6.1. A three-digit number is 56 times greater than its last digit. How many times greater is it than its first digit? Justify your answer.

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# Solution:

Method 1. The last digit is such that when multiplied by 6, the resulting number ends with the same digit. By exhaustive search, we confirm that this can be any even digit (and only it). Therefore, this three-digit number is either 112, 224, 336, or 448 (the option with the last digit 0 is not valid, as it would not be a three-digit number). In all cases, the answer is 112.

Method 2. Let the number be of the form $\overline{a b c}$. Then, by the condition, $100 a + 10 b + c = 56 c$, from which $20 a + 2 b = 11 c$. The number $11 c$ is therefore divisible by 2, meaning the digit $c$ is even, $c = 2d (d \leqslant 4)$. Thus, $10 a + b = 11 d$. But $10 a + b = \overline{a b}$. Since this number is divisible by 11 without a remainder, $a = b$. Then $d = a$ and the original number has the form $\overline{a a (2a)} = a \cdot \overline{112}$, meaning it is 112 times its first digit.

Answer: 112.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct answer obtained and its uniqueness justified | 7 points |
| Correct answer obtained with justification, but none of the numbers implementing it are provided | points are not deducted |
| All possible numbers (112, 224, 336, and 448) found, but it is not justified that there are no other numbers | 3 points |
| Not all cases are considered in the exhaustive search method | 2 points |
| Some of the numbers 112, 224, 336, and 448 are found or the correct answer is given without justification | 1 point |