# Find the minimum positive integer such that it is divisible by A and sum of its digits is equal to B in Python

Suppose we have two numbers A and B, we have to find the minimum positive number M so that M is divisible by A and the sum of the digits of M is same as B. So, if there is no such result, then return -1.

So, if the input is like A = 50, B = 2, then the output will be 200 as this is divisible by 50 and sum of its digit = 2 + 0 + 0 = 2.

To solve this, we will follow these steps −

• Define one element type container, that contains two numbers a and b and one string

• que := a new list

• elem := a new element with (0, 0, blank string)

• visited[0, 0] := 1

• insert elem at the end of que

• while size of que > 0, do

• temp_elem := delete first element from que

• if temp_elem.a is 0 and temp_elem.b is b, then

• return integer of temp_elem.string

• for i in range 0 to 9, do

• x :=(temp_elem.a * 10 + i) mod a

• y := temp_elem.b + i

• if y <= b and visited[x, y] is False, then

• visited[x, y] := 1

• insert new element with x, y and temp_elem.string concatenate i into que

• return -1

## Example

Let us see the following implementation to get better understanding −

Live Demo

visited = [[0 for x in range(501)] for y in range(5001)]
class Element:
def __init__(self, a, b, string):
self.a = a
self.b = b
self.string = string
def get_number(a, b):
que = []
elem = Element(0, 0, "")
visited[0][0] = 1
que.append(elem)
while len(que) > 0:
temp_elem = que.pop(0)
if temp_elem.a == 0 and temp_elem.b == b:
return int(temp_elem.string)
for i in range(0, 10):
x = (temp_elem.a * 10 + i) % a
y = temp_elem.b + i
if y <= b and visited[x][y] == False:
visited[x][y] = 1
que.append(Element(x, y, temp_elem.string + str(i)))
return -1

a, b = 50, 2
print(get_number(a, b))

## Input

50, 2

## Output

200

Updated on: 20-Aug-2020

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