Chứng tỏ rằng:

a, ${16}^5+2^{15}$ chia hết cho 33

b, $8^8+4^{10}$ chia hết cho 17

a, Ta có:

$2+2^2+2^3+2^4+...+2^{99}+2^{100}$

= $\left(2+2^2+2^3+2^4+2^5\right)$ +...+ $\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)$

= 2.$\left(1+2+2^2+2^3+2^4\right)$ +...+ $2^{96}\left(1+2+2^2+2^3+2^4\right)$

= $2.31+2^6.31+...+2^{96}.31$

= $\left(2+2^6+...+2^{96}\right).31$ chia hết cho 31

b, Ta có:

$5+5^2+5^3+5^4+5^5+5^6+...+5^{149}+5^{150}$

= $\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+...+\left(5^{149}+5^{150}\right)$

= $5\left(1+5\right)+5^3\left(1+5\right)+5^5\left(1+5\right)+...+5^{149}\left(1+5\right)$

= $5.6+5^3.6+5^5.6+...+5^{149}.6$

= $(5+5^3+5^5+...+5^{149}).6$ chia hết cho 6

Ta lại có:

$5+5^2+5^3+5^4+5^5+5^6+...+5^{149}+5^{150}$

= $\left(5+5^2+5^3+5^4+5^5+5^6\right)$+...+$\left(5^{145}+5^{146}+5^{147}+5^{148}+5^{149}+5^{150}\right)$ (có đúng 25 nhóm)

= $\left[\right(5+5^4)+(5^2+5^5)+(5^3+5^6\left)\right]$ + ... + $\left[\left(5^{145}+5^{148})+(5^{146}+5^{149})+(5^{147}+5^{150}\right)\right]$

= $\left[5\right(1+5^3)+5^2(1+5^3)+5^3(1+5^3\left)\right]$ + ... + $\left[5^{145}\left(1+5^3)+5^{146}(1+5^3)+5^{147}(1+5^3\right)\right]$

= $(5.126+5^2.126+5^3.126)$ + ... + $(5^{145}.126+5^{146}.126+5^{147}.126)$

= $(5+5^2+5^3).126$ + $(5^7+5^8+5^9).126$ +  ... + $(5^{145}+5^{146}+5^{147}).126$

= 126.[$(5+5^2+5^3)$ + $(5^7+5^8+5^9)$ + ... + $(5^{145}+5^{146}+5^{147})$] chia hết cho 126.

Vậy $5+5^2+5^3+5^4+5^5+5^6+...+5^{149}+5^{150}$ vừa chia hết cho 6, vừa chia hết cho 126