Find the smallest natural number that, when divided by 7, gives a remainder of 1, and when divided by 8

Find the smallest natural number that, when divided by 7, gives a remainder of 1, and when divided by 8, gives a remainder of 2.

Let’s denote the result of dividing without a remainder by 7 through x, then the number can be written as:

7x + 1;

Let’s denote the result of dividing without a remainder by 8 through y, then the number can be written as:

8y + 2;
Since they are the same number, then:
7x + 1 = 8y + 2;
Let us express x:
7x = 8y + 1;
x = (8y + 1) / 7;

x must be an integer, which means that the expression 8y + 1 must be a multiple of 7;
The smallest number that satisfies this condition y = 6;

Then:
x = (8 x 6 + 1) / 7 = 7;

The required number is
7 x 7 + 1 = 50;

8 x 6 + 2 = 50; – it turned out the same number;

Answer: The smallest natural number that, when divided by 7, gives a remainder of 1, and when divided by 8, a remainder of 2, is 50;