Mehul 's activity

Algebra Level 3

One day, Mehul wrote the integers 1, 2, 3, ... ,19, 20 on a blackboard. Then he erases any arbitrary numbers $a,b$ and replaces them with a new number of value $a+b-1$ . And he repeats it for another 18 times.

Show that there is only one number remain. What is the value of that number?

The answer is 191.

1 solution

Nihar Mahajan
Jun 8, 2015

Denote $S_n$ as the sum of numbers at $n^{th}$ operation.Since $a,b$ are replaced by $a+b-1$ , the sum gets decreased by 1 every successive operation.So we have:

$S_0=1+2+3+\dots+19+20 = \dfrac{20\times 21}{2}=210 \\ S_1=209 \\ S_2=208 \\ \dots\\ \dots \\ S_{19}=210-19=\boxed{191}$

Moderator note:

Yes, another way to see it is that every time we erase and replace a new number, the sum of these numbers decreases by 1. So if we done it 19 times, the sum is decreased by 19. Thus, we just need to evaluate the value of $(1+2+3+\ldots+20)-19$ .

Amazing! Have you made this question?

Archit Boobna - 6 years ago

No , I took it from a book.

Nihar Mahajan - 6 years ago

Mehul 's activity

The answer is 191.

1 solution

Moderator note:

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