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Question Number 113005 by Aina Samuel Temidayo last updated on 10/Sep/20

Two different two−digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four−digit number so formed, the number obtained is 5481. What is the sum of the two−digit numbers?

$$\mathrm{Two}\:\mathrm{different}\:\mathrm{two}−\mathrm{digit}\:\mathrm{natural} \\ $$$$\mathrm{numbers}\:\mathrm{are}\:\mathrm{written}\:\mathrm{beside}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{larger}\:\mathrm{number} \\ $$$$\mathrm{is}\:\mathrm{written}\:\mathrm{on}\:\mathrm{the}\:\mathrm{left}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{absolute}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{numbers}\:\mathrm{is}\:\mathrm{subtracted}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{four}−\mathrm{digit}\:\mathrm{number}\:\mathrm{so}\:\mathrm{formed},\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{obtained}\:\mathrm{is}\:\mathrm{5481}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}−\mathrm{digit}\:\mathrm{numbers}? \\ $$

Answered by floor(10²Eta[1]) last updated on 11/Sep/20

n_1 =10a+b n_2 =10c+d n_1 >n_2 n_3 =abcd=10^3 a+10^2 b+10c+d=100n_1 +n_2 n_3 −(n_1 −n_2 )=5481 n_1 +n_2 =? 100n_1 +n_2 −n_1 +n_2 =99n_1 +2n_2 =5481 ⇒n_1 is odd∴b∈{1, 3, 5, 7, 9} since n_2 ≥10: 5481=99n_1 +2n_2 ≥99n_1 +20 n_1 ≤55 n_2 ≤98: 5481=99n_1 +2n_2 ≤99n_1 +196 53.3≤n_1 ⇒n_1 ≥54 but we know that n_1 is odd so the only possible case is n_1 =55 n_1 =55⇒5445+2n_2 =5481 ⇒n_2 =18 so n_1 +n_2 =73

$$\mathrm{n}_{\mathrm{1}} =\mathrm{10a}+\mathrm{b} \\ $$$$\mathrm{n}_{\mathrm{2}} =\mathrm{10c}+\mathrm{d} \\ $$$$\mathrm{n}_{\mathrm{1}} >\mathrm{n}_{\mathrm{2}} \\ $$$$\mathrm{n}_{\mathrm{3}} =\mathrm{abcd}=\mathrm{10}^{\mathrm{3}} \mathrm{a}+\mathrm{10}^{\mathrm{2}} \mathrm{b}+\mathrm{10c}+\mathrm{d}=\mathrm{100n}_{\mathrm{1}} +\mathrm{n}_{\mathrm{2}} \\ $$$$\mathrm{n}_{\mathrm{3}} −\left(\mathrm{n}_{\mathrm{1}} −\mathrm{n}_{\mathrm{2}} \right)=\mathrm{5481} \\ $$$$\mathrm{n}_{\mathrm{1}} +\mathrm{n}_{\mathrm{2}} =? \\ $$$$\mathrm{100n}_{\mathrm{1}} +\mathrm{n}_{\mathrm{2}} −\mathrm{n}_{\mathrm{1}} +\mathrm{n}_{\mathrm{2}} =\mathrm{99n}_{\mathrm{1}} +\mathrm{2n}_{\mathrm{2}} =\mathrm{5481} \\ $$$$\Rightarrow\mathrm{n}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{odd}\therefore\mathrm{b}\in\left\{\mathrm{1},\:\mathrm{3},\:\mathrm{5},\:\mathrm{7},\:\mathrm{9}\right\} \\ $$$$\mathrm{since}\:\mathrm{n}_{\mathrm{2}} \geqslant\mathrm{10}: \\ $$$$\mathrm{5481}=\mathrm{99n}_{\mathrm{1}} +\mathrm{2n}_{\mathrm{2}} \geqslant\mathrm{99n}_{\mathrm{1}} +\mathrm{20} \\ $$$$\mathrm{n}_{\mathrm{1}} \leqslant\mathrm{55} \\ $$$$\mathrm{n}_{\mathrm{2}} \leqslant\mathrm{98}: \\ $$$$\mathrm{5481}=\mathrm{99n}_{\mathrm{1}} +\mathrm{2n}_{\mathrm{2}} \leqslant\mathrm{99n}_{\mathrm{1}} +\mathrm{196} \\ $$$$\mathrm{53}.\mathrm{3}\leqslant\mathrm{n}_{\mathrm{1}} \Rightarrow\mathrm{n}_{\mathrm{1}} \geqslant\mathrm{54} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{n}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{odd}\:\mathrm{so}\:\mathrm{the}\: \\ $$$$\mathrm{only}\:\mathrm{possible}\:\mathrm{case}\:\mathrm{is}\:\mathrm{n}_{\mathrm{1}} =\mathrm{55} \\ $$$$\mathrm{n}_{\mathrm{1}} =\mathrm{55}\Rightarrow\mathrm{5445}+\mathrm{2n}_{\mathrm{2}} =\mathrm{5481} \\ $$$$\Rightarrow\mathrm{n}_{\mathrm{2}} =\mathrm{18} \\ $$$$\mathrm{so}\:\mathrm{n}_{\mathrm{1}} +\mathrm{n}_{\mathrm{2}} =\mathrm{73} \\ $$

Commented by Aina Samuel Temidayo last updated on 11/Sep/20

Ok. Thanks.

$$\mathrm{Ok}.\:\mathrm{Thanks}. \\ $$

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