$N$ is a 6 digit Natural number such that its sum of the digits is $43$.
Find $N$ if Exactly One of the statements below is False:
$(1)$ $N$ is a Perfect Square
$(2)$ $N$ is a Perfect Cube
$(3)$ N $\lt 500000$
My Try: The least 6 digit number with Sum of the digits $43$ is $169999$ and the Highest number is $999970$.
Case(i):-> Let Statement 2 is False, That is $N$ is not Perfect Cube.
Now since $N$ $\lt 500000$ and $N$ Last digit can only be either of 0,1,4,5,6 and 9.
Since Least number with sum of digits $43$ is $169999$ and Highest number $\lt 500000$ with sum of digits $43$ is $499993$. Now we need to check a Perfect Square in the Interval $[169999\: 499993]$
I tried to list out numbers in the above interval , but its becoming very Lengthy…Help Required. Thanks
Best Answer
First of all, statement $2$ is false :
Since the sum of the digits of $N$ is $43\equiv 7\pmod9$, we know that $N\equiv 7\pmod9$. However, since there is no $m\in\mathbb N$ such that $m^3\equiv 7\pmod9$, we know that statement $2$ is false.
As a result, we have to find $N$ which satisfies both statement $1$ and $3$. Here, note that $$m^2\equiv 7\pmod9\iff m\equiv 4,5\pmod9.$$ This fact will enable you to eliminate many cases.
So, we have $66$ cases to check if the sum of the digits of $m^2$ is $43$ : $$m=418+9t, 419+9t\ \ (t=0,1,\cdots,32).$$