Final answer:
The 2-digit number that is 10 times the sum of its digits, with the tens digit being 2 greater than the units digit, is 20.
Step-by-step explanation:
The given problem is to find a 2-digit number that is 10 times the sum of its digits, with the condition that the tens digit is 2 greater than the units digit. To solve this, let's denote the tens digit as 'x' and the units digit as 'y'. Thus, the number can be written as 10x + y.
According to the problem, we have two equations:
1. 10x + y = 10(x + y) (The number is 10 times the sum of its digits)
2. x = y + 2 (The tens digit is 2 greater than the units digit)
Substituting the second equation into the first equation gives us:
10(y + 2) + y = 10((y + 2) + y)
10y + 20 + y = 10(2y + 2)
11y + 20 = 20y + 20
Solving for 'y', we get:
y = 0.
Using the second equation x = y + 2, we find that:
x = 0 + 2
x = 2.
So the tens digit is 2 and the units digit is 0, hence the number is 20.