# Tag Info

19

19

First Way Second Way Third Way

17

Five $8$s

15

Hmm, let's see.

12

The number is because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively: Adding the first two gives Also from the first equation, So the number must be

10

To calculate the answer, all we need are some bounds on the order of magnitude of $x$ and $y$. (In particular, we won't be looking at the answer options, nor will we need to resort to any kind of reasoning "from the assumed uniqueness of the answer", which is questionable at best.) Here's all it takes: The starting number has 1998 digits. Therefore ...

9

9

A solution is Proof:

9

Given OP's deleted answer it looks like concatenation is allowed and not all numbers must be used. Here are the numbers I was able to do (26/40)

9

All possible ways to start using one operation (two instances of $x$) are: $0,1,2x,x^2,x^x,11x$. I've built on top of $x^x$ to get a four digit number:

9

Assuming we can use the floor function (denoted here with brackets $\lfloor\cdot\rfloor$): So we can make Then we could simply do: and Edited to add: If floor function is allowed we can make each of the following using a single four: and the following using two fours: then we could do:

8

I have found a number can be achieved by using the same number of another digit, kind of based on OP's example Stretching this a bit, the best I've managed so far for the bonus is

8

1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 5 + 1 7 = 5 + 2 8 = 5 + 3 9 = 5 + 4 10 = 5 + 4 + 1 11 = 5 + 4 + 2 12 = 5 + 4 + 3 13 = 5 + 4 + 3 + 1 14 = 5 + 4 + 3 + 2 15 = 5 + 4 + 3 + 2 + 1 16 = 4 ⋅ (1 + 3) 17 = (5 ⋅ 4) - 3 18 = (5 ⋅ 4) - 2 19 = (5 ⋅ 4) - 1 20 = (5 ⋅ 4) 21 = (5 ⋅ 4) + 1 22 = (5 ⋅ 4) + 2 23 = (5 ⋅ 4) + 3 24 = (...

8

A few more possibilities:

7

Without allowing any tricks, especially not: Then here are all the solutions: Justification: import itertools import operator d1 = lambda x,y: 10*x + y d2 = lambda x,y: x + y/10 d3 = lambda x,y: x + y/100 for f,g in itertools.permutations([operator.add, operator.sub, operator.mul, operator.truediv, operator.floordiv, operator.pow, d1,d2,d3, operator.xor]*2,...

7

Under a suitable interpretation of "you can also use decimals", another answer is

7

I tried a computer program to solve this problem. I got 425 different expressions giving 2016. So I added restrictions. I used only addition and multiplication. I stil get 26 expressions. You'll find them below. To answer the question about "no brackets" and "not ending in a product":

7

6

6

Here is seven 8s

5

Interpreting the constraints as purely typographic: or (also rather mathsy): or, bending the rules a tiny bit (there are many kinds of parentheses):

5

This works: or or or

5

I think that the constant hiding behind all of this is Notice first that

5

Complete answer(thanks to Ross Millikan and BAWS):

5

Assuming the available numbers are because of Note that all numbers not mentioned in hexomino's answer use concatenation of results (denoted here by $\#$):

5

Here's six 8s:

5

"a guy" has covered the initial question here is the bonus We can even go a bit further, without much difficulty Here are a few more

4

Here is 19

4

One option:

4

Here are a few: And very many more. There are $2,762$ solutions in this form (without using parentheses).

Only top voted, non community-wiki answers of a minimum length are eligible