Find the required digit

Question

What is the digit at the $50^{th}$ place from left of $(\sqrt50 +7)^{50}$

I thought of binomial expansion but it was way too lengthy. Can anyone suggest any other way?

Answer 1

Consider the number $N=(7+\sqrt{50})^{50} +(7-\sqrt{50})^{50}$. By looking at the binomial expansion of each term, or in several other ways, one can verify that $N$ is an integer.

The number $7-\sqrt{50}$ has absolute value about $0.071$. The $50$-th power is well under $10^{-50}$, indeed less than $4\times 10^{-58}$.

Thus the number we were given, plus a positive number well under $10^{-50}$, is the integer $N$. It follows that the $50$-th digit after the decimal point is $9$.

Remark: Whenever $7+\sqrt{50}$ has a problem, its friend (conjugate) $7-\sqrt{50}$ is ready to help.

Answer 2

The number is nearly an integer as a calculation with PARI shows :

? u=(sqrt(50)+7)^50;v=u-truncate(u);print(v);print(truncate(v*10^50)) 0.999999999999999999999999999999999999999999999999999999999616604333890863500765 4542881453840381770764819910027343314351354431719554458557486254486946758447 99999999999999999999999999999999999999999999999999 ?

So, 57(!) nines follow after the decimal point, so the required digit is a nine.

The number u is

? u %6 = 2608271528336322863765068032332051824400732172980700124997.9999999999999999 99999999999999999999999999999999999999999616604333890863500765454288145384038177 0764819910027343314351354431719554458557486254 ?

Answer 3

Note that $a=7+\sqrt{50}$ and $b=7-\sqrt{50}$ satisfy $a+b=14$ and $ab=-1$. Note that $a\gt 14$ so that $a^n \gt 10^n$ (by some margin) so that $a^{50}\gt 10^{50}$ and $b^{50}\lt 10^{-50}$

Note that if $Y_n=a^n+b^n$ we have $Y_n=14Y_{n-1}+Y_{n-2}$ so the $Y_r$ are integers. ($a$ and $b$ satisfy $x^2-14x-1=0$, $Y_0=2, Y_1=14$)

Conclude.