Each Digit Once in Powers of N

From 0, 1, 2, ..., 9, there are 10 digits. It is not possible for both $N^3$ and $N^4$ to have 5 digits each, because $N$ would be less than 10 in that case. However, if $N$ is less than 10, then $N^3$ could not have 5 digits. It is not possible for $N^3$ to have at most 3 digits and $N^4$ to have at least 7 digits, because N would be at least 1001.001, which is not possible since $N^3$ has only 3 digits. Thus, $N^3$ has 4 digits and $N^4$ has 6 digits. For $N^3$ to have 4 digits, the range of $N$ is from 10 to 21. For $N^4$ to have 6 digits, the range of $N$ is from 18 to 31. Thus, the only possible range of N is from 18 to 21.

If $N$ is 20,then the digit 0 would be repeated (as the units digit in both) If $N$ is 21, then the digit 1 would be repeated (as the units digit in both) If $N$ is 19, then $N^3=6859$ and $N^4=130321$ . 3 is repeated, so that is not possible. If $N$ is 18, then $N^3=5832$ and $N^4=104976$ . Each of 0, 1, ..., 9 occurs exactly once in $N^3$ and $N^4$ . Thus, $N$ is 18 and $N^2=324$ .

[LaTeX edits - Calvin]

let the no.of digits in N^3 and N^4 be n and m respectively then we have n+m=10 that is possible for the numbers 18,19,20 and 21

19^3=6859 and 19^4=130321. so N (not equal to)19

in 20, if we square, the last two digits will be zero if we cube the last three digits will be zero and so on

for 21, the last digit of its cube and its fourth power(and fifth and sixth powers and so on) will be equal and that is 1 so the required value of N is 18 and 18^2=324 :)

If you think about this question, you find that $N^3$ has 4 digits and $N^4$ has 6 digits. This means that the lowest value of $N$ is 18 and the maximum value is 21. By using trial and error you can figure out that the answer is 18.

Since $N^3$ and $N^4$ together contain each of the digits exactly once, they contain 10 digits altogether. It is easy to see that $N^3$ thus has 4 digits and $N^4$ has 6 digits. This means that $N^3<10000$ and $N^4 \geq 100000$ , thus $17<\sqrt[4]{100000} \leq N<\sqrt[3]{10000}<22$ , so the only possible values for N are 18, 19, 20 or 21. Trying each of them shows that $N=18$ , thus $N^2=324$

$N^3$ & $N^4$ cannot be of same length,so as to comprise all digits exactly once... Let $N^3$ be $\overline{abcd}$ & $N^4$ be $\overline{efghij}$ ... Now maximum value of N for which $N^4$ is 6-digit & $N^3$ is 4 digit is 21,& minimum value is 18. We observe that for N=18,the condition holds... So required solution is $18^2$ =324

Firstly for there to be 10 digits occurring only once, the digits of N^3 and N^4 must total 10, therefore it may be considered that under the conditions required N^3 must have four digits and N^4 must have six digits. Taking the number 10, it's given powers are 1000 and 10000, the total is 9 digits, so the solution is greater than 10. Taking the number 20, it's given powers are 8000 and 160000, the digits total 10 so the solution must be near 20. For 22 in place of N 11 digits are produced, so the solution must be lower than 22. Trying 21^4, using fourth root as there is a greater chance of double digits with more numbers, there are double digits, trying 19 in the same way also gives double digits. Though trying 18 in place of N gives 5832 and 104976 which when placed in order gives the numbers 0 1 2 3 4 5 6 7 8 9 so 18 must be the solution.

Suppose $\mathfrak{N}(N)$ denotes number of digits of $N$ , then it is necessary to have $f(n) = \mathfrak{N}(N^3) + \mathfrak{N}(N^4) = 10$ The function $f(n)$ is clearly non-decreasing, note that $f(16) = 9\text{ and } f(22) = 11$ Hence we need $N \in \{17,18,19,20,21\}$ , trying these $5$ values we find that $N = 18$ works, hence the answer is $N^2 = 18^2 = \boxed{324}$

Either $N^3$ has $4$ digits and $N^4$ has $6$ digits or they both have $5$ digits. If they both had $5$ digits then clearly $N <10$ but $9^3 = 729 < 10000$ and so does not have $5$ digits, therefore $N^3$ must have $4$ digits and $N^4$ must have $6$ digits.

Therefore we can say with confidence that $N^3 < 10000 \Rightarrow N< \sqrt[3]{10000}$

$21^3 < 10000 < 22^3$ so we have $N < 22$

We also have $99999 < N^4 \Rightarrow \sqrt[4]{99999} < N$

$17^4 < 99999 < 18^4$ and so we get $17 < N < 22$ meaning $N \in \{ 18,19,20,21\}$

Upon checking we see immediately that $N=18 \Rightarrow N^3 = 5832, N^4 = 104796$ which satisfies the conditions and thus, $N^2 = 18^2 = 324$

There are 10 distinct digits between N^3 and N^4 You can't have it 5 digits each as this would mean that N^4/N^3 < 10 and 10^3 is not composed of 5 digits. 3 and 7 is not possible as it would mean N^4/N^3 < 1000 and 1000^3 is not 3 digits long.

This would mean they have 4 and 6 digits respectively.

We could further reduce our options by looking for numbers whose 4th power is 5 digits long, this would be our lower bound as N^4 must have 6 digits. N^4 > 99,999 N > 17.78

Similarly, and upper bound can be made by looking for a 5 digits N^3 < 10,000 N < 21.54

So our values would be within the range: 18 <= n <= 21

19 cannot be the answer as the last digit is repetitive 9, 1, 9, 1 20 cannot be the answer as the last digit is always 0 21 cannot be the answer as the last digit is always 1

This leads us to the answer (N) being 18 The problems asks for N^2 which is equal to 18^2 = 324

Because 18^3 = 5832 and 18^4 = 104976 also consist number 0,1,2,3,4,5,6,7,8,9 so, 18 is solution. Thus, 18^2 = 324(ANSWER)