Four Identical Digits to Target of 0-10; Almost Any Integer Has a 3 In It

How would you arrange all four of the same digits (such as 1 1 1 1 or 2 2 2 2 or 3 3 3 3 ... 9 9 9 9) so they will equal the numbers 0-10? 

Like Target Number and 24, you need to use all four numbers and you can use all four binary operations (+, -, x, and/or /) as well as exponents (b to the power of m or b^m means b times itself m times: 2^3 = 2x2x2 = 8) and factorials (n! means n(n-1)(n-2)...(1): 4! = 4x3x2x1 = 24).

Show your method by using the operations and parentheses to make clear what you did first, second, and your final calculation. 

Will any group of four digits have possible solutions for all of the numbers 0-10?

Remember your properties of zero, one, and exponents:

8 + 0 = 8 8 x 1 = 8 8^0  = 1

8 -- 0 = 8 8 / 1 = 8 8^1  = 8

0 x 8 = 0 8 / 8 = 1 8^2  = 8 x 8 = 64

0 / 8 = 0 8^1  = 8 8^3  = 8 x 8 x 8 = 512

8 / 0 = O Undefined

Please remember never to multiply your base (in the examples above, 8 is the base) by your exponent (the 0,1,2,3 above)

NOTE: I have failed to find solutions for all numbers for each set of four numbers with the exception of one set of four numbers. Please help me find solutions for some of my failed attempts.

During the class a few students came up with solutions to these mysteries using factorials. 3 factorial or 3!=3x2x1=6 or 4!=4x3x2x1=24.

Almost any integer has a 3 in it.

What a crazy statement! No, actually, we proved it with the 5-6th graders.

This is an engagingly sounding sentence that begs some clarification. 

Of course, no one would claim that every number has a digit 3 in it. But what is almost? There are infinitely many integers, and, among them, there are obviously infinitely many integers that have at least one digit equal to 3. For example, we have an infinite sequence 3, 33, 333, 3333, and so on. As easily, we can produce an example of an infinite sequence of integers that do not have a digit 3: 1, 11, 111, 1111, and so on.

Still, the claim is not altogether frivolous. Let's consider in turn integers less than 10, 100, 1000, etc. And, for every such group of integers, let's count how many of them have at least one digit equal to 3. For convenience, let's start with 0.

Among 10 numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 , 9 written with a single digit, only 1 has digit 3 - the number 3 itself. Among 100 numbers 0, 1, ... ,9, 10, 11, ..., 98, 99 written with 1 or 2 digits, the following include digit 3: 3, 13, 23, 30, 31, ..., 38, 39, 43, 53, ..., 93. 19 numbers in all. 

When we look at the numbers below 1000, it is easy to notice that there are 10 groups of 100 consecutive integers each. The first, as we just found contains 19 integers with a digit 3. And the same is true of 8 other groups of 100 integers: those that start with 1, 2, 4, 5, 6,7 ,8 , or 9. All 100 integers that start with 3 obviously have at least one 3. So, below 1000, there are 9·19 + 100 integers with a digit 3 in them.

I was able to work this proof out to 38 iterations and reached 98.1% of all integers from 1-10^38 have a 3 in it. If I had gone another 20 iterations, my conjecture is that I would have reached 99%; then another 40 iterations and we reach 99.9% of integers between 1-10^100 (googol).