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Question Number 20693 by Tinkutara last updated on 31/Aug/17

$$\mathrm{Integers}1,2,3,....,n,\mathrm{where}n>2,\mathrm{are}$$ $$\mathrm{written}\mathrm{on}\mathrm{a}\mathrm{board}.\mathrm{Two}\mathrm{numbers}m,k$$ $$\mathrm{such}\mathrm{that}1<m<n,1<k<n\mathrm{are}$$ $$\mathrm{removed}\mathrm{and}\mathrm{the}\mathrm{average}\mathrm{of}\mathrm{the}$$ $$\mathrm{remaining}\mathrm{numbers}\mathrm{is}\mathrm{found}\mathrm{to}\mathrm{be}17.$$ $$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{two}$$ $$\mathrm{removed}\mathrm{numbers}?$$

Answered by ajfour last updated on 31/Aug/17

$${(m+k)}_{maximum}=51$$ $$\frac{n(n+1)}{2}-(m+k)=17(n-2)$$ $$\Rightarrow m+k=\frac{n(n+1)}{2}-17(n-2)⩽2n-2$$ $$firstletussee$$ $$\frac{n^2}{2}+\frac{n}{2}-17n+34-2n+2⩽0$$ $$\Rightarrow n^2-37n+72⩽0$$ $${(n-\frac{37}{2})}^2⩽{\left(\frac{37}{2}\right)}^2-72$$ $$\mid n-\frac{37}{2}\mid ⩽\frac{1369-288}{4}$$ $$\mid n-\frac{37}{2}\mid ⩽\sqrt{270.25}(\approx 16.48)$$ $$\Rightarrow n-18.5⩽16.48$$ $$orn=34$$ $$\mathit{m}+\mathit{k}=\frac{\mathit{n}(\mathit{n}+1)}{2}-17(\mathit{n}-2)$$ $$=17\times 35-17\times 32$$ $$=51.$$

Commented byTinkutara last updated on 01/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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