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Suppose we have a number n. We have another two numbers a and b. We have to check whether we can generate a number using a and b that divides n.

So, if the input is like n = 115, a = 3, b = 2, then the output will be True as 115 is divisible by 23 which is made of 2 and 3.

To solve this, we will follow these steps −

- Define a function util() . This will take temp, a, b, n
- if temp > n, then
- return False

if n is divisible by temp, then

- return True

- return true when at least one of util(temp * 10 + a, a, b, n) or util(temp * 10 + b, a, b, n) is true otherwise false
- From the main method return true when at least one of util(a, a, b, n) or util(b, a, b, n) is true otherwise false

Let us see the following implementation to get better understanding −

def util(temp, a, b, n): if temp > n: return False if n % temp == 0: return True return util(temp * 10 + a, a, b, n) or util(temp * 10 + b, a, b, n) def solve(n, a, b): return util(a, a, b, n) or util(b, a, b, n) n = 115 a = 3 b = 2 print(solve(n, a, b))

115, 2, 3

True

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