Question

The positive integers 1,2,3,4,5....,20 are written on paper.
Here certain operations are performed 19 times repeatedly based on the following conditions.
In each step, any two numbers (X and Y) currently on the paper removed, and a new number X+Y-
1 will be written. What will be the number left on paper at the end of all these operations?​

Answer 1

Given :- The positive integers 1,2,3,4,5, ____ 20 are written on paper. Here certain operations are performed 19 times repeatedly based on the following conditions. In each step, any two numbers (X and Y) currently on the paper removed, and a new number X + Y- 1 will be written.

To Find :- What will be the number left on paper at the end of all these operations ?

Answer :-

1, 2, 3, 4, 5, 6, 7, 8, 9, _____________ 20 .

Let we take out (3 , 4) pair and then adds (3 + 4 - 1) = 6 back .

if we take (5 , 8) pair we will add (5 + 8 - 1) = 12 back .

conclusion :-

• sum is reducing by (-1) each time .
• In 19 times operations sum is reduced by 19 .
• In 19 times operations we will add back all the numbers .

therefore,

→ The number left on the paper = sum of all integers - 19

→ Required Left number = (1 + 2 + 3 + ____ 20) - 19

→ Required Left number = (20 * 21)/2 - 19

→ Required Left number = 210 - 19

→ Required Left number = 191 (Ans.)

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