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Note : need to setup following java heap size -Xms128M -Xmx512M The sum of all the minimal product-sum numbers for 2>= k <=12000 is 7587457 ("Total time taken:"+ ((System.currentTimeMillis()-start_time)/1000)+"sec") ("The sum of all the minimal product-sum numbers for 2≤ k ≤12000 is "+data) Long start_time= System.currentTimeMillis() įor(int index=2 productSum.size()> iterator=uniqueProductSum.entrySet().iterator() Public static void computeKValue(int k,int data) InnerFactor.add(new FactorData(product_factor*value,sum_of_factors+factdata.product_sum,((no_of_factors-1)+factdata.no_of_factors),actual_factors+""+factdata.actual_factors) ) Int tempk=(product_factor*value)-(sum_of_factors+factdata.product_sum)+((no_of_factors-1)+factdata.no_of_factors) InnerFactor.add( new FactorData(product_factor*value,sum_of_factors+value,no_of_factors,actual_factors+""+value)) įor (FactorData factdata:factorList.get(value)) K=(product_factor*value)-(sum_of_factors+value)+no_of_factors Public static List factorization(int data)
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Static HashMap primeList = new HashMap() Static HashMap uniqueProductSum= new HashMap() Static HashMap productSum= new HashMap() There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive that is,0 b), the sum is 61.What is the sum of all the minimal product-sum numbers for 2>= k > factorList = new HashMap>() The points P (x1, y1) and Q (x2, y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to