Let the integer be \(n\) with last digit \(d\in\{0,\dots,9\}\). The condition says \(d=\tfrac{n}{2016}\), hence \(n=2016d\). Consistency of the last digit requires \(2016d\equiv d\pmod{10}\), so \(2015d\equiv 0\pmod{10}\Rightarrow 5d\equiv 0\pmod{10}\), giving even \(d\): \(d\in\{0,2,4,6,8\}\). Since \(n>0\), \(d\neq 0\). Therefore \( n\in\{4032,8064,12096,16128\}. \)

Let the number be \(N\). After prefixing 1 we have \(100+N\). Appending 8 gives \[ M=10(100+N)+8=1008+10N. \] Since \(M=28N\), \[ 1008+10N=28N \Rightarrow 18N=1008 \Rightarrow N=56. \] Check: \(56\cdot 28=1568\).

Let \(N=100a+10b+c\). After the digit increases, \( 100(a+3)+10(b+2)+(c+1)=4N. \) Hence \( 100a+10b+c+321=400a+40b+4c \Rightarrow 321=300a+30b+3c. \) Dividing by \(3\) gives \( 107=100a+10b+c, \) so \(N=107\). Check: the new number is \(428\), and \(4\cdot107=428\).

\[ 2^{20}\cdot 5^{17}=2^{3}\cdot 10^{17}=8\cdot 10^{17}. \] It is \(8\) followed by \(17\) zeros, so the sum of digits is \(8\).

Let \(P(n)\) denote the product of all non-zero digits of \(n\). Take \(n=299\). Then \(P(299)=2\cdot 9\cdot 9=162\), while \(P(300)=3\) (zeros are ignored). Hence \(\dfrac{P(299)}{P(300)}=\dfrac{162}{3}=54\). Therefore the answer is “yes”; an example is \(299\) and \(300\).

Assume there exists a natural number \(n = n_k n_{k-1} \ldots n_1\). In this case, the number \(1n2\) must be divisible by 12, which means it must also be divisible by 4. According to the divisibility rule for 4, the number \(n12\) must also be divisible by 4. Similarly, if \(2n4\) is divisible by 24, it follows that \(n14\) must be divisible by 4 as well. Consequently, the difference \(2 = n14 - n12\) must be divisible by 4, leading to a contradiction.

Next, let's examine which values of \(n\) are appropriate. For \(n = 1, 2, 3, 4, 5\), only \(n = 3\) meets the condition. We will now demonstrate that for \(n \geq 6\), the sum of the digits of the number \(5n\) is less than \(2n\). The number \(5n\) can have at most \(n\) digits, so the sum of its digits cannot exceed \(9n\). In contrast, \(2n\) will be at least \(9n\). This assertion holds true for \(n = 6\), and as \(n\) increases by one, the right side of the inequality rises by 9, while the left side increases by at least 64.

Assume it is possible to divide all natural numbers from 1 to \(k\) into two groups, such that the numbers in each group can be arranged consecutively to produce two identical sequences. It is evident that \(k\) must be greater than 10, since the digits from 1 to 9 do not allow for any repetitions.

Next, let us identify the largest power of ten, represented as \(10^n\), that does not exceed \(k\). The digits of the number \(10^n\) can fully fit into one of the resulting sequences. However, this means that the same sequence of one followed by \(n\) zeros must also appear in the second sequence.

This sequence of digits could not have originated from the combination of two or more natural numbers, as natural numbers cannot start with a zero. Therefore, it must be part of a single number that is different from \(10^n\). The smallest number that differs from \(10^n\) and contains this set of digits is \(10^n + 1\).

Consequently, we reach a contradiction because \(10^n\) is defined as the largest power of ten that does not exceed \(k\).

Two zeros among pairwise differences mean at least two equal pairs or a triple of equal digits. With \(3\) or \(4\) digits this cannot coexist with both differences \(2\) and \(7\). For \(5\) digits, to realize \(7\) we need \((1,8)\) or \((0,7)\). The multiset \(\{0,0,1,1,8\}\) yields only \(1\) and \(8\), not \(2\). The multiset \(\{1,1,1,3,8\}\) yields \(0,0\) (from \(1\)–\(1\)), \(2\) (from \(3\)–\(1\)), and \(7\) (from \(8\)–\(1\)). The smallest number with these digits is \(11138\).

Answer: \(N=29\). Single-digit numbers share no digit with other single-digit numbers, so \(N>9\). Look at \(9\): both neighbors must contain digit \(9\). The smaller such neighbor is at least \(19\); the other (distinct) is at least \(29\), hence \(N\ge 29\). For \(N=29\) the arrangement 1, 11, 10, 20, 21, 12, 2, 22, 23, 3, 13, 14, 4, 24, 25, 5, 15, 16, 6, 26, 27, 7, 17, 18, 8, 28, 29, 9, 19 works, so the minimum is \(29\).

Let's consider two three-digit numbers, which we will denote as \(n_1 = 10^{100} - 1\) and \(n_2 = 5 \cdot 10^{99} - 1\). The first number \(n_1\) is equal to 99...9 (specifically, it consists of 100 nines). This number is divisible by \(10^{100}\), which implies that its decimal representation ends with \(n_1\).

Next, we examine the second number \(n_2\). It is also divisible by \(10^{100}\), meaning that its decimal representation ends with \(n_2\) as well.

Now, we need to verify that the decimal representations of \(n_1^2\) and \(n_2^2\) do not end with \(n_1\) and \(n_2\), respectively. For \(n_1^2\), we have \(n_1^2 = (10^{100} - 1)^2 = 10^{200} - 2 \cdot 10^{100} + 1\). The last digit of this expression is 1, which indicates that it is not divisible by 5. Therefore, the decimal representation of \(n_1^2\) does not end with \(n_1\).

Similarly, for \(n_2^2\), we calculate \(n_2^2 = (5 \cdot 10^{99} - 1)^2 = 25 \cdot 10^{198} - 10^{100} + 1\). The last digit of this number is also 1, which means it is not divisible by 5. Consequently, the decimal representation of \(n_2^2\) does not end with \(n_2\).

In conclusion, we have demonstrated that both numbers \(n_1\) and \(n_2\) are unusual three-digit numbers. Furthermore, it can be asserted that there are additional unusual three-digit numbers as well.

Let us denote the three natural numbers as \(a, b, c\). According to the problem, the last digit of the sum of any two of these numbers is equal to the last digit of the third number. We can express this with the following equalities: \(a + b \equiv c \mod 10\), \(b + c \equiv a \mod 10\), and \(c + a \equiv b \mod 10\). From these equalities, we can deduce that the expressions \(a + b - c\), \(b + c - a\), and \(c + a - b\) are all divisible by 10. As a result, the total sum of the three numbers, denoted as \(s = a + b + c\), is also divisible by 10. Additionally, from the equality \(s = (a + b - c) + 2c\) and the condition of divisibility, we can conclude that the last digit of the sum \(s\) matches the last digit of the number \(2c\). This implies that the number \(c\) must end in either 0 or 5. Similarly, the numbers \(a\) and \(b\) must also end in 0 or 5. Furthermore, since the sum \(s\) is even, at least one of the numbers \(a, b, c\) must be even. This leads us to conclude that one of these numbers is divisible by 10, while the other two are divisible by 5. Consequently, the product \(abc\) is divisible by 250, which means that the last three digits of the product can only be 250, 500, 750, or 000.

Here are examples of triples of numbers that satisfy the conditions and yield the specified last digits: \(10 \cdot 10 \cdot 1000\); \(5 \cdot 5 \cdot 10 = 250\); \(5 \cdot 5 \cdot 20 = 500\); \(5 \cdot 5 \cdot 30 = 750\).

Let the resulting number be \(N=1000A+k\) with \(0\le k\le 999\). The condition gives \(N=\frac{A(A+1)}{2}\). Hence \( \frac{A(A+1)}{2}=1000A+k \;\Longrightarrow\; A^2-1999A=2k. \) Since \(0\le 2k\le 1998\), we get \( 0\le A(A-1999)\le 1998. \) If \(A\le 1998\) the left side is negative; if \(A\ge 2000\) it is at least \(2000\). Therefore the only possibility is \(A=1999\), yielding \(k=0\) (appended “000”). Thus \(A=1999\).

Since \(2018=224\cdot 9+2\), the smallest such number is the unique \(225\)-digit number starting with 2 and followed by \(224\) nines. Next come the \(225\)-digit numbers starting with 3. The remaining sum is \(2015=223\cdot 9+8\), so among the remaining \(224\) positions we place \(223\) nines and one eight, giving \(224\) numbers. The largest of these has the 8 in the units place: \(3\underbrace{99\ldots 9}_{223}8\). All numbers starting with \(\ge 4\) are larger, hence this is the 225th.

Let \(S(k)\) represent the sum of the digits of the natural number \(k\). A natural number \(a\) is termed \(n\)-good if there exists a sequence of natural numbers \(a_0, a_1, \ldots, a_n\) such that \(a_n = a\) and \(a_{i+1} = a_i - S(a_i)\) for all \(i = 0, 1, \ldots, n - 1\). We want to determine if, for any natural number \(n\), there exists a natural number that is \(n\)-good but not \((n + 1)\)-good.

For natural numbers \(n\) and \(k\), we define the functions \(f(n) = n - S(n)\) and \(f_k(n) = f(f(...(n)...))\) (where \(f\) is applied \(k\) times). When \(n\) is increased by 1, the sum of its digits \(S(n)\) either increases by 1 (if \(n\) does not end with 9) or decreases. This indicates that the function \(f\) is non-decreasing, and the inequality \(f(n + 10) > f(n)\) holds for all \(n\).

First method. Choose a natural number \(d\) such that \(10d > 20d(n + 1)\), and set \(k = 10d\). Define \(b_0 = 10k - 1\) and \(c_0 = 10k - k\), along with \(b_i = f^i(b_0)\) and \(c_i = f^i(c_0)\). We need to demonstrate that the inequality \(b_n > c_n > b_{n+1}\) holds. (*)

Since \(S(c_0) \leq 9k\), we can use induction to show that \(c_i \geq 10k - k - 9ki\). For \(i \leq n + 1\), we have \((9i + 1)k \leq 10ki \leq 10d + 1(n + 1) < 10^{2d}\). This implies that in the number \(c_i\), at least \(k - 2d\) of the leading digits are nines. Therefore, \(S(c_i) \geq 9(k - 2d)\), which leads us, again by induction, to conclude that \(c_i \leq 10k - k - 9(k - 2d)i\). Thus, we obtain the bounds: \(10k - (9i + 1)k \leq c_i \leq 10k - k - 9(k - 2d)i\).

Similarly, we can derive bounds for \(b_i\): \(10k - 9ki - 1 \leq b_i \leq 10k - 1 - 9(k - 2d)i\). To establish the inequality \(c_n < b_n\), it suffices to verify that \(10k - k - 9(k - 2d)n < 10k - 9kn - 1\), which simplifies to \(k > 18dn + 1\). This is satisfied by our choice of \(d\). To show that \(b_{n+1} < c_n\), we need to check that \(10k - 1 - 9(k - 2d)(n + 1) < 10k - (9n + 1)k\), which is equivalent to \(8k + 1 > 18d(n + 1)\), and this is also true.

Since \(c_n = f^n(c_0)\), the number \(c_n\) is \(n\)-good. Furthermore, if \(x \leq b_0\), then \(f^{n+1}(x) \leq f^{n+1}(b_0) = b_{n+1} < c_n\). If \(x > b_0\), then \(f(x) \geq f(10k) = b_0\), which implies \(f^{n+1}(x) \geq f^n(b_0) = b_n > c_n\). Thus, \(c_n\) is not \((n+1)\)-good.

Second method. Assume the contrary: that every \(n\)-good number \(x\) is also \((n+1)\)-good. This means that \(x = f^{n+1}(y)\) for some \(y\). Consequently, the number \(f(y)\) is also \(n\)-good, and thus \((n+1)\)-good; this implies that \(x\) is \((n+2)\)-good. Similarly, we can show that any \(n\)-good number is \((n+k)\)-good for all natural \(k\); we will refer to such numbers simply as good.

Let us select a natural number \(k > 3 \cdot 10^n\) and estimate the number \(D_k\) of good \(k\)-digit numbers in two different ways.

1. For each number \(y \in [2 \cdot 10^{k-1}, 10^k)\), the number \(g(y) = f^n(y)\) is good. Moreover, \(g(y) \geq y - 9kn \geq y - 10^{k-1} \geq 10^{k-1}\), indicating that \(g(y)\) is a good \(k\)-digit number. On the other hand, the equation \(f(x) = a\) has at most 10 solutions for any \(a\), hence the equation \(g(y) = a\) has no more than \(10^n\) solutions.
2. Let \(x\) be a good \(k\)-digit number. Then \(x = f^{10^k}(y)\) for some \(y\). Since \(f^{10^k}(y) \leq y - 10^k\), the number \(y\) must be at least \((k+1)\)-digit. Let \(s\) be the smallest number for which \(f^s(y)\) is \(k\)-digit. Then \(f^{s-1}(y) \geq 10^k\), which implies \(f^s(y) = f(f^{s-1}(y)) \geq f(10^k) = 10^k - 1\). Thus, \(f^s(y) = 10^k - 1\), meaning that any \(k\)-digit good number can be expressed as \(f^t(10^k - 1)\) for some \(t\).

Now we will demonstrate that the number \(f^t(10^k - 1)\) can be \(k\)-digit only under specific conditions, leading to a contradiction with the estimate from the previous point. Let \(b_0 = 10^k - 1\) and \(b_i = f^i(b_0)\). It suffices to show that \(b_{t_0} < 10^k - 1\).

Assume the contrary; then all numbers \(b_i\) for \(i \leq t_0\) are \(k\)-digit. We will estimate the number of such indices \(i < t_0\) for which \(b_i - b_{i+1} < k\) (i.e., \(S(b_i) < k\)). The digits of any such number \(b_i\) form a sequence of \(k\) non-negative integers whose sum does not exceed \(k - 1\). However, there are exactly \(\binom{k - 1 + k - 1}{k - 1}\) such sequences (this can be easily derived from problem 30719 b). Thus, there are no more than \(\binom{k - 1 + k - 1}{k - 1}\) required indices.

Consequently, in the sequence \(b_0, b_1, \ldots, b_{t_0}\), the next number is less than the previous one by at least \(k\) in at least \(t_0 - 22k - 1\) cases. Note that \(k \cdot 22k - 1 \leq 10^k\). Therefore, \(b_{t_0} \leq b_0 - (t_0 - 22k - 1)k \leq 10^k - 10^k = 0\). This leads to a contradiction.

Answer: \(k=2\). Lower bound: \(k\ge2\). Given any single nondecreasing \(N\), pick \(g\) not among its digits; \(ggggg\) differs in all positions. Sufficiency: take \(N_1=13579\) and \(N_2=12468\); any nondecreasing five-digit number matches at least one position with \(N_1\) or \(N_2\).

Answer: yes. Let the last two digits of \(n\) be \(b8\) and let \(S\) be the digit sum of the higher part. If \(b\le 8\), then \( s(n)=S+b+8,\ s(n+1)=S+b+9,\ s(n+2)=S+b+1,\ s(n+3)=S+b+2, \) and \(n+2=10T\), \(n+3=10T+1\). Hence \(s(n+2)\mid 10T\) and \(s(n+3)\mid(10T+1)\) with \(s(n+3)=s(n+2)+1\ge 9\), making them incompatible. Therefore \(b\neq 0,\dots,8\). For \(b=9\) we get compatible divisibilities (e.g., \(n=60398\)).

Let's consider an example of a number that satisfies the given conditions. Let \(n = 13 \cdot 1...1 = 14...43\), where the number of ones will be specified later. If we swap the one and the three, we get the number \(34...41 = 31 \cdot 1...1\). Additionally, if our number consists solely of ones and is divisible by \(13 \cdot 31 = 403\), then both numbers will share the same prime factors.

Next, we will demonstrate that for any number \(d\) that is not divisible by 2 or 5, there exists a number whose decimal representation consists only of ones and is divisible by \(d\).

Consider the sequence of numbers that consist only of ones: \(1, 11, 111, \ldots\). Since there are infinitely many such numbers, there will be at least two numbers among them that yield the same remainder when divided by \(d\). Let's denote these two numbers as \(x\) and \(y\). The difference between these two numbers, \(A = x - y\), will take the form \(A = 1...10...0\), meaning it consists of several ones followed by zeros. Furthermore, the number \(A\) is divisible by \(d\). Since \(d\) is coprime to 10, the number formed solely of ones obtained from \(A\) by removing the zeros will also be divisible by \(d\).

Let \(N\) be the symmetric number formed by writing the integers from 1 to \(n\) in sequence (where \(n > 1\)). Assume that \(N\) has \(m\) digits, with \(m > 18\). We define \(A\) and \(B\) as the numbers created by the first and last \(k\) digits of \(N\), respectively, where \(k = \lfloor m/2 \rfloor\).

Next, let \(10^p\) be the largest power of ten that divides \(A\) completely. From this, we can conclude that \(n < 2 \cdot 10^{p+1}\). This implies that \(n\) can have at most \(p + 2\) digits.

Moreover, the number \(A\) includes a segment of the form \(99\ldots9\) (with \(p\) nines) followed by \(100\ldots0\) (with \(p\) zeros) and ending with \(1\). Consequently, the number \(B\) will contain a segment of the form \(100\ldots0\) (with \(p\) zeros) followed by \(199\ldots9\) (with \(p\) nines). It is clear that these segments cannot match when the number is read symmetrically, making it impossible for \(N\) to be a palindrome.