Some natural number AA was divided with the remainder by 3, 18 and 24. The sum of these three remainders

Some natural number AA was divided with the remainder by 3, 18 and 24. The sum of these three remainders was equal to 27. Find the remainder after dividing the number AA by 3.

Let us denote the remainders of dividing the number A by 3, 18 and 24 through r1, r2, r3.

Then the number A can be represented as follows:

А = 3 * k + r1, where k is a natural number,

A = 18 * m + r2, where m is a natural number,

A = 24 * n + r3, where n is a natural number.

Let us add the resulting dash of equality:

3 * A = 3 * k + r1 + 18 * m + r2 + 24 * n + r3 =

= 3 * k + 18 * m + 24 * n + r1 + r2 + r3.

By the condition of the problem, it is known that:

r1 + r2 + r3 = 27. Therefore, we have:

3 * A = 3 * k + 18 * m + 24 * n + 27,

A = k + 6 * m + 8 * n + 9 = 3 * k + r1,

r1 = -2 * k + 6 * m + 8 * n + 9 = 2 * (-k + 3 * m + 4 * n + 4) + 1.

Therefore, r1 is an odd number. But r1 is the remainder of division by 3,

those. or 0, or 1, or 2. This implies that r1 = 1.