Question

$(11{)}_2+(22{)}_3+(33{)}_4+(44{)}_5+(55{)}_6+(66{)}_7+(77{)}_8+(88{)}_9=(?{)}_{10}$

• A. 396
• B. 276
• C. 250
• D. 342

Answer 1

The correct option is B 276

$(11{)}_2=(1\times 2^1+1\times 2^0{)}_{10}=\left(3{)}_{10}\right(22{)}_3=(2\times 3^1+2\times 3^0{)}_{10}=(8{)}_{10}(33{)}_4=(3\times 4^1+3\times 4^0{)}_{10}=\left(15{)}_{10}\right(44{)}_5=(4\times 5^1+4\times 5^0{)}_{10}=(24{)}_{10}(55{)}_6=(5\times 6^1+5\times 6^0{)}_{10}=\left(35{)}_{10}\right(66{)}_7=(6\times 7^1+6\times 7^0{)}_{10}=(48{)}_{10}(77{)}_8=(7\times 8^1+7\times 8^0{)}_{10}=\left(63{)}_{10}\right(88{)}_9=(8\times 9^1+8\times 9^0{)}_{10}=(80{)}_{10}sum=(276{)}_{10}$
$\Rightarrow$ Option (b) is correct answer.