Calculate the 146th digit after the decimal point of $ \frac{1}{293} $ The question is: Calculate the 146th digit after the decimal point of $\frac{1}{293}$
1 / 293 = 0,00341296928.., so e.g., the fifth digit is a 1.
We know that 293 is a prime, probably this would help us.
I think an equation involving modulos has to be solved, but I am not sure how to tackle this.
Any help is appreciated! Could perhaps someone give a general method to solve these kind of problems?
EDIT: You are supposed to solve this without using a computer.
 A: Consider a rational number $r = \frac{1}{d}$ and suppose it has a decimal expansion of the form $0.\underbrace{d_1d_2\ldots d_m}_{\text{non-recurring}}\underbrace{d_{m+1}d_{m+2}\ldots d_{m+n}}_{\text{recurring}}d_{m+n+1}\ldots$, that is $d_{k+n} = d_{k}$ for $\forall k > m$.
Let $A$ be the integer formed by the first $m$ digits $A = d_1d_2\ldots d_m$, and $B$ be the integer formed by the next $m$ digits $B = d_{m+1} d_{m+2} \ldots d_{m+n}$. Then we have
$$
    10^m r - A = 0.d_{m+1}d_{m+2}\ldots d_{m+n} d_{m+1} d_{m+2} \ldots 
$$
Multiplying both sides with $10^n$ we get
$$
  10^n (10^m r - A) = B + (10^m r - A)
$$
Or, equivalantly
$$
    10^m \left(10^n - 1 \right) = \left( B - A \left(10^n - 1 \right)\right) d
$$
This implies that $10^m \left(10^n - 1 \right)  \bmod d = 0$. 
Since in the case at hand $d$ is relatively prime to $10$, the smallest solutions for $m$ and $n$ are $m=0$ and $n = \operatorname{ord}_{d}(10)$. The multiplicative order is defined as a smallest exponent $n$ such that $10^n \equiv 1 \mod d$. Multiplicative order $\operatorname{ord}_d(10)$ is a divisor of the Euler totient function $\phi(d)$. Since $d = 293$ is prime
$$
   \phi(293) = 293-1 = 292 = 2^2 \cdot 73
$$
thus we should try $n = 73$, $n=146$ and then $n=292$. It is not hard to see that $10^{73} = - 1 \bmod 293$, thus $n= 146$.
Having determined that, the 146-th digits equals $B \bmod 10$. 
$$
   (10^n-1) = B d
$$
meaning that $B \bmod 10 = (-1) d^{-1} \bmod 10 = 3$.
A: The following is a small variant of the methods of Sasha and Lopsy.  We show that $10$ is a quadratic residue of $293$. This enables us to conclude that $10^{146} \not\equiv -1 \pmod{293}$.
It is just a Legendre symbol calculation. For ease of typing we denote the Legendre symbol by $(a/p)$ instead of $\left(\frac{a}{p}\right)$.
Note that $(10/293)=(2/293)(5/293)$.  Since $293$ is of the form $8k+5$, we have $(2/293)=-1$. 
To calculate $(5/293)$ we use  Quadratic Reciprocity. Since one, and indeed both, of $5$ and $293$ are of the shape $4k+1$, 
$$(5/293)=(293/5)=(3/5).$$ 
One could continue with Quadratic Reciprocity, but by inspection $(3/5)=-1$. 
Thus $(10/293)=(-1)(-1)=1$, and we conclude that $10$ is indeed a quadratic residue of $293$.
A: The number $\dfrac{1}{293}$ is the following in its decimal form:

Image Courtesy: Wolfram | Alpha
A: Let $r$ be the remainder when $10^{146}$ is divided by $293$. Then the answer is given by the last digit of $(10^{146} - r) / 293$.
Why is this true? Since $1/293$ is a positive number less than one, it is of the form 
$1/293 = 0.a_1a_2a_3a_4 \ldots$ 
and thus
$10^{146}/293 = a_1a_2 \ldots a_{145}a_{146}.a_{147}a_{148} \ldots$
On the other hand, by the division algorithm $10^{146}/293 = q + r/293$, where $q$ is the quotient and $0 \leq r < 293$ is the remainder. Since $q$ is an integer and $0 \leq r/293 < 1$, it follows that $q = a_1a_2 \ldots a_{145}a_{146}$ and $r/293 = 0.a_{147}a_{148} \ldots$.
Then $(10^{146} - r)/293 = q = a_1a_2 \ldots a_{145}a_{146}$.
Thus we can apply modular arithmetic to solve the problem. Notice that 
$(10^{146} - r) \cdot 293^{-1} \equiv -r \cdot 293^{-1} \equiv -r \cdot 3^{-1} \equiv -r \cdot 7 \equiv 3r \mod 10$. 
Therefore the last digit is equal to $3r \mod 10$. 
What remains is to calculate $r$. For this particular case I don't know of any better way than direct calculation. Repeated squaring works, but you might want to use a calculator. It turns out that $10^{146} \equiv 1 \mod 293$, and thus the answer is $3$.
A: This is a 3.  The basic period of the decimal expansion looks like this.
0  .0003412969
1.  2832764505
2.  1194539249
3.  1467576791
4.  8088737201
5.  3651877133
6.  1058020477
7.  8156996587
8.  0307167235
9.  4948805460
10. 7508532423
11. 2081911262
12. 7986348122
13. 8668941979
14. 5221843

I computed it using this little program.
apple = 1000
biter = 293
out = ".000"
for k in range(1000):
    out += str(apple//biter)
    apple %= biter
    apple *=10
print out

