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10 years, 1 month ago
For a positive integer n, let $$P_n$$ denote the product of the digits of n, and $$S_n$$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $$P_n$$ + $$S_n$$ = n is
10 years, 1 month ago
Let n can be a 2 digit or a 3 digit number. First n be a 2 digit number. So n = 10x + y and Pn = xy and Sn = x + y Now, Pn + Sn = n ∴ xy + x + y = 10x + y , we have y = 9 . Hence there are 9 numbers 19, 29,.. ,99, so 9 cases . Now if n is a 3 digit number. Let n = 100x + 10y + z,So Pn = xyz and Sn = x + y + z Now, for Pn + Sn = n ; xyz + x + y + z = 100x + 10y + z ; so. xyz = 99x + 9y . For above equation there is no value for which the above equation have an integer (singlt digit) Value. Hence option D.
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