TS ICET 24 May 2018 Shift 1 Paper

Question : 109 of 200

Marks: +1, -0

342_(7) + 243_(7)

Solution:

342_(7)=3 × 7^{2}+4 × 7^{1}+2 × 7^{0}

243_(7)=2 × 7^{2}+4 × 7^{1}+3 × 7^{0}

342_(7)+243_(7)=(5 × 7^{2}+8 × 7^{1}+5 × 7^{0})

=(5 × 7^{2}+(7+1) × 7^{1}+5 × 7^{0})

=(6 × 7^{2}+1 × 7^{1}+5 × 7^{0})

=(615)_7

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