For a two digit number, the number on the one digit is three times that on the ten digit number. If the number on the tens of digits is exchanged with the number on the one digit, the two digit number obtained is 54 times larger than the original number. How to find the original two digit number?

For a two digit number, the number on the one digit is three times that on the ten digit number. If the number on the tens of digits is exchanged with the number on the one digit, the two digit number obtained is 54 times larger than the original number. How to find the original two digit number?

Let the original ten digit number be x and the individual digit number be 3x. From the meaning of the question, we can get: (3x × 10 + x) - (10x + 3x) = 54. The solution is: x = 3, the original number is 3 × 10 + 9 = 39. A: the original two digit number is 39

A two digit number, the one digit number is three times of the ten digit number, if the one digit number and the ten digit number are exchanged, then the new number is 54 times larger than the original number, and the equivalent relationship is obtained

Let X be the number in ten, then 3x be the number in one
The original number is 10x + 3x
The new number is 10 * 3x + X
If the new number is 54 larger than the original number, the equivalent relation is: new number = original number + 54
10*3x+x=10x+3x+54
31x=13x+54
18x=54
x=3
3*3=9
The original number is 3 * 10 + 9 = 39

A two digit number is 5 more than a ten digit number,
Then the sum of the new two digit number and the original two digit number is 121. Find the original two digit number

The number of one digit is 5 more than that of ten digit, that is, after the transposition of one digit and ten digit, it is less than the original number: 5 × 9 = 45,
（121+45）÷2=83
That is, the original number is 83