Evaluate and rigurously complete in a more justifi...

Evaluate and rigurously complete in a more justifiable anc correct manneer what's lacking in the third problem: My Mathematical Working-Out on the Five S’s of Nonnegative Integers
Simple
Call a nonnegative integer simple if its only nonzero digits are ones.
(a) Which two-digit integers can be expressed as the ratio of two simple numbers?
Before diving into it, I want to point out my inclusion of 1 as an actual simple number. Although the problem states that a simple number adopts only “ones” as its nonzero digits, it does not restrict a single 1 from being considered a simple number itself.

So, I'll be representing the two-digit integers as T and the ratio of two simple numbers as N/D. Particularly, we need to find an N/D that equals T. On the one hand, T is, by definition, a particular number within 10 to 99 because otherwise it would not be constrained to the two-digit rule. On the other hand, N/D must be a positive integer and within such range. Hence, the numerator needs to be greater enough than the denominator to ensure a positive integer of at least 10. And, the result must be neither greater nor fewer than two digits to ensure the range boundaries.

Therefore, we shall select a suitable N and D. But which exactly? To crack this, we need to picture what N and D largely look like: They are any positive integer whose nonzero digits are 1s only. These include: 1, 10, 11, 100, 101 and so forth. So, by adopting these alternative values of N and D we can discard certain pairs based on weather they give a remainder different to 0. That is, a decimal quotient which our T can not take the form of because it must be an integer. Illustratively, 111/11 is discarded given that the remainder is 1.

However, the ratios that do yield an integer are 11/1 and 10/1 (or equivalently = 100/10) because 1 doesn't change the already 2-digit numerator value in both cases. Evidently, 11/1 results in 11 and 10/1 in 10. Additionally, 1001/11 constitutes another valid ratio because the numerator and denominator produce a perfectly coordinated carry-over. To explain it, when we multiply 11 to its most pertinent digit during the first division step: [11*9] it yields a remainder of 1 [100-99]. Then, by combining it with the subsequent 1 in the hundreds place of the dividend [11] we turn out dividing the divisor by itself [11/11]. Therefore, giving a final remainder of zero and quotient T of 91.
My final answers for T are 11, 10 and 91.

Symmetric
(a) Suppose S is simple, and R is the number obtained by reversing the order of its digits. If 901·S is simple, 901·R must also be simple. Why?
Let S be a simple number [One that all its nonzero digits are 1s] and R the number obtained by reversing the digits of S.
IF 901 multiplied by S is simple WHY would 901 multiplied by R also be simple?

Short answer: For 901 × S to be simple S can only take the form of 111 or any longer string of consecutive ones (1111, 11111, etc). Consequently, since all S digits are 1s reversing it for R does not change the nature of 901 × R as simple.

Long answer: For 901S to be simple we need to figure out how S could make it actually simple. Here, I appreciated 901 as a number itself: And, noticed 901S can yield a simple number (0s and 1s) only if S manages to get rid of the 9 by adding a 1 to it and turning it into a 10 during the division process. That way, it gets the simple number nature.

Having that on mind, we may select those S values which don't and do make this happen. For example, 1 and 11 don't have exact carries for eliminating 9 and turn it into 10. Hence, I discarded them. However, through examination I realised that this exact-carry system could only be achieved if I used 111, 1111 .... and any of its consecutive strings of only 1s .... because they were the closest to precisely: Maintain the 1 and turn the 9 into a 10 when multiplying by 901. In this sense, although R would adopt the S digits in the opposite order it'd remain the same because of the 1-only composition.

My final answer: The values of the products do not change from their simple nature because reversing all the digits of S to obtain R does not change its value. Thus, 901×R remains as simple.

Special
Call a nonnegative integer special if it can replace 901 in the observation from part (b).
(a) Find a nonnegative integer that isn’t special.
In block 2 we found how 901·S could be simple while simultaneously having 901·R as simple. Now, we need to find a number that replaces 901 – which we will call it special represented by E – such that when multiplied with S [E.S] remains simple but its reversal [E*R] no.

To tackle this, I created a big question: Which properties must E have so that when multiplied by a simple number the resulting product remains simple, while the product with the digit-reversed number does not?

At this stage, I realised that if a symmetric simple number multiplies with any E to produce a simple product, then its reversed form would necessarily produce the same result, due to the invariance of symmetric numbers under digit reversal. That's why, I discarded 11, 111, 1111 and other analogous cases. Instead, we now know that our simple numbers S and R should include 0s, such as: 10, 101, 1001, 1100, etc.

Additionally, at this point I found something interesting evaluating this stance: "ES must be simple while its reversing ER not" However, if ES was simple in the first place ER would necessarily be forced to be simple as well, since digit reversal would still yield a different but nevertheless simple number.

For this reason, it is necessary to identify a number E such that, when multiplied by any of the previously identified admissible simple numbers, the resulting product in ES fails to be simple and thus fails too in ER.

Mathematically, these are those ones:

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