Just substitute one by one

Algebra Level 1

All the boxes below must be filled with the same digit. Which digit is it?

$\boxed{\phantom0}^\boxed{\phantom0} = 4\boxed{\phantom0}\boxed{\phantom0}5\boxed{\phantom0}$

Note: The right hand side a 5-digit integer.

4 5 7 6

1 solution

Naren Bhandari
Feb 9, 2018

$N^N$ has exactly $N-1$ digits for $1< N< 8$ .

Here we have $\begin{aligned}N -1 = 5\Rightarrow N = \boxed{6}\end{aligned}$ .

Therefore $6^6 = 46,656$ is true for right side hidden digits.

Let us prove that $N^N$ has exactly $N-1$ digits within the range of $1 < N<8$ .

Proof : To determine the number of digits we have the formula as $\begin{aligned}& p = \left\lfloor{\log_{10} q}\right\rfloor +1 \\& N-1 = \left\lfloor{\log_{10}(N^N)}\right\rfloor +1\\& N-2 = \left\lfloor{N\log_{10} (N) }\right\rfloor\\& N-2 = \left\lfloor{N \log_{10} (N)}\right\rfloor \end{aligned}$ Therefore $N\neq 10$ (from above equation).

Let us write $\left\lfloor {N\log_{10} (N)}\right\rfloor = n + r$ where $n$ is the integer and $r$ is the fractional part .Then $\begin{aligned} & N - 2 =n+r \\& N =n+r+2\\& N =n +2 \end{aligned}$ The range of $7\log(7)$ is $5\leq 7\log(7) <6$ and $8\log(8)$ is $7\leq 8\log8\ < 8$ .

Hence the equation above can only be true iff $\left\lfloor{n} \right\rfloor = 5$ .This shows that $N^N$ has exactly $N-1$ digits within the range of $1<N<8$ .

Is this correct sir @Pi Han Goh ?

Naren Bhandari - 3 years, 3 months ago

No, this is not a valid proof. While it is true that $N^N$ has $N-1$ digits for $1<N<8$ , your proof is not correct. This is because you can't just substitute $(p,q) = \left(N-1, N^N\right)$ from the start and hope that there's no error.

For the following line: I don't understand your reasoning. Why do you need to express $\left\lfloor {N\log_{10} (N)}\right\rfloor$ as the sum of another 2 floor functions of 2 new expressions? Plus, $\lfloor n \rfloor = N \log_{10} N$ is an absurd equation. The left hand side of the equation is an integer, but the right hand side of the equation is only true for powers of 10, which contradicts your constraints.

Let us write $\left\lfloor {N\log_{10} (N)}\right\rfloor =\left\lfloor{ n}\right\rfloor +\left\lfloor{r}\right\rfloor$ where $\left\lfloor {n}\right\rfloor = N\log_{10}(N)$ and $\left\lfloor{r}\right\rfloor = \log_{10}(N)= 0$ .

I don't even understand how you managed to get $\lfloor n \rfloor = 5$ in the first place.

Your proof is unfortunately, wrong. Good try though.

The simplest way to prove this claim is just by trial and error (since there's only 6 numbers to test).

Pi Han Goh - 3 years, 3 months ago

Just substitute one by one

1 solution

0 pending reports