I am a 3-digit number. My first digit is thrice my last digit. The sum of my last two digit is one less than my last digit.
6 -1 and 2 ?
612
Aashifa
To find the solution to the problem, we need to follow a step-by-step approach.
Let's assign variables to the unknowns in the problem. We'll use the variable "x" to represent the hundreds digit, "y" to represent the tens digit, and "z" to represent the ones digit.
From the given information, we can form the following equations:
1. "I am a 3-digit number": This tells us that x cannot be zero.
2. "My first digit is thrice my last digit": This can be expressed as "x = 3z".
3. "The sum of my last two digits is one less than my last digit": This can be expressed as "y + z = z - 1".
Now, let's solve the equations:
From equation 2, we have x = 3z.
Substituting this value of x in equation 3, we get y + z = z - 1.
Rearranging this equation, we have y = -1.
Now, we can substitute the value of y in equation 3 to get -1 + z = z - 1.
Simplifying this equation, we find -1 = -1, which is true but does not provide any additional information.
Now, let's find the value of z. Since we only have one equation involving z (equation 2), we can use it to find z.
From equation 2, we have x = 3z, which implies z = x/3.
Since z represents the ones digit, it must be an integer. Therefore, x must be a multiple of 3.
Now, let's consider the possible values of x:
If x = 3, then z = 3/3 = 1, and the corresponding number is 311.
If x = 6, then z = 6/3 = 2, and the corresponding number is 622.
If x = 9, then z = 9/3 = 3, and the corresponding number is 933.
Therefore, there are three possible numbers: 311, 622, and 933, which satisfy the given conditions.