Let the consecutive odd numbers be x and (x+2).x+(x+2)=68⇒2x+2=68⇒2x=68−2⇒x=663321⇒x=33∴The required numbers are 33 and (33+2), i.e., 35.
Let one of the smallest odd no be x therefore another consecutive odd no would be x+2 .given sum of no is 68 iex+(x+2)=682x+2=682x=66x=33
therefore two no are 33 and 35
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