Find a three-digit positive integers such that the sum of all three digit is i4, the tens digit is two more than ones and if the digit is reversed, the number is unchanged.
To find a three-digit positive integer that satisfies the given conditions, let's break down the problem step by step.
Condition 1: The sum of all three digits is 14.
Let's denote the three digits as hundreds (H), tens (T), and ones (O). According to the condition, we have the equation: H + T + O = 14.
Condition 2: The tens digit is two more than the ones digit.
In mathematical terms, this can be expressed as: T = O + 2.
Condition 3: The number is unchanged when its digits are reversed.
The original number can be expressed as 100 * H + 10 * T + O, and the number with reversed digits is 100 * O + 10 * T + H. According to the condition, these two numbers are the same: 100 * H + 10 * T + O = 100 * O + 10 * T + H.
Now, let's solve these equations simultaneously:
From Condition 2, we have T = O + 2. Substituting this into the other equations, we get:
H + (O + 2) + O = 14
2H + 2O + 2 = 14
2H + 2O = 12
H + O = 6
From Condition 3, we have 100H + 10T + O = 100O + 10T + H. Simplifying this equation, we get:
99H = 99O
H = O
Since H = O, we can substitute this back into the equation H + O = 6:
H + H = 6
2H = 6
H = 3
Therefore, the hundreds digit (H) is 3. Substituting this back into T = O + 2, we get:
T = 3 + 2
T = 5
Now we have H = 3, T = 5, and O = 3.
Therefore, the three-digit positive integer that satisfies all the given conditions is 353.
let the unit be be x
then the tens digit is x+2
let the hundreds digit by y
x + x+2 + y = 14
2x + y = 12
number: 100y + 10(x+2) + x
= 100y + 11x + 20
number reversed:
100x + 10(x+2) + y
= 110x + y + 20
but they are to be equal
100y + 11x + 20 = 110x + y + 20
99y = 99x
y = x
back into 2x+y = 12
2x+x=12
x = 4, then y = 4
the number is 454