Using base 10, when 12 is squared and the order reversed, the positive square root is the reverse of the first number. 13 also has this property. What is the name for this property? Are there any integers that also have this property and (clearly) are of more than two digits? (Sorry that I have changed the subject a bit!)
Asked
Active
Viewed 76 times
1
-
Do you want to include numbers ending in zero, such as 10? – DominikS Oct 04 '23 at 08:59
-
7They are called skinny numbers – jjagmath Oct 04 '23 at 09:17
2 Answers
1
This property is referred to as skinny (thanks @jjagmath for the reference).
There seem to be many numbers with this property, just a few examples (I wrote some code): 1121, 22011, 110102, 130011, 310011. In total there are 611 such numbers between $10$ and $10^6$ (including ones with trailing zeros).
Observation: If $n$ has that property, the reverse of $n$ will have the property too.
Conjecture: The digits from any number with this property are exclusively from the set $\{0, 1,2,3\}$. This, and more properties are stated in reference, albeit without proof
DominikS
- 2,510
1
For a positive integer $n$, define the function $r(n)$ to be the number whose base-10 digits are reversed. Then consider the function $$f(n) = r(n^2) - (r(n))^2. \tag{1}$$ Then $n$ meets the conditions stated in your question if and only if $f(n) = 0$. We will exclude all cases where $n$ is divisible by $10$, as well as cases where $n$ is a palindrome satisfying $f(n) = 0$.
A computer search for all $n \le 10^6$ yields numerous solutions, e.g. $121201$. However, in all cases, the set of distinct digits of $n$ belong to one of the following: $$\{0,1\}, \{0,2\}, \{1,2\}, \{1,3\}, \{0,1,2\}, \{0,1,3\}.$$ So for instance, we don't see any $n$ that use digits greater than $3$, and we also don't see any $n$ that use $\{2,3\}$. I think this could provide some ideas about proving that no such $n$ violating at least one of these two conditions exists.
heropup
- 135,869
-
1The square of a palindromic number does not need to be a palindrome. Hence $f(n) = 0$ does not hold for all palindromic numbers. – DominikS Oct 04 '23 at 09:05
-
@DominikS Thanks for catching that. I've revised my response accordingly. – heropup Oct 04 '23 at 09:08
-
No worries! I wonder if it might be worth including the palindromes in the result - they do seem to satisfy the same restriction on digits. – DominikS Oct 04 '23 at 09:11
-
1
-
$(10^n \times a+b)^2=(10^{2n} \times a^2+2 \times 10^n \times a \times b+b^2)$ ; if $a=2$ and $b=3$, $2ab=12$ which is greater than $10$, so it generates problem. – Lourrran Oct 04 '23 at 09:52