Question:
The integers \(1, 2 ... 40\) are written on the blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number \(a+b-1\) is written. What will be the number left on the board at the end?
The integers \(1, 2 ... 40\) are written on the blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number \(a+b-1\) is written. What will be the number left on the board at the end?
Updated On: Oct 11, 2024
- 781
- 780
- 819
- 820
- 821
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The Correct Option is A
Solution and Explanation
Given that the integers \(1, 2, 3 … 40\) are written on the blackboard. Any two numbers, say a and b, are erased and a new number \(a + b – 1\) is written. This operation is repeated 39 times.
In this manner, the numbers \(1, 2, 3 … 40\) are added and 1 is subtracted 39 times.
So, number left on the blackboard
=\(1, 2, 3 … 40-(39\times1)\)
\(=\frac{40\times41}{2}-39\)
\(=20\times41-39\)
\(=820-39=781\)
So,The correct answer is A.
In this manner, the numbers \(1, 2, 3 … 40\) are added and 1 is subtracted 39 times.
So, number left on the blackboard
=\(1, 2, 3 … 40-(39\times1)\)
\(=\frac{40\times41}{2}-39\)
\(=20\times41-39\)
\(=820-39=781\)
So,The correct answer is A.
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