The sum of 1006 different positive integers is 1019057. If none of them is
greater than 2012, what is the minimum number of these integers which must be
odd?

• answer is 3 odd integers...
  we have to calculate minimum odd integers...
  so,first consider sum of all even integers up to 2012..
  that is 2+4+6+...2012=2(1+2+3+....+1006)=2*((1006)*(1007)/2))=1006*1007=1013042
  now remaining sum required=1019057-1013042=6015
  that must be formed by odd numbers sine all even integers are over..
  now consider odd numbers from back 2011,2009,2007,2005,2003,2001...
  by using 3 odds like (2011+2003+2001) and (2009+2005+2001) and so on...
  we can get the result...
  so max odd integers required are 3...
• 12 years agoHelpfull: Yes(4) No(1)
• Since the sum of 1006 different integers = 1019057,so it is clear that the no. of odd integers in 1006 integers must be odd.
  Since we are to find the minimum no. of odd integers in 1006 integers,so, we begin assuming that the minimum no. of odd integers be 1. So, it implies that the number of even integers =(1006-1)= 1005.
  Since it is given that no integer exceeds 2012,so max. value of an integer can be 2012.Among the 1st 2012 +ve integers ,the no. of even integers = 1006.
  Due to our assumption,we first consider the sum of 1st 1005 +ve even integers.
  2+4+6+...+2010=2*(1+2+3+…+1005)=1006*1005=1011030; the reaming odd integer must be (1019057 - 1011030) =8027,but which violates the condition of the problem.
  Now, if we include 2012 in place of 2 while selecting the 1005 even integers from 1006 even integers, Then ,we get the sum : 4+6+...+2010+2012=
  =2*(2+3+…1005+1006)=1008*1005=1013040,and this is also ofcourse the highest sum if 1005 even integers are selected from 1st 2012 +ve integers.
  So,we need (1019057 - 1013040)=6017,more; So, the odd integer required in this case =6017.
  So, it is clear that the minimum no. of odd integers in 1006 integers summing to 1019057,can’t be 1.
  So, now let’s assume that minimum no. of odd integers in the selected 1006 integers is = 3.
  At this stage we should observe that the sum of the highest 3 odd integers among the 1st 2012 +ve integers = 2011+2009+2007=6027.
  Now, previously we’ve seen 4+6+...+2010+2012=1013040,which is 6017 less than the required sum.
  Again,in this case we chose 1005 even numbers leaving the provision of selecting only one odd integer.
  Now, if we exclude 4 & 6 from the above mentioned even numbers ,then the sum will be 10 less,i,e,
  8+10+...+2010+2012=1013040-10=1013030,which is 6027 less that the required sum,and since we have excluded 4,6,i.e,2 even numbers ,so now we have selected 1003 even numbers,i.e,we still have the provision of selecting 3 odd integers, and the only requirement is that the sum of the 3 chosen odd integers must be 6027.But we’ve recently observed that
  2011+2009+2007=6027.
  So, the minimum no. of odd integers among the 1006 integers will be 3 & moreover they will be 2007,2009,2011.
  So, the 1006 integers are :[8,10,12,14,16,……,2007,2008,2009,2010,2011,2012] ,whose sum is: 1019057 .
  So,the minimum no. of integers among these 1006 integers ,which must be odd = 3.
  
  Thanks To All..:-):-)
• 12 years agoHelpfull: Yes(3) No(2)