If the order of three digits is reversed, the sum of the three digits and the original three digits is 1171, and the three digits can be calculated

If the order of three digits is reversed, the sum of the three digits and the original three digits is 1171, and the three digits can be calculated

Let X be the number of ten, then the number of one is 3x-2, and the number of hundred is x + 1, so 100 (x + 1) + 10x + (3x-2) + 100 (3x-2) + 10x + (x + 1) = 1171. The solution is: x = 3. A: the original three digit number is 437

If the order of three digits is reversed, the sum of the three digits and the original three digits is 1171, and the three digits can be calculated

Let X be the number of ten, then the number of one is 3x-2, and the number of hundred is x + 1, so 100 (x + 1) + 10x + (3x-2) + 100 (3x-2) + 10x + (x + 1) = 1171. The solution is: x = 3. A: the original three digit number is 437

For a three digit number, the number on the hundred digit is twice as large as the number on the one digit, and the number on the ten digit number is 0. If the order of the three digits is reversed, the three digit number will be obtained
The sum of the number and the original number is 1414, ask for the original number? (more detailed) elder brother and elder sister help!

If the single digit is x, the hundred digit is 2x + 1,
[100(2x+1)+x]+[100x+(2x+1)]=1414
Solution: x = 13 / 3, this is impossible, X must be a positive integer, you check the problem again, wrong