Find the smallest natural number that, when divided by 6, gives a remainder of 5, and when divided by 7

Find the smallest natural number that, when divided by 6, gives a remainder of 5, and when divided by 7 gives a remainder of 2.

Any natural number x, which, when divided by 6, gives a remainder of 5, can be represented as:

x = 6 * k + 5,

where k is some integer.

Going through the values of k, starting from the smallest k = 1, we will look for the smallest natural number of the form 6 * k + 5, which, when divided by 7, gives a remainder of 2.

For k = 1, we get x = 6 * 1 + 5 = 11. The number 11 when divided by 7 gives a remainder of 4.

With k = 2, we get x = 6 * 2 + 5 = 17. The number 11 when divided by 7 gives a remainder of 3.

With k = 3 we get x = 6 * 3 + 5 = 23. The number 23 when divided by 7 gives a remainder of 2.

Therefore, the number 23 is the desired number.

Answer: the required number is 23.