There is a three-digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?
Question
There is a three-digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?
Solution
To solve this problem, we need to set up a few equations based on the information given.
Let's denote the three digits as follows:
- The first digit is X
- The second digit is Y
- The third digit is Z
From the problem, we know that:
- Y = 4Z (The second digit is four times as big as the third digit)
- X = Y - 3 (The first digit is three less than the second digit)
Now, we need to find the values of X, Y, and Z that satisfy these equations.
Since Y is four times Z, Y can be 4, 8. But since X is three less than Y, Y must be 4 (because if Y was 8, X would be 5, but there is no digit 5 that is three less than 8). So, Y = 4.
Substituting Y = 4 into the equations, we get:
- Z = 1 (from Y = 4Z)
- X = 1 (from X = Y - 3)
So, the three-digit number is 141.
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