A number theory problem by أحمد الحلاق

Note first that $8^{10} = (7 + 1)^{10}$ . In the binomial expansion of this last expression, all the terms will be divisible by $7^{2} = 49$ except for the last two, namely

$\dbinom{10}{1}*7^{1}*1^{9} + \dbinom{10}{0}*7^{0}*1^{10} = 10*7 + 1 = 71$ .

Thus $8^{10} - 1 \equiv (71 - 1) \pmod{49} = \boxed{21}$ .

8=1 (mod 7)

8^7=1 (mod 7^2)

8^10=8^3 (mod 49)

8^10-1=512-1 (mod 49)

8^10-1=511=21 (mod 49)

Therefore, required remainder=21.

Sorry for using '=' instead of the 'congruent to' symbol.

$8^{10} - 1 = (7+1)^{10} - 1 = \sum_{n=0}^{10} \left(\begin{array}{c} 10 \\ n\end{array}\right) 7^n - 1 = 49 \cdot \left[\sum_{n=2}^{10} \left(\begin{array}{c} 10 \\ n\end{array}\right) 7^{n-2}\right] + 10\cdot 7 + 1 - 1 = 49\left[\cdots\right] + 70 \equiv 21\ \text{mod}\ 49.$