7.2. Find the largest natural number in which each digit, starting from the third, is equal to the sum of all the previous digits of the number.
Answer: 101248.

Solution: Let the first digit of the number be $a$, the second digit be $b$. Then the third digit is $(a+b)$, the fourth digit is $(2a+2b)$, the fifth digit is $(4a+4b)$, and the sixth digit is $(8a+8b)$. There cannot be a seventh digit, because if it exists, it would be equal to $(16a+16b)$, but this expression can only be a digit if $a=b=0$. Therefore, the desired number is a six-digit number. Now we need to find the maximum possible value of $a$ such that the sixth digit remains a "digit," i.e., the inequality $8a+8b<10$ is satisfied. This is possible when $a=1, b=0$, so the desired number is 101248.

Criteria: Correct answer - 4 points (may be without explanation). Correct reasoning that the desired number is a six-digit number, but the example is not constructed - 3 points. 7 points are given if there is a proof that numbers greater than 101248 cannot exist (a correct method of constructing the example can be considered as a proof).