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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-di




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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