i am a three digit decimal whose digit in the tenths place is 4 times the digit in the hundredths place and 4 more than the digit in the thousandths place. who am i?
56.4
The digit in the tenths place must be 4.
Let's break down the information given:
- The digit in the tenths place is 4 times the digit in the hundredths place.
- The digit in the tenths place is also 4 more than the digit in the thousandths place.
- We are looking for a three-digit decimal number.
Let's solve this step by step:
Step 1: Let's represent the digit in the hundredths place as "x."
Step 2: The digit in the tenths place is 4 times "x," so it is 4x.
Step 3: The digit in the tenths place is also 4 more than the digit in the thousandths place, so it is "4 more than x," which can be represented as x + 4.
So, the decimal number can be written in the form x + 4x + (x + 4).
Adding the terms together, we get:
x + 4x + (x + 4) = 6x + x + 4
= 7x + 4
Now, we know that the decimal number is a three-digit number, so the digit in the hundredths place (x) cannot be zero.
Let's try some values for x and see if we can find a solution:
Let x = 1:
7(1) + 4 = 11 → Not a three-digit number.
Let x = 2:
7(2) + 4 = 18 → Not a three-digit number.
Let x = 3:
7(3) + 4 = 25 → Not a three-digit number.
Let x = 4:
7(4) + 4 = 32 → A three-digit number!
Therefore, the solution is 432.
To find the three-digit decimal number that meets the given conditions, we'll break it down into the three place values: the hundredths place, the tenths place, and the thousandths place.
Let's denote the digit in the hundredths place as A, the digit in the tenths place as B, and the digit in the thousandths place as C.
According to the given conditions:
The digit in the tenths place (B) is 4 times the digit in the hundredths place (A).
B = 4A
The digit in the tenths place (B) is also 4 more than the digit in the thousandths place (C).
B = C + 4
Using this information, we can solve for A, B, and C.
Since B = 4A and B = C + 4, we can substitute B from the first equation into the second equation:
4A = C + 4
Rearranging the equation, we have:
C = 4A - 4
Now, let's consider the possible values of A.
Since A, B, and C are digits, they can only take values from 0 to 9. However, the condition B = 4A indicates that B is a multiple of 4. Therefore, A must be a multiple of 1/4, which limits our options further.
The possible values for A are: 1/4, 2/4 (which is equal to 1/2), and 3/4.
Let's substitute these values and determine the corresponding values for B and C:
1) If A = 1/4:
B = 4A = 4(1/4) = 1
C = B - 4 = 1 - 4 = -3 (Not possible as C cannot be negative)
2) If A = 1/2:
B = 4A = 4(1/2) = 2
C = B - 4 = 2 - 4 = -2 (Not possible as C cannot be negative)
3) If A = 3/4:
B = 4A = 4(3/4) = 3
C = B - 4 = 3 - 4 = -1 (Not possible as C cannot be negative)
Since none of the values for A yield valid values for B and C, we conclude that there is no three-digit decimal number that satisfies the given conditions.