Question

राइट ए प्रोग्राम इन जवा टो असाइन 3 इंटिगर एंड प्रिंट थे सम एंड प्रोडक्ट​

Answer 1

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1);

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1); double diff = floorD1 - d1;

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1); double diff = floorD1 - d1; if (diff==0)

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1); double diff = floorD1 - d1; if (diff==0) System.out.println(“The double is a whole number”);

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1); double diff = floorD1 - d1; if (diff==0) System.out.println(“The double is a whole number”); else

It is a tricky problem. So a trick that can be used is to determine the floor of the number and compare it with the original number. Suppose the original double is d1.double floorD1 = Math.floor(d1); double diff = floorD1 - d1; if (diff==0) System.out.println(“The double is a whole number”); else System.out.println(“The double is NOT a whole number”);