Is it possible to get 521?

Algebra Level 3

Suppose the numbers 1, 2, 3, …, 100 are written on a paper. At each step, we select any two numbers and replace them with either their sum or their difference. Is it possible to have 521 in the end as the only remaining number?

Yes, there is a unique way to do it. Yes, there are infinitely many ways to do it. Yes, there are finitely many and more one way to do it. No, it is not possible.

2 solutions

Chan Lye Lee
May 24, 2020

Firstly, we have the numbers 1, 2, …, 100 written on paper. The sum of these numbers is 5050, call it $S$ . We select two numbers, says, $a$ and $b$ and replace them with either their sum $a + b$ or their difference $a - b$ .

If we replace these two numbers by their sum, then the sum of all numbers on paper remains the same.

If we replace these two numbers by their difference, then the sum of all numbers on paper is $S - (a + b) + (a - b) = S - 2b$ , still an even number.

So, for each step, the sum of the numbers on paper still remains as an even number.

Hence, it is $\textcolor{#D61F06} {\text{NOT POSSIBLE}}$ to have 521 in the end as the only remaining number.

Nice! good solution

Mahdi Raza - 1 year ago

I got it wrong

Spriha Basir - 1 year ago

(100+99)+(97+98)+(50+51)+(8+9)+(5+4)=521. I’m not understanding the question may be. 🙂

Debasish Roy - 1 year ago

Or you could easily say. 50even number sum or diff is even. And other 50odd numbers sum and diff is also even. So final result can’t be 521 or nay odd number.

Debasish Roy - 1 year ago

Pop Wong
May 29, 2020

At each step, we have below possible outcome

Picked Even	Picked Odd	Return Number	No. of even number changed	No. of odd number changed
1	1	Odd	-1	0
2	0	Even	-1	0
0	2	Even	+1	-2

Initial, we have 50 even numbers and 50 odd numbers.

The changes of no. of odd number is always even. Thus, there is no way to have 1 even and 1 odd for the last step to return an odd number (521).

Is it possible to get 521?

2 solutions

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