a two-digit counting number has a value that is 13 greater than 3 times the sum of the digits. If the units digit is 1 greater than the tens digit, what is the number
To find the two-digit number, we'll consider the given conditions:
1. The number is 13 greater than 3 times the sum of its digits.
2. The units digit is 1 greater than the tens digit.
Let's start by assigning variables to the digits of the number. Let's call the tens digit "x" and the units digit "y".
Since the question states that the units digit is 1 greater than the tens digit, we can write the equation: y = x + 1.
Now, we can represent the value of the two-digit number using the equation: 10x + y.
According to the first condition, the number should be 13 greater than 3 times the sum of its digits. The sum of the digits is x + y, so we can write the equation: 10x + y = 3(x + y) + 13.
Now, let's solve this equation to find the values of x and y:
10x + y = 3x + 3y + 13
Simplifying the equation:
10x + y = 3x + 3y + 13
10x - 3x + y - 3y = 13
7x - 2y = 13
Now, we need to find integer values for x and y that satisfy this equation.
Let's substitute different values of y and solve for x. We'll use trial and error by plugging in values that satisfy the equation, one by one, until we find a solution.
If we substitute y = x + 1 into the equation, it becomes:
7x - 2(x + 1) = 13
7x - 2x - 2 = 13
5x = 15
x = 3
Now we have the value for x, which is 3. We can substitute this value back into the equation for y:
y = x + 1
y = 3 + 1
y = 4
So, the tens digit (x) is 3, and the units digit (y) is 4. Therefore, the two-digit number is 34.