Question

Consider the addition of numbers with different bases,

$(X{)}_7+(Y{)}_8+(W{)}_{10}+(Z)=(K{)}_9$

If X = 36, Y = 67, W = 98 and K = 241 then, Z is equal to

• A. $(19{)}_{10}$
• B. $(34{)}_5$
• C. $(13{)}_{16}$
• D. $(34{)}_{10}$

Answer 1

Answer: (A), (B), (C)

$(13{)}_{16}$

$(36{)}_7=(27{)}_{10}$

$(67{)}_8=(55{)}_{10}$

$(98{)}_{10}=(98{)}_{10}$

$(241{)}_9=(199{)}_{10}$

$\left(Z\right)=(199{)}_{10}-(27{)}_{10}-(55{)}_{10}-(98{)}_{10}$

$\left(Z\right)=(19{)}_{10}$



$\therefore Z=(19{)}_{10}=(34{)}_5=(13{)}_{16}$