write bit 0; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010100.
  16. Confirm the bit weights sum back to 340.
  17. Answer: 340₁₀ = 0101010100₂.

Final answer: 0101010100

Foundations · Problem 0022

Express decimal 341 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 341.
  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 341 is below 512; write bit 0; new remainder 341.
  5. Weight 256: take it because remaining 341 is at least 256; write bit 1; new remainder 85.
  6. Weight 128: skip it because remaining 85 is below 128; write bit 0; new remainder 85.
  7. Weight 64: take it because remaining 85 is at least 64; write bit 1; new remainder 21.
  8. Weight 32: skip it because remaining 21 is below 32; write bit 0; new remainder 21.
  9. Weight 16: take it because remaining 21 is at least 16; write bit 1; new remainder 5.
  10. Weight 8: skip it because remaining 5 is below 8; write bit 0; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010101.
  16. Confirm the bit weights sum back to 341.
  17. Answer: 341₁₀ = 0101010101₂.

Final answer: 0101010101

Foundations · Problem 0023

Express decimal 342 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 342.
  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 342 is below 512; write bit 0; new remainder 342.
  5. Weight 256: take it because remaining 342 is at least 256; write bit 1; new remainder 86.
  6. Weight 128: skip it because remaining 86 is below 128; write bit 0; new remainder 86.
  7. Weight 64: take it because remaining 86 is at least 64; write bit 1; new remainder 22.
  8. Weight 32: skip it because remaining 22 is below 32; write bit 0; new remainder 22.
  9. Weight 16: take it because remaining 22 is at least 16; write bit 1; new remainder 6.
  10. Weight 8: skip it because remaining 6 is below 8; write bit 0; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010110.
  16. Confirm the bit weights sum back to 342.
  17. Answer: 342₁₀ = 0101010110₂.

Final answer: 0101010110

Foundations · Problem 0024

Express decimal 343 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 343.
  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 343 is below 512; write bit 0; new remainder 343.
  5. Weight 256: take it because remaining 343 is at least 256; write bit 1; new remainder 87.
  6. Weight 128: skip it because remaining 87 is below 128; write bit 0; new remainder 87.
  7. Weight 64: take it because remaining 87 is at least 64; write bit 1; new remainder 23.
  8. Weight 32: skip it because remaining 23 is below 32; write bit 0; new remainder 23.
  9. Weight 16: take it because remaining 23 is at least 16; write bit 1; new remainder 7.
  10. Weight 8: skip it because remaining 7 is below 8; write bit 0; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101010111.
  16. Confirm the bit weights sum back to 343.
  17. Answer: 343₁₀ = 0101010111₂.

Final answer: 0101010111

Foundations · Problem 0025

Express decimal 344 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 344.
  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 344 is below 512; write bit 0; new remainder 344.
  5. Weight 256: take it because remaining 344 is at least 256; write bit 1; new remainder 88.
  6. Weight 128: skip it because remaining 88 is below 128; write bit 0; new remainder 88.
  7. Weight 64: take it because remaining 88 is at least 64; write bit 1; new remainder 24.
  8. Weight 32: skip it because remaining 24 is below 32; write bit 0; new remainder 24.
  9. Weight 16: take it because remaining 24 is at least 16; write bit 1; new remainder 8.
  10. Weight 8: take it because remaining 8 is at least 8; write bit 1; new remainder 0.
  11. Weight 4: skip it because remaining 0 is below 4; write bit 0; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011000.
  16. Confirm the bit weights sum back to 344.
  17. Answer: 344₁₀ = 0101011000₂.

Final answer: 0101011000

Foundations · Problem 0026

Express decimal 345 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 345.
  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 345 is below 512; write bit 0; new remainder 345.
  5. Weight 256: take it because remaining 345 is at least 256; write bit 1; new remainder 89.
  6. Weight 128: skip it because remaining 89 is below 128; write bit 0; new remainder 89.
  7. Weight 64: take it because remaining 89 is at least 64; write bit 1; new remainder 25.
  8. Weight 32: skip it because remaining 25 is below 32; write bit 0; new remainder 25.
  9. Weight 16: take it because remaining 25 is at least 16; write bit 1; new remainder 9.
  10. Weight 8: take it because remaining 9 is at least 8; write bit 1; new remainder 1.
  11. Weight 4: skip it because remaining 1 is below 4; write bit 0; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011001.
  16. Confirm the bit weights sum back to 345.
  17. Answer: 345₁₀ = 0101011001₂.

Final answer: 0101011001

Foundations · Problem 0027

Express decimal 346 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 346.
  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 346 is below 512; write bit 0; new remainder 346.
  5. Weight 256: take it because remaining 346 is at least 256; write bit 1; new remainder 90.
  6. Weight 128: skip it because remaining 90 is below 128; write bit 0; new remainder 90.
  7. Weight 64: take it because remaining 90 is at least 64; write bit 1; new remainder 26.
  8. Weight 32: skip it because remaining 26 is below 32; write bit 0; new remainder 26.
  9. Weight 16: take it because remaining 26 is at least 16; write bit 1; new remainder 10.
  10. Weight 8: take it because remaining 10 is at least 8; write bit 1; new remainder 2.
  11. Weight 4: skip it because remaining 2 is below 4; write bit 0; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011010.
  16. Confirm the bit weights sum back to 346.
  17. Answer: 346₁₀ = 0101011010₂.

Final answer: 0101011010

Foundations · Problem 0028

Express decimal 347 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 347.
  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 347 is below 512; write bit 0; new remainder 347.
  5. Weight 256: take it because remaining 347 is at least 256; write bit 1; new remainder 91.
  6. Weight 128: skip it because remaining 91 is below 128; write bit 0; new remainder 91.
  7. Weight 64: take it because remaining 91 is at least 64; write bit 1; new remainder 27.
  8. Weight 32: skip it because remaining 27 is below 32; write bit 0; new remainder 27.
  9. Weight 16: take it because remaining 27 is at least 16; write bit 1; new remainder 11.
  10. Weight 8: take it because remaining 11 is at least 8; write bit 1; new remainder 3.
  11. Weight 4: skip it because remaining 3 is below 4; write bit 0; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011011.
  16. Confirm the bit weights sum back to 347.
  17. Answer: 347₁₀ = 0101011011₂.

Final answer: 0101011011

Foundations · Problem 0029

Express decimal 348 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 348.
  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 348 is below 512; write bit 0; new remainder 348.
  5. Weight 256: take it because remaining 348 is at least 256; write bit 1; new remainder 92.
  6. Weight 128: skip it because remaining 92 is below 128; write bit 0; new remainder 92.
  7. Weight 64: take it because remaining 92 is at least 64; write bit 1; new remainder 28.
  8. Weight 32: skip it because remaining 28 is below 32; write bit 0; new remainder 28.
  9. Weight 16: take it because remaining 28 is at least 16; write bit 1; new remainder 12.
  10. Weight 8: take it because remaining 12 is at least 8; write bit 1; new remainder 4.
  11. Weight 4: take it because remaining 4 is at least 4; write bit 1; new remainder 0.
  12. Weight 2: skip it because remaining 0 is below 2; write bit 0; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011100.
  16. Confirm the bit weights sum back to 348.
  17. Answer: 348₁₀ = 0101011100₂.

Final answer: 0101011100

Foundations · Problem 0030

Express decimal 349 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 349.
  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 349 is below 512; write bit 0; new remainder 349.
  5. Weight 256: take it because remaining 349 is at least 256; write bit 1; new remainder 93.
  6. Weight 128: skip it because remaining 93 is below 128; write bit 0; new remainder 93.
  7. Weight 64: take it because remaining 93 is at least 64; write bit 1; new remainder 29.
  8. Weight 32: skip it because remaining 29 is below 32; write bit 0; new remainder 29.
  9. Weight 16: take it because remaining 29 is at least 16; write bit 1; new remainder 13.
  10. Weight 8: take it because remaining 13 is at least 8; write bit 1; new remainder 5.
  11. Weight 4: take it because remaining 5 is at least 4; write bit 1; new remainder 1.
  12. Weight 2: skip it because remaining 1 is below 2; write bit 0; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011101.
  16. Confirm the bit weights sum back to 349.
  17. Answer: 349₁₀ = 0101011101₂.

Final answer: 0101011101

Foundations · Problem 0031

Express decimal 350 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 350.
  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 350 is below 512; write bit 0; new remainder 350.
  5. Weight 256: take it because remaining 350 is at least 256; write bit 1; new remainder 94.
  6. Weight 128: skip it because remaining 94 is below 128; write bit 0; new remainder 94.
  7. Weight 64: take it because remaining 94 is at least 64; write bit 1; new remainder 30.
  8. Weight 32: skip it because remaining 30 is below 32; write bit 0; new remainder 30.
  9. Weight 16: take it because remaining 30 is at least 16; write bit 1; new remainder 14.
  10. Weight 8: take it because remaining 14 is at least 8; write bit 1; new remainder 6.
  11. Weight 4: take it because remaining 6 is at least 4; write bit 1; new remainder 2.
  12. Weight 2: take it because remaining 2 is at least 2; write bit 1; new remainder 0.
  13. Weight 1: skip it because remaining 0 is below 1; write bit 0; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011110.
  16. Confirm the bit weights sum back to 350.
  17. Answer: 350₁₀ = 0101011110₂.

Final answer: 0101011110

Foundations · Problem 0032

Express decimal 351 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 351.
  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 351 is below 512; write bit 0; new remainder 351.
  5. Weight 256: take it because remaining 351 is at least 256; write bit 1; new remainder 95.
  6. Weight 128: skip it because remaining 95 is below 128; write bit 0; new remainder 95.
  7. Weight 64: take it because remaining 95 is at least 64; write bit 1; new remainder 31.
  8. Weight 32: skip it because remaining 31 is below 32; write bit 0; new remainder 31.
  9. Weight 16: take it because remaining 31 is at least 16; write bit 1; new remainder 15.
  10. Weight 8: take it because remaining 15 is at least 8; write bit 1; new remainder 7.
  11. Weight 4: take it because remaining 7 is at least 4; write bit 1; new remainder 3.
  12. Weight 2: take it because remaining 3 is at least 2; write bit 1; new remainder 1.
  13. Weight 1: take it because remaining 1 is at least 1; write bit 1; new remainder 0.
  14. Continue until the ones column; the final remainder must be zero.
  15. Reading the recorded bits left to right gives 0101011111.
  16. Confirm the bit weights sum back to 351.
  17. Answer: 351₁₀ = 0101011111₂.

Final answer: 0101011111