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
29.10.2023 13:59
(169,256)
29.10.2023 14:28
Yes n = 256 and m = 169 Confirm
29.10.2023 14:55
Yeah , got the same but did a lot of caseworks
29.10.2023 15:36
me too bro me too, I proved $n \leq 36^2$ and then bashed out all the cases
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
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
30.10.2023 06:30
Here's a different kind of solution from my side.
Note that we hardly have to deal with any casework here.
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.
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 $n \leq 36^2$?
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 $(m,n)=(169,256)$.
18.10.2024 21:42
I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video
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.
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?