 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says find the product by suitable rearrangement. Here we have given six problems but before finding the product let us learn the associative law of multiplication. If A, B and C are three whole numbers then the product of A into B with C is equal to the product of B into C with A. So this is the key idea we are going to use in this problem to find the product of the given six problems. So this property shows that the grouping of numbers does not affect the product. So we group the numbers in such a way that the calculation becomes easier. So starting with the first one which is 2 into 1768 into 50. So first we will find the product of 2 and 50. So on rearranging 2 into 50 into 1768. So the product of 250 is 100. Then we have 1768 and on multiplying 100 with 1768 we get 176800. Hence the product of the first part is 176800. So this completes the first part and now proceeding on to the second part which is 4 into 166 into 25. So again on rearranging these numbers first we will multiply 4 with 25 and then to the product we will multiply 166. On multiplying 4 with 25 we get 100. Then we have 166 and on multiplying 166 with 100 we get 16600. And hence the product is equal to 16600. So this completes the second part and now proceeding on to the next part which is 8 into 291 into 125. Now first we will multiply 8 with 125. So making suitable arrangements 8 into 125 into 291. On multiplying 8 with 125 we get 1000. Then we have 291 and on multiplying 291 with 1000 we get 291000. And hence the product of third part is 291000. So this completes the third part and now proceeding on to the next part which is 625 into 279 into 16. Now on observing these three numbers we find that the product of 625 and 16 will give us a number whose units digit is 0. So on suitable arrangement 625 into 16 into 279. First we will multiply these two numbers and on multiplying 625 with 16 we get 10000 into 279 and on multiplying 279 with 10000 we get 2790000. And thus the product is equal to 2790000. So this completes the fourth part and now let us proceed on to the next part where we have to find the product of 285, 5 and 60. So first let us multiply 5 with 60 so that we get a number whose unit digit is 0. So 285 as it is and on multiplying 5 with 60 we get 300. Now on multiplying 285 with 300 we get 85500. So the required product is 85500 which completes the eighth part and now proceeding on to the last part which is 125 into 40 into 8 into 25. So first let us make a suitable rearrangement such that multiplying any two numbers of these four numbers gives the number whose unit digit is 0. So on multiplying 125 with 40 gives 5000 and on multiplying 8 with 25 gives 200 and on multiplying 5200 we get 5 into 2 is 10 and then we have 2 3 4 5 0s. 1 2 3 4 5 and hence the product is equal to 100000. So this completes the session. Hope you enjoyed the session. Take care and bye for now.