How many natural number are possible between 10 to 600 for which the number it self equal to the sum of the product of its digits and the sum of the digits
Answers
Given : natural number between 10 to 600
number it self equal to the sum of the product of its digits and the sum of the digits
To Find : How many possible
Solution:
10 to 600 there are two digits and 3 digits numbers
1st Consider 2 digit numbers
ab where 1 ≤ a 9 and 0 ≤ b ≤ 9
10a + b = ab + a + b
=> 9a = ab
=> 9 = b ∵ a ≠ 0
Possible number 9
19 , 29 , 39 , 49 , 59 , 69 , 79 , 89 , 99
Three digit number
abc where 1 ≤ a ≤ 5 and 0 ≤ b , c ≤ 9
100a + 10b + c = abc + a + b + c
=> 99a + 9b = abc
=> 9 (11a + b) = abc
=> 9x11 * a + b = abc
9x11 * a > abc ∵ bc maximum value = 9*9 = 81
Hence no possible solution
So no such 3 digits numbers possible
Possible number 9
19 , 29 , 39 , 49 , 59 , 69 , 79 , 89 , 99
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