The integers 1,2,...... 40 are written on a 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 is the number left on the board at the end? (CAT 2007)

820

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821

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781

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819

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780

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Solution

The correct option is C 781
option (c) 781

Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be $S_{40} - 39 = 781.$

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The integers 1,2,...... 40 are written on a 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 is the number left on the board at the end? (CAT 2007)

The correct option is C 781option (c) 781 Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be S40−39=781.

The correct option is C 781
option (c) 781

Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be $S_{40} - 39 = 781.$