Find the maximum three-digit number that is a multiple of 5.10 and 14. what digit can be written instead

Find the maximum three-digit number that is a multiple of 5.10 and 14. what digit can be written instead of an asterisk in the number 681 * so that it is divisible by 9, divisible by 5, divisible by 6. Find the volume of the rectangular parallel of the pipid, the dimensions of which are 4m, 3m and 5 dm.

1) Note that 1008 is a multiple of 14. Subtract 14 from it until we get a round number. It will be the maximum three-digit number, which is simultaneously divisible by 5, 10 and 14:
1008 – 14 = 994 – not round;
994 – 14 = 980 – round.
Answer: 980.
2) a) For a number to be divisible by 9, the sum of its digits must be divisible by 9.
6 + 8 + 1 = 15.
The fourth digit can be from 0 to 9, and the only option is when the sum of the digits is divisible by 9: the fourth digit is 3 (the sum is 18).
Answer: 6813
b) For a number to be divisible by 5, there must be 5 or zero digits at its end.
Answer: 6810 or 6815.
c) For a number to be divisible by 6, it must be simultaneously divisible by 2 (that is, be even) and 3 (that is, the sum of its digits must be divisible by 3). The following options are possible: 6810, 6816.
Answer: 6810, 6816.
3) Let’s convert all measurements to decimeters and multiply:
40 * 30 * 5 = 6000 cubic decimeters.
Answer: 6000 cubic decimeters.