sweetshiva9963
  • sweetshiva9963
Write each of the following (base-10) integers in base 2, base 4, and base 8. a) 137 b) 6243 c) 12,345
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
137 in base 2 is 10001001. if you want to study I will guide you (just one example) and you can find out the leftover.
whpalmer4
  • whpalmer4
In base 10, we have the following place values (in reverse order): 1, 10, 100, 1,000, 10,000, etc. In base 2, instead of powers of 10, we have powers of 2: \[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192\] In base 4, we have powers of 4: \[1, 4, 16, 64, 256, 1024, 4096, 16384\] In base 8, we have powers of 8: \[1, 8, 64, 512, 4096, 32768\] To convert, we need to find the combination of powers of the base that equals our number. For example, with 137 written in base 2, the largest power of 2 we can extract is \(2^7=128\). \(137-128=9\), so the next largest power of \(2\) we can extract is \(2^3=8\). \(9-8=1\), so the final power of \(2\) we can extract is \(2^0=1\). We write \(2^7+2^3+2^0\) as \(10001001_2\). For another example, I'll write \(1000_{10} = x_8\) \(1000-512 = 488\) so \(8^3\) is our first power \(488-64 > 64\) so we can take out multiple \(8^2\) \(488 \text{ div } 64 = 7\) so we can take out \(7*8^2\) leaving \(488-448=40\) \(40-8 > 8\) so we can take out multiple \(8^1\), \(5\) in all, leaving \(40-40=0\) So \(1000_{10}=1750_{8}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.