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Question Number 114295 by mathdave last updated on 18/Sep/20

Answered by mr W last updated on 19/Sep/20

say the number is [abcde]  let A=a+c+e  B=b+d  A+B=41 (as given)  A−B=0 or ±11, ±22, ...  ⇒A=20, B=21 (impossible, since B<2×9=18)  ⇒A=26, B=15  ⇒A=37, B=4  (impossible, since A<3×9=27)  so we have only one possibility:  A=26, B=15  ⇒A=9+9+8  ⇒B=9+6  ⇒smallest number is 86999  ⇒largest number is 99968

saythenumberis[abcde]letA=a+c+eB=b+dA+B=41(asgiven)AB=0or±11,±22,...A=20,B=21(impossible,sinceB<2×9=18)A=26,B=15A=37,B=4(impossible,sinceA<3×9=27)sowehaveonlyonepossibility:A=26,B=15A=9+9+8B=9+6smallestnumberis86999largestnumberis99968

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