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Question Number 113005 by Aina Samuel Temidayo last updated on 10/Sep/20
Twodifferenttwo−digitnaturalnumbersarewrittenbesideeachothersuchthatthelargernumberiswrittenontheleft.Whentheabsolutedifferenceofthetwonumbersissubtractedfromthefour−digitnumbersoformed,thenumberobtainedis5481.Whatisthesumofthetwo−digitnumbers?
Answered by floor(10²Eta[1]) last updated on 11/Sep/20
n1=10a+bn2=10c+dn1>n2n3=abcd=103a+102b+10c+d=100n1+n2n3−(n1−n2)=5481n1+n2=?100n1+n2−n1+n2=99n1+2n2=5481⇒n1isodd∴b∈{1,3,5,7,9}sincen2⩾10:5481=99n1+2n2⩾99n1+20n1⩽55n2⩽98:5481=99n1+2n2⩽99n1+19653.3⩽n1⇒n1⩾54butweknowthatn1isoddsotheonlypossiblecaseisn1=55n1=55⇒5445+2n2=5481⇒n2=18son1+n2=73
Commented by Aina Samuel Temidayo last updated on 11/Sep/20
Ok.Thanks.
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