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The sequence $(x_n)$ is given recursively with $x_1 = 188$, and $x_{n+1}$ is obtained from $x_n$ by adding twice the sum of the digits of the number $x_n$.
Prove that the numbers 2008 and 2106 are not terms of this sequence.

Attempt:
Experimentally , using Python , I agree that these two numbers are really not terms of this sequence.

How to solve this task analytically , since we always get a new number and therefore we observe the new sum of the digits.

Note: $x_1=188, x_2=188+2(1+8+8)=222, ...$

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    $\begingroup$ I don't understand the sequence, do you have $x_{n+1} = f(x_n)$, where $f(x_n)$ is twice the sum of the digits, or $x_{n+1} = f(x_n) + x_n$? Because, if it's the first one, i don't see how could the sequence have a number bigger than $x_1$ $\endgroup$
    – Lucio Rosi
    Commented Jun 18 at 12:17
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    $\begingroup$ "Attempt: Using python," <<< When observing the sequence with python, what other property of the sequence did you notice? $\endgroup$
    – Stef
    Commented Jun 18 at 12:21
  • $\begingroup$ I have added a note. $\endgroup$
    – user1316790
    Commented Jun 18 at 12:31
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    $\begingroup$ Hint: think about this modulo $9$. (In particular, remember the divisibility test for $9$.) $\endgroup$ Commented Jun 18 at 12:38
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    $\begingroup$ If you combine "the sequence is increasing" with "I wrote a python program" that is actually a valid proof. $\endgroup$
    – DanielV
    Commented Jun 18 at 13:17

3 Answers 3

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Define $S(X_n) = \text{the sum of the digits of number $X_n$}$. Then $X_{n + 1} = X_n + 2S(X_n)$.

Notice that $X_{n + 1} = 3X_n + 2(S(X_n) - X_n)$, so it's obvious that $3|X_{n+1}$ due to $9 \mid S(X_n) - X_n$. And we'll find $9\mid X_n \text{ for $n \ge 3$}$. So since $9 \nmid 2008$, $2008$ isn't a term of this sequence.

For $2106$, we assume that $X_i = 2106$. Then $S(X_{i-1}) = 9$, $18$ or $27$, and so $X_{i-1} = 2088$, $2070$ or $ 2052$. But these aren't right.

When $X_{i-1} = 2088$, $X_i = 2124$.

When $X_{i-1} = 2070$, $X_i = 2088$.

When $X_{i-1} = 2052$, $X_i = 2070$.

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When we have $x_2=222$ , we can see that $x_3=222+2(2+2+2)=234$ & all new terms will be multiples of $3$.
Check : $x_4=234+2(2+3+4)=252$ , $x_5=252+2(2+5+2)=270$

Now , $2008$ is not a multiple of $3$ (or rather not a multiple of $9$) , hence it can not occur there.

You can try Similar Argument with $2106$ , though it is not Same Argument.


Completing my answer with Inspiration from user Savior who has got it right earlier while Acknowledging the comment from user Heropup who tried the modulus method :

Now , the element after $2000=1998+2$ must be a multiple of $9$ & must be EVEN.
It must be within $2000$ & $2054$ , because that is the largest gap between terms here , which we might get when assuming $1998$ is occurring (it will indeed occur , though that is not necessary to know)

Hence atleast one of $2016$ , $2034$ , $2052$ must occur.
Continuing the Sequence with $2016$ gives $2034$ & then $2052$ : No matter which terms occur , $2052$ must occur.
Hence $2070$ must occur.
Hence $2088$ must occur.
Hence $2124$ must occur , which exceeds the target $2106$.
Conclusion : $2106$ can not occur.


Sequence , for refernce , is :
$188\ , 222\ , 234\ , 252\ , 270\ , 288\ , 324\ , 342\ , 360\ , 378\ , 414\ , 432\ , 450\ , 468\ , 504\ , 522\ ,$
$540\ , 558\ , 594\ , 630\ , 648\ , 684\ , 720\ , 738\ , 774\ , 810\ , 828\ , 864\ , 900\ , 918\ , 954\ , 990\ ,$
$1026\ , 1044\ , 1062\ , 1080\ , 1098\ , 1134\ , 1152\ , 1170\ , 1188\ , 1224\ , 1242\ , 1260\ , 1278\ ,$
$1314\ , 1332\ , 1350\ , 1368\ , 1404\ , 1422\ , 1440\ , 1458\ , 1494\ , 1530\ , 1548\ , 1584\ , 1620\ ,$
$1638\ , 1674\ , 1710\ , 1728\ , 1764\ , 1800\ , 1818\ , 1854\ , 1890\ , 1926\ , 1962\ , 1998\ , 2052\ ,$
$2070\ , \color{purple}{{[[2088]]}}\ , \color{purple}{{[[2124]]}}\ , 2142\ , 2160\ , 2178\ , 2214\ , 2232\ , 2250\ , 2268\ , 2304\ , 2322\ ,$
$2340\ , 2358\ , 2394\ , 2430\ , 2448\ , 2484\ , 2520\ , 2538\ , 2574\ , 2610\ , 2628\ , 2664\ , 2700\ ,$
$2718\ , 2754\ , 2790\ , 2826\ , 2862\ , 2898\ , 2952\ , 2988\ , 3042\ , 3060\ , 3078\ , 3114\ , 3132\ ,$
$3150\ , 3168\ , 3204\ , 3222\ , 3240\ , 3258\ , 3294\ , 3330\ , 3348\ , 3384\ , 3420\ , 3438\ , 3474\ ,$
$3510\ , \cdots$

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    $\begingroup$ But 2106 is divisible by 3. How to argument it here then? I now see that i get $100a+10b+c+2(a+b+c)=102a+12b+3c$ (and same in general), which is indeed always divisible by 3. But so is 2106. Is it maybe because that expression is not divisible by 9 and 2106 is? $\endgroup$
    – user1316790
    Commented Jun 18 at 12:46
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    $\begingroup$ That was why I said it is not Same Argument. You have to use Similar Argument to think over that. $\endgroup$
    – Prem
    Commented Jun 18 at 12:49
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    $\begingroup$ The choice of modulus for $2106$ is not straightforward, unlike for $2008$. The smallest modulus I found was $25$, for which $2106$ is congruent to $6$, but the smallest $x_n$ that is congruent to $6$ modulo $25$ is $x_{823} = 29106$. I suspect there is not a modulus $m$ for which the sequence $x_n$ is never congruent to $2106$ modulo $m$. $\endgroup$
    – heropup
    Commented Jun 18 at 13:05
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    $\begingroup$ Nice to know that the modulo way goes very widely off target , @heropup , I have updated the answer accordingly with alternate thinking !! $\endgroup$
    – Prem
    Commented Jun 18 at 15:39
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First answer on here, so here goes nothing:

The numbers 2008 and 2106 are not terms of the sequence defined by $x_1 = 188$ and $x_{n+1} = x_n + 2 \times \text{sum of digits of } x_n$.

Proving That 2008 Is Not a Term of the Sequence

Given the sequence defined by:

$x_1 = 188$

$x_{n+1} = x_n + 2 \times S(x_n)$

where $S(x_n)$ is the sum of the digits of $x_n$.

We notice that the sequence starts with $x_1 = 188$, which is a multiple of 3. Each subsequent term, $x_{n+1}$, is calculated by adding to $x_n$ twice the sum of its digits. The operation $2 \times S(x_n)$ preserves the divisibility by 3, because the sum of digits modulo 3 remains consistent with $x_n$ modulo 3.

Therefore, every term in the sequence from $x_2$ onward is divisible by 3, since 2008 is not divisible by 3 (2008 mod 3 = 2), it cannot be a term in this sequence.

Proving That 2106 Is Not a Term of the Sequence

For $x_i = 2106$, to explore the feasibility:

$2106 = x_{i-1} + 2 \times S(x_{i-1})$

this implies:

$x_{i-1} = 2106 - 2 \times S(x_{i-1})$

Given the range of possible sums of digits, $S(x_{i-1})$ could logically be around 9, 18, or higher but staying realistic, based on the nature of digits...

Calculating for some reasonable $S(x_{i-1})$ values:

  • For $S(x_{i-1}) = 9$: $x_{i-1} = 2106 - 18 = 2088$
  • For $S(x_{i-1}) = 18$: $x_{i-1} = 2106 - 36 = 2070$

However, further calculating $x_i$ from these $x_{i-1}$ values shows none of these lead to 2106 when the true sum of digits for each $x_{i-1}$ is used in the sequence formula. Thus, 2106 also cannot be reached by the sequence.

Proof using Python

Using Python to generate the sequence and check for the presence of 2008 and 2106, we find that neither 2008 nor 2106 are terms of the sequence defined by $x_1 = 188$ and $x_{n+1} = x_n + 2 \times \text{sum of digits of } x_n$:

>>> def sum_of_digits(n):
...     return sum(int(digit) for digit in str(n))
...
>>> # Initialize the sequence
>>> x = 188
>>> sequence = [x]
>>> terms_to_test = {2008, 2106}
>>> found_terms = set()
>>>
>>> # Generate sequence and check for the terms in question
>>> while x <= 2106:  # Generate until the larger of the two numbers
...     x = x + 2 * sum_of_digits(x)
...     sequence.append(x)
...     if x in terms_to_test:
...         found_terms.add(x)
...
>>> # Analyze and print the results
>>> is_2008_in_sequence = 2008 in found_terms
>>> is_2106_in_sequence = 2106 in found_terms
>>> is_2008_in_sequence, is_2106_in_sequence, sequence[-1]  # Show the last element to ensure sufficient range
(False, False, 2124)

Here we see the following:

  • 2008: The sequence was generated up to 2124, and 2008 was not found among the terms. This confirms that 2008 is not a term because it is not divisible by 3, a property that all terms from $x_2$ onward in the sequence must satisfy due to the sequence's structure.

  • 2106: Similarly, 2106 was not found among the terms up to 2124. The additional checks for feasible previous terms $x_{i-1}$ for $x_i = 2106$ did not yield any valid results leading to 2106.

Thus, the Python results align with the explanation and confirm that neither 2008 nor 2106 can be terms of the sequence.

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