To solve this problem, we need to find the number of sets \((a, b, c)\) of distinct positive integers that satisfy the conditions \(b - a = c - b\) and \(a + b + c = 12\).
First, let's express the condition \(b - a = c - b\).
From this, we can deduce: \(b - a = c - b \implies b - a = b - b + x \implies c = 2b - a\), where \(x\) is the common difference.
Now substitute this expression for \(c\) into the equation \(a + b + c = 12\):
\(a + b + (2b - a) = 12\)
Simplifying gives:
\(3b = 12 \implies b = 4\)
With \(b = 4\), substitute back to find \(a\) and \(c\):
\(a = 4 - x\\) and \(c = 4 + x\), ensuring distinct positive integers.
Substituting into \(a + b + c = 12\):
\((4 - x) + 4 + (4 + x) = 12\)
Simplifying confirms the equality for all values of \(x\) satisfying integer conditions.
Finding distinct positive integers for them, try \(a = 3, b = 4, c = 5\); \(a = 2, b = 4, c = 6\); and \(a = 1, b = 4, c = 7\):
| Set | a | b | c |
|---|---|---|---|
| 1 | 3 | 4 | 5 |
| 2 | 2 | 4 | 6 |
| 3 | 1 | 4 | 7 |
Thus, there are 3 potential sets \((a, b, c)\) satisfying the conditions.
Therefore, the correct answer is 3.
Important Questions from Number System
Consider the following statements :
1. (25)! + 1 is divisible by 26
2. (6)! + 1 is divisible by 7
Which of the above statements is/are correct ?
If the sum S is divided by 8, what is the remainder ?
If the sum S is divided by 60, what is the remainder ?
What is the Highest Common Factor of 2 3× 3 5and 3 3× 5 2?
Four prime numbers are arranged in ascending order. The product of the first three numbers is 255 and that of the last three is 1955. The largest prime number is:
