A number theory problem by dewita sonya

This is the same as the sum of the first $2014$ digits after the decimal point in $\frac{6^{2014}}7$ . Since $6^{2014} \equiv 1$ mod $7$ , this is the same as the sum of the first $2014$ digits after the decimal point in $\frac17 = 0.\overline{142857}$ .

This is $335(1+4+2+8+5+7) + 1+4+2+8 = \fbox{9060}$ .

But if it is $\frac{60^{2013}}{7}$ instead of $\frac{60^{2014}}{7}$ , the answer given by Patrick Corn doesn't work. It works because $2014$ is even. By observing the first few powers of $60$ we will know that.

The following shows $n$ and $\left\lfloor \frac{60^n}{7} \right\rfloor$ (or the integer part). It can be seen that when $n$ is odd, the $n$ digits before the decimal point starts with $857142$ and when $n$ is even then, $142857$ .

$\space n \quad \left\lfloor \frac{60^n}{7} \right\rfloor$

$\space 1 \quad 8$

$\space 2 \quad 514$

$\space 3 \quad 30857$

$\space 4\quad 1851428$

$\space 5\quad 111085714$

$\space 6\quad 6665142857$

$\space 7 \quad 399908571428$

$\space 8 \quad 23994514285714$

$\space 9 \quad 1439670857142857$

$10 \quad 86380251428571428$

Since $2014$ is even, the $2014$ digits before the decimal point starts with $142857$ repeats for $335 (335\times 6 = 2010$ times and with a remainder of $4$ .

Therefore, the sum of the $2014$ digits before the decimal point:

$S=335\times (1+4+2+8+5+7)+(1+4+2+8) = \boxed{9060}$