The above equation was written on a chalkboard to represent an example of a correctly answered multiplication problem. However, the class prankster went when the teacher wasn't looking and smudged many of the numbers. What did the original problem read?
(In reply to Answer
by K Sengupta)
It is obvious from the respective last digits of the multiplier and the product that the last digit of the multiplicand must be 9.
Since 7*3(mod 10) = 1, it is obvious that the last two digits of the number at the third step is 51, so that the last two digits of the multiplier is 17.
Since 17*9 = 153, it is obvious that the first two digits of the number in the first step must possess the form 1(mod 9), so that the said first two digits can be 10, 20, ..., 90. of these, only 10 works, since 117*9 = 1053.
Thus, the number in the first step = 1053, and the multiplier = 117, giving the number in the third step as 3*117 = 351.
We now note that, the number at the second step consists of three digits with the middle digit being 1. considering all three digit multiples of 117, we observe that the middle digit of such a multiple can be 1, only when the number in the second step is 117, so that the multiplicand is 319.
Since the multiplier is 117 and the multiplicand is 319, we are now in a position to complete the muliplication as follows:
Edited on January 22, 2009, 1:23 pm