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Exercise 1.1 · Q4

Q.Simplify the following and write in binary notation: (1111000110000)2×5642371(1111000110000)_2 \times 5642371.

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The binary number is first converted to decimal, multiplied by the given decimal factor, and the product is converted back to binary by successive division by 2.

[!FORMULA]

Binary→\todecimal: (dn…d0)2=∑dk2k(d_n\ldots d_0)_2=\sum d_k2^k. Decimal→\tobinary: repeatedly divide by 22, recording remainders; the binary digits are the remainders read from LAST to FIRST.

  1. Convert (1111000110000)2(1111000110000)_2 to decimal. 1-bits are at positions 12,11,10,9,5,412,11,10,9,5,4: 212+211+210+29+25+24=4096+2048+1024+512+32+16=77282^{12}+2^{11}+2^{10}+2^{9}+2^{5}+2^{4}=4096+2048+1024+512+32+16=7728.
  2. Multiply: 7728×5,642,3717728\times 5{,}642{,}371. Split 7728=8000−2727728=8000-272: 5,642,371×8000=45,138,968,0005{,}642{,}371\times 8000 = 45{,}138{,}968{,}000. 5,642,371×272=5,642,371×200+5,642,371×72=1,128,474,200+406,250,712=1,534,724,9125{,}642{,}371\times 272 = 5{,}642{,}371\times 200+5{,}642{,}371\times 72 = 1{,}128{,}474{,}200+406{,}250{,}712=1{,}534{,}724{,}912. Product =45,138,968,000−1,534,724,912=43,604,243,088=45{,}138{,}968{,}000-1{,}534{,}724{,}912=43{,}604{,}243{,}088.
  3. Cross-check by the other split 7728=7700+287728=7700+28: 5,642,371×7700=43,446,256,7005{,}642{,}371\times7700=43{,}446{,}256{,}700 and 5,642,371×28=157,986,3885{,}642{,}371\times28=157{,}986{,}388; sum =43,604,243,088=43{,}604{,}243{,}088 — matches step 2. …

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