In a three digit number, the hundreds digit is twice the units digit. If the digits are reversed, the new number is 396 less than the original number. Find the number.
let the unit digit be x
then the hundred digit is 2x
let the tens digit by y
so the number is 100(2x) + 10y + x
= 201x + 10y
the number reversed is
100(x) + 10y + 2x
= 102x + 10y
201x+10y - (102x+10y) = 396
99x = 396
x = 4
looks like y can be any digit from 0 to 9
the original number could be
814, reverse = 418 , difference = 396
874, reverse = 478, difference = 396
etc
246
Let's solve this step-by-step:
Let's assume the digits of the three-digit number are a, b, and c.
According to the problem, the hundreds digit (a) is twice the units digit (c). We can write this as:
a = 2c (equation 1)
If the digits are reversed, the new number is 396 less than the original number. We can write this as:
100c + 10b + a = 100a + 10b + c - 396
Simplifying this equation, we get:
99c - 99a = -396
Dividing both sides of the equation by 99, we get:
c - a = -4 (equation 2)
Now, we have two equations:
a = 2c (equation 1)
c - a = -4 (equation 2)
We can substitute equation 1 into equation 2 to eliminate 'a':
c - 2c = -4
Simplifying this equation, we get:
-c = -4
Multiplying both sides of the equation by -1, we get:
c = 4
Now, substitute the value of 'c' (4) back into equation 1 to find 'a':
a = 2(4)
Simplifying this equation, we find:
a = 8
So, we have found that 'a' is 8 and 'c' is 4.
Since the digits are reversed, the new number is formed by 'c', 'b', and 'a'.
So, the new number is 48a = 484.
Thus, the three-digit number is 484.
To solve this problem, we need to use algebra and solve for the unknown digits in the number. Let's assume the number is represented by ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.
According to the given information, we know that the hundreds digit is twice the units digit. So we can write the equation: A = 2C.
We are also given that when the digits are reversed, the new number is 396 less than the original number. So the new number is represented by CBA and can be written as: 100C + 10B + A.
Since the new number is 396 less than the original number, we can write the equation: 100C + 10B + A = ABC - 396.
Now, let's substitute the value of A from the first equation into the second equation:
100C + 10B + 2C = (2C)BC - 396.
Simplifying the equation, we get:
102C + 10B = 200C + 10B - 396.
We can now cancel out the 10B terms from both sides of the equation:
102C = 200C - 396.
Next, let's isolate the C term by subtracting 200C from both sides:
102C - 200C = -396.
Simplifying, we get:
-98C = -396.
To solve for C, we divide both sides of the equation by -98:
C = -396 / -98.
Cancelling out the negatives, we get:
C = 4.
Now that we know the units digit, we can substitute this value back into the first equation to find the value of A:
A = 2C = 2 * 4 = 8.
Lastly, since the hundreds digit is 8 and the units digit is 4, we can find the value of the tens digit by subtracting the sum of the other two digits from 9:
B = 9 - (A + C) = 9 - (8 + 4) = 9 - 12 = -3.
But we know that digits cannot be negative, so we made a mistake somewhere along the way. It seems like there is no solution to this problem. Please recheck the provided information and the problem statement to ensure we have all the correct information.