## Task B-1.4.

Among all natural numbers divisible by 8, determine those for which the sum of the digits is 7, and the product of the digits is 6.
## Solution.

Notice that the numbers do not contain the digit 0, as this would make the product of the digits equal to 0.

For the product of the digits to be 6, the number must have:

- one digit 2 and one digit 3, with the rest being 1s, or
- one digit 6, with the rest being 1s.

Since the sum of the digits is also 7, the desired number can be:

- a two-digit number with digits 1 and 6, or
- a four-digit number with digits 1, 1, 2, 3.

Multiples of 8 are divisible by 4, which means the last two digits of the desired number must be divisible by 4. Therefore, the desired number ends with 12, 16, or 32.

If the number is a two-digit number with digits 1 and 6, the only possible number is 16.

If the number is a four-digit number (with digits 1, 1, 2, 3), the possible numbers are 1132, 1312, or 3112, of which only the first is not divisible by 8, and the remaining two are.

Therefore, all solutions are the numbers 16, 1312, and 3112.

Note: A student earns 4 points based on the recorded conclusions expressed in words or symbols that show:

- that 0 cannot be a digit (1 point)
- that they use the fact that the product of the digits is 6, meaning the only possible digits are 1, 6, or 1, 2, 3 (1 point)
- that they use the fact that the sum of the digits is 7, so the number must be a two-digit number with digits 1, 6 or a four-digit number with digits 1, 1, 2, 3 (2 points)

Listing the possibilities and checking the divisibility by 8 to find the desired numbers, 16, 1312, and 3112, is worth 2 points.

If a student has the correct solution but lacks the reasoning or process showing how they concluded these are the only possibilities, they receive 2 points.