If $\lfloor\log _71\rfloor+\lfloor\log _72\rfloor+\lfloor\log_73\rfloor+\dots+\lfloor\log_7N\rfloor=N$ then find $N$ 
If $\lfloor\log _71\rfloor+\lfloor\log _72\rfloor+\lfloor\log_73\rfloor+\dots+\lfloor\log_7N\rfloor=N$ then find $N$.

Honestly speaking,this time i do not have any clue to move forward.
I was thinking to apply only one logarithmic rule -:

$\lfloor\log _71\rfloor+\lfloor\log _72\rfloor+\lfloor\log_73\rfloor+\dots+\lfloor\log_7N\rfloor=\lfloor\log_71×2×3×\dots×N\rfloor$

please help me out
 A: Let's say $7^k \le N < 7^{k+1}$.
Then $[\log_7 1]..... [\log_7 6] = 0$ Sum for $0 \le N \le 6$ is: $0<N$.
$[\log_7 7]...... [\log_7 48] = 1$ Sum for $7\le N \le 48$ is $1*(N-6)<N$. Sum for $N=48$ is $42$
$[\log_7 49]......[\log_7 (7^3-1)] = 2$ So sum for $49 \le (7^3 -1)$ is $42 + 2(N- 48)$.
If $42+ 2(N - 48) = N$ then $N = 2*48 - 42= 54$.
For $N > 54$ each incremental $[\log_7 N] > 1$ so the sum will surpass $N$ and increase at a faster rate so the sum equalling $N$ will never occur again.
A: Because the logarithms are enclosed in floors, the multiplication rule is not valid, but the question also reduces to one of whole numbers.
For $1\le k\le 6$, $x=\lfloor\log_7k\rfloor=0$. For $7\le k\le48$, $x=1$ and the sum of floored logarithms to this point is 42, so we continue with the numbers for which $x=2$:
$$N=49\implies\Sigma=44$$
$$N=50\implies\Sigma=46$$
$$N=51\implies\Sigma=48$$
$$N=52\implies\Sigma=50$$
$$N=53\implies\Sigma=52$$
$$\color{blue}{N=54\implies\Sigma=54}$$
Thus $N=54$, i.e. $\sum_{k=1}^{54}\lfloor\log_7k\rfloor=54$.
