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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?

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From the left end, or after the decimal point? –  André Nicolas Feb 19 at 19:28
    
@AndréNicolas from the left end –  MathGod Feb 19 at 19:31

3 Answers 3

up vote 4 down vote accepted

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.

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Thanks! It was very helpful. –  MathGod Feb 19 at 19:42
    
You are welcome. –  André Nicolas Feb 19 at 19:44

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.

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I didn't fill in all the steps, because it is interesting to discover them yourself. There are a number of questions a bit like this where the answer is "nearly an integer" which arise when there is some integer recurrence involving conjugates of some kind (e.g. change the sign of the square root), where one conjugate has absolute value less than $1$, so its powers are vanishingly small. The Fibonacci Sequence is another example. Learn when to look at the conjugate ... –  Mark Bennet Feb 19 at 19:52

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 ?

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Do you have any elegant way to do this or any other way that an be done with pen and paper? –  MathGod Feb 19 at 19:37

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