There is a two digit number, the number on the tenth digit is 7 times larger than the number on the single digit, and the two digit number is 9 times the sum of the number on the single digit and the number on the tenth digit

There is a two digit number, the number on the tenth digit is 7 times larger than the number on the single digit, and the two digit number is 9 times the sum of the number on the single digit and the number on the tenth digit

Let X be the number of ten digits, then there is: 10x + X-7 = 9 (x + X-7), the solution is: x = 8, then X-7 = 1. That is, the two digit number is 81

For a two digit number, its ten digit number plus seven times of the one digit number is still equal to the two digit number. Such a two digit number has ()
A. One B. two C. three D. four

Let ten digits be x and one digit be y. according to the meaning of the question, we get that x + 7Y = 10x + y, that is, 3x = 2Y, while 1 ≤ x ≤ 9, 0 ≤ y ≤ 9, and X and y are all integers. According to the conditions, the values of X and y can be divided into three groups: 2, 3; 4, 6; 6 and 9, that is, there are three such two digits: 23, 46 and 69

If there is a two digit number, the number on the tenth digit is 5 times larger than the number on the single digit, and the number is 8 times the sum of the two digits, find the two numbers

Let X be the number of tens and y be the number of ones
X-Y=5
10X+Y=8(X+Y)
X=7 Y=2
So the number is 72