Kvant Math Problem 729
The statement asks for a natural number with a specified property, not for all such numbers.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m39s
Source on kvant.digital
Problem
Find a natural number with the following property: if one writes its square and its cube side by side, and then reverses the order of the written digits, the result is the sixth power of that number.
N. B. Vasiliev
Exploration
The statement asks for a natural number with a specified property, not for all such numbers.
Trying small values,
$$1^2=1,\qquad 1^3=1,$$
gives the string $11$, whose reversal is $11\neq 1^6$.
For $n=2$,
$$2^2=4,\qquad 2^3=8,$$
giving $48$, whose reversal is $84\neq 64=2^6$.
For $n=3$,
$$3^2=9,\qquad 3^3=27.$$
Writing them side by side gives $927$. Reversing the digits gives $729$, and
$$729=3^6.$$
Thus $n=3$ already satisfies the requirement.
Since the problem asks for a number with the property, finding one valid example is sufficient.
Problem Understanding
We must find a natural number $n$ such that when the decimal representations of $n^2$ and $n^3$ are written consecutively and the resulting string of digits is reversed, the obtained number equals $n^6$.
This is a Type D problem, an existence problem.
The candidate discovered in the exploration is
$$n=3.$$
The only task is to verify directly that it has the required property.
Proof Architecture
The proof consists of a single verification.
First compute $3^2$ and $3^3$.
Next concatenate their decimal representations.
Then reverse the resulting digits and compute $3^6$.
Finally check that the two numbers coincide.
Solution
Take
$$n=3.$$
Then
$$3^2=9,\qquad 3^3=27.$$
Writing the square and the cube side by side produces the decimal number
$$927.$$
Reversing the order of its digits gives
$$729.$$
On the other hand,
$$3^6=(3^3)^2=27^2=729.$$
Hence the number obtained by reversing the digits of the concatenation of $3^2$ and $3^3$ is exactly $3^6$.
Thus the required natural number is
$$\boxed{3}.$$
Verification of Key Steps
The concatenation step gives $927$, not $117$, $297$, or any other arrangement, because the square $9$ is written first and the cube $27$ second.
Reversing the digits of $927$ yields $729$. Computing independently,
$$27^2=729,$$
so indeed
$$3^6=729.$$
The defining property is satisfied exactly.
Alternative Approaches
No substantially different argument is needed. Since the problem asks only for the existence of a natural number with the stated property, a direct construction is the shortest method. Testing small values immediately produces $n=3$, and a straightforward computation verifies the condition.