For any natural number $n$, expressed in base 10 , let $s(n)$ denote the sum of all its digits. Find all natural numbers $m$ and $n$ such that $m<n$ and $$ (s(n))^2=m \text { and }(s(m))^2=n . $$
Problem
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Tags: number theory
math_comb01
29.10.2023 13:59
(169,256)
ATGY
29.10.2023 14:28
Yes n = 256 and m = 169 Confirm
User2411
29.10.2023 14:55
Yeah , got the same but did a lot of caseworks
everythingpi3141592
29.10.2023 15:36
me too bro me too, I proved $n \leq 36^2$ and then bashed out all the cases
math_comb01
29.10.2023 16:08
everythingpi3141592 wrote: me too bro me too, I proved $n \leq 36^2$ and then bashed out all the cases I did exactly the same thing lol
Sg48
29.10.2023 19:14
everythingpi3141592 wrote: me too bro me too, I proved $n \leq 36^2$ and then bashed out all the cases But $ n \leq 1296 \implies s(n) \leq 27 \implies m \leq 729 \implies s(m) \leq 24 \implies n \leq 576 \implies s(n) \leq 23 \implies m \leq 529$ 13 less cases to bash
SouradipClash_03
30.10.2023 06:30
Here's a different kind of solution from my side.
Interpret the question this way.
Consider an infinite sequence of numbers $\{a_n, a_{n-1}, \dots,\} $ and so on such that:
$(s(a_n))^2 = a_{n-1}$
$(s(a_{n-1}))^2 = a_{n-2}$ follows
...
and so on.
We wish to find some $a_k$ in this sequence, such that $(s(a_k))^2 = a_{k-1}$ and $(s(a_{k-1}))^2 = a_k$. Note that the sequence will just loop around. Thus in other words, we want a sequence that can generate a two fold loop, and the terms in the loop will give our desired $m,n$.
Now, we do casework on $a_n$ modulo $9$.
(i) If $3 | a_n$, then the sequence of $s(a_i)$ will be $\{s(a_n), s(a_{n-1}), s(a_{n-2}), \dots, 9, \dots , 9\}$ and it will keep looping back at $9$, which is a one fold loop. (Notice the sequence is decreasing in nature for any $a_n \geq 27$). Thus we have no solutions.
(ii) If $a_n = \pm 1 (\text{mod}$ $9)$, the sequence of $s(a_i)$ will be $\{s(a_n), s(a_{n-1}), \dots, 1 \dots, 1\}$ and it will keep looping back at $1$, again a one fold loop. Thus, no solutions again.
(iii) If $a_n = \pm 2/$$\pm4 (\text{mod}$ $9)$, the sequence of $s(a_i)$ mod $9$ will be loop around $4$ and $7$, and thus the sequence of $s(a_i)$ will be $\{s(a_n), s(a_{n-1}), \dots, 13, 16, \dots, 13, 16\}$ for $a_n \geq 27$. Even if $a_n \leq 27$, we can check a few cases and we will again arrive at our $13,16$ loop. This is the only two fold loop that we obtain, thus our answer is $\{m,n\} = \{169,256\}$.
Notice that any other loop arriving before any sequence would contradict the existence of our obtained loops, so we have no other solutions.
Note that we hardly have to deal with any casework here.
polynomialian
30.10.2023 06:43
s(n) for a four digit square <31 by observation m<961 => n < 1000 and similarly we can do for n>10000. So n<1000 Then bash n<1000 to get the answer.
CrimsonBlade273
30.10.2023 13:09
everythingpi3141592 wrote: me too bro me too, I proved $n \leq 36^2$ and then bashed out all the cases Umm, like how is $n \leq 36^2$?
The_Great_Learner
31.10.2023 16:25
Sol:-Suppose $n$ has $r$ digits. $s(m) \geq 10^{\frac{r}{2}} \implies s(n)^2=m \geq 10^{\frac{10^{r/2}}{9}} \implies 9r \geq s(n) \geq 10^{\frac{10^{r/2}}{18}} \implies r \leq 3$.So $m,n$ are perfect squares which have atmost $3$ digits which implies $s(m),s(n) \leq 27$ and hence $m,n \in \{1^2,2^2,...,27^2\}$. Also note that the highest sum of digits in this set is $19$. So $m,n \in \{1^2,2^2,...,19^2\}$ If $3|m$ then $9|m$ since $m$ is perfect square and then $9|s(m)$ since $m \equiv s(m) \pmod 9$ which implies $81|n$ and hence $9|s(n)$ which implies $81|m$. Similarly if $3|n$ then $81|m , 81|n$. In such case the only possibility is $(m,n)=(9,18)$ which clearly doesn't satisfy the equation. So $m,n \in \{1^2,2^2,4^2,5^2,...,17^2,19^2\}$.Now we could just make a table as follows:- \begin{tabular}{ |c|c|c|c|c|c| } \hline $s(n)$&$m=s(n)^2$ & $s(m)$ & $n=s(m)^2$ & $s(n)$ \\ \hline 1 & 1 & 1 & 1 & 1 \\ 2 & 4 & 4 & 16 & 7 \\ 4 & 16 & 7 & 49 & 13 \\ 5 & 25 & 7 & 49 & 13 \\ 7 & 49 & 13 & 169 & 16 \\ 8 & 64 & 10 & 100 & 1 \\ 10 & 100 & 1 & 1 & 1 \\ 11 & 121 & 4 & 16 & 7 \\ 13 & 169 & 16 & 256 & 13 \\ 14 & 196 & 16 & 256 & 13 \\ 16 & 256 & 13 & 169 & 16 \\ 17 & 289 & 19 & 361 & 10 \\ 19 & 361 & 10 & 100 & 1 \\ \hline \end{tabular}We can observe that only $3$ rows are non absurd out of which $m<n$ is only satisfied by $(m,n)=(169,256)$.
little-fermat
18.10.2024 21:42
I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video
guruguha9
02.11.2024 13:16
polynomialian wrote: s(n) for a four digit square <31 by observation m<961 => n < 1000 and similarly we can do for n>10000. So n<1000 Then bash n<1000 to get the answer. how many marks did you get for writing this solution? in the exam.
guruguha9
02.11.2024 13:17
guruguha9 wrote: polynomialian wrote: s(n) for a four digit square <31 by observation m<961 => n < 1000 and similarly we can do for n>10000. So n<1000 Then bash n<1000 to get the answer. how many marks did you get for writing this solution? in the exam?