Exploring Digit Properties: Solving for Two-Digit Numbers

Mathematics often presents intriguing problems that involve the properties of digits within numbers. In this blog post, we'll explore a problem where we need to find all two-digit numbers \( N \) that satisfy a specific equation involving the product and sum of their digits. Let's dive into the details and solve this problem step-by-step.

Let `P(n)` and `S(n)` denote the product and sum, respectively, of the digits of a positive integer `n`. Determine all two-digit numbers `N` that satisfy the equation:

`P(N) + 2S(N) = N` 

For example, `P(23)=2\times 6` and `S(23)=2+3=5`. 

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Answer: `14, 36, 77`

Explanation: 

•  Let `N = 10a + b`, where `a` and `b` are the tens and units digits of `N`, respectively.
  Here, `a` and `b` are integers such that `1 \leq a \leq 9` and `0 \leq b \leq 9`.
• Then ,
• The product of the digits `P(N)` is
  
  `P(N) = a \cdot b`
• The sum of the digits `S(N)` is:
  
       `S(N) = a + b`
• Then, the equation `P(N) + 2S(N) = N` becomes

     `a \cdot b + 2(a + b) = 10a + b`

     `a \cdot b + 2a + 2b = 10a + b`

     `a \cdot b + 2a + 2b - b = 10a`

     `a \cdot b + 2b - b = 10a - 2a`

     `a \cdot b + b = 8a`

     `b(a + 1) = 8a`

`b = \frac{8a}{a + 1}`

• For `b` to be an integer, `\frac{8a}{a + 1}` must be an integer.
• Check values of `a` from 1 to 9 to find valid solutions:
• `a = 1`: `b = \frac{8 \cdot 1}{1 + 1} = 4`
• `a = 2`: `b = \frac{8 \cdot 2}{2 + 1} = \frac{16}{3}` (not an integer)
• `a = 3`: `b = \frac{8 \cdot 3}{3 + 1} = 6`
• `a = 4`: `b = \frac{8 \cdot 4}{4 + 1} = \frac{32}{5}` (not an integer)
• `a = 5`: `b = \frac{8 \cdot 5}{5 + 1} = \frac{40}{6}` (not an integer)
• `a = 6`: `b = \frac{8 \cdot 6}{6 + 1} = \frac{48}{7}` (not an integer)
• `a = 7`: `b = \frac{8 \cdot 7}{7 + 1} = 7`
• `a = 8`: `b = \frac{8 \cdot 8}{8 + 1} = \frac{64}{9}` (not an integer)
• `a = 9`: `b = \frac{8 \cdot 9}{9 + 1} = 7.2` (not an integer)
•    The valid pairs `(a, b)` are `(1, 4)`, `(3, 6)`, and `(7, 7)`.
•    Therefore, the valid two-digit numbers `N` are

`N = 10a + b \implies 14, 36, 77`