The greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q - p), (r - p) and (r - q)).For the given problem, p = 83, q = 125 and r = 209.Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 - 83, 209 - 83, 209 - 125) = 42.
Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Let the numbers be x, x + 2, x + 4, x + 6, and x + 8Avg = x + x + 2 + x + 4 + x + 6 + x + 8/5 = 40Therefore, 5x + 20 = 200Therefore, x = 180/5 = 36Therefore, r = x + 4 = 40 s = x + 6 = 42Therefore, Product = 1680
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