You need to be exact

Number Theory Level 2

Can you use exactly $(486^{56} - 622^{12})$ 1's and just 2 mathematical operators/symbols to make a total of $100$ ?

Details and assumptions:
- Concatenating digits (ofcourse) is allowed.
- No other digit must be used.
- Two symbols of the same type will count as two different entities.

No; but it is possible with 4 operators. No; but it is possible with 5 operators. Yes No.

3 solutions

Tran Hieu
Mar 30, 2016

Because the number of 1s is even and large than 4, we could have the following (A11-11)/A = 100, in which A = 11...1, the number of 1s in A is half of the big number minus 2.

Kay Xspre
Mar 29, 2016

Wolfram Alpha told me that number is something approximately $2.828\times10^{150}$ , and is even number. Let $\zeta = 486^{56}-622^{12}$ , We can just then make it $\sum_{n=1}^{\zeta-100}(-1)^{n+1}+\underbrace{(1+1+1+1+\dots+1)}_{100\:1's} = 100$

Although a good one but you have used around 100 mathematical symbols. The question clearly states that you can use just 2 mathematical symbols. Did you misunderstand it for 2 types of symbols? Also, thanks for making me realise that some edits are needed so that the question is more clear.

Yatin Khanna - 5 years, 2 months ago

Yatin Khanna
Mar 29, 2016

OF COURSE, IT IS POSSIBLE. THE LARGE NUMBER IS JUST FOR CONFUSING.
AS IS CLEARLY VISIBLE THAT THE RESULTANT NUMBER IS EVEN; THEREFORE WE DIVIDE THE 1's IN 2 GROUPS WITH EACH CONTAINING HALF THE NUMBER MENTIONED.
THEN;
1111........................11111/1111.......................11111% = 100

You need to be exact

3 solutions

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