3. The number 2458710411 was written 98 times in a row, resulting in a 980-digit number. From this number, it is required to erase 4 digits. What is the number of ways this can be done so that the newly obtained 976-digit number is divisible by 6?
Solution. If a number is divisible by 6, then it is divisible by 3 and 2. A number is divisible by 2 if and only if its last digit is even. The 980-digit number given in the condition ends with 2458710411, i.e., it has the form ...2458710411. For the number to be divisible by 2, it is necessary to strike out the last two ones so that the number ends with an even digit. The resulting number will have the form 2458710411...245871041124587104. Now it is necessary for this number to be divisible by 3. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The sum of the digits of the resulting number is:

$(2+4+5+8+7+1+0+4+1+1) * 97 + (2+4+5+8+7+1+0+4) = 33 * 97 + 31 = 3232$.

The remainder of the division of the number 3232 by 3 is 1. This means that the numbers must be struck out in such a way that their sum is divisible by 3 and leaves a remainder of 1. Then, upon subtraction, a number will be obtained whose sum of digits will be exactly divisible by 3. The maximum sum that can be obtained by striking out digits is $16 (8+8)$. Therefore, it is necessary to consider all numbers from 0 to 16 that are divisible by 3 and leave a remainder of 1, and all possible combinations of their formation. These numbers are: $1, 4, 7, 10, 13, 16$. The combinations by which they can be formed are as follows:

$1 (0+1); 4 (0+4, 2+2); 7 (2+5, 7+0); 10 (5+5, 2+8); 13 (8+5); 16 (8+8)$.

It remains to calculate the number of ways in which each combination can be obtained. The first combination can be obtained by striking out one 1 and one 0. There are $97 * 3 + 1 = 292$ ones in our number, and 98 zeros. Therefore, the number of ways in which this can be done is $(97 * 3 + 1) * 98 = 28616$. For subsequent combinations, we get:

$(0+4): 98 * 98 = 9604$

$(2+2): 98 * 97 / 2 = 4753$

$(2+5): 98 * 98 = 9604$

$(7+0): 98 * 98 = 9604$

$(5+5): 98 * 97 / 2 = 4753$

$(2+8): 98 * 98 = 9604$

$(5+8): 98 * 98 = 9604$

$(8+8): 98 * 97 / 2 = 4753$.

Summing all the obtained combinations, we get the number of ways 90895. However, it should be noted that one of the proposed ways is not suitable. If we strike out the last digits 0 and 4, then our number will end with 1, i.e., it will not be divisible by 2, and therefore, it will not be divisible by 6. Therefore, the number of combinations is $90895 - 1 = 90894$ (ways).

Answer: 90894 ways.

| Criteria for evaluating the performance of the task | Points |
| :--- | :---: |
| A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are correctly performed. The correct answer is obtained. | **10** |
| A correct sequence of all steps of the solution is provided. Gaps in the justification of the choice of number combinations (for example, one extra solution (0+4) is not taken into account, pairs of digits are not divided by 2, etc.) or a computational error or typo that does not affect the further course of the solution are allowed. As a result of this error or typo, an incorrect answer may be obtained. | **6** |
| The task is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, part of the combinations is taken into account and calculated correctly); |
| - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, most combinations are not taken into account or extra combinations are taken into account, combinations are calculated incorrectly, cases of striking out the same digit (2+2, |
| 5+5, 8+8) are not taken into account, etc.); |

All cases of solutions that do not meet the above criteria for scoring 3 and 6 points.