a.

\(B=(32.34.36...60)(31.33.35....59)\)

\(=(2.16.2.17.2.18...2.30)(31.33.35...59)\)

\(=2^{15}(16.17.18...30)(31.33.35...59)\)

\(=2^{15}(16.18...30)(17.19.21...29)(31.33.35...59)\)

\(=2^{15}(2.8.2.9....2.15)(17.19..29)(31.33...59)\)

\(=2^{15}.2^8(8.9.10...15)(17.19...29)(31.33...59)\)

\(=2^{23}(8.10.12.14)(8.11.13.15).(17.19...29)(31.33...59)\)

\(=2^{23}.(8.10.12.14).T=2^{23}(2^3.2.5.2^2.3.2.7).T\)

\(=2^{23}.(2^7.105)T=2^{30}.105T\vdots 2^{30}\)

b.

\(31\equiv -30\pmod {61}\)

\(32\equiv -29\pmod {61}\)

\(33\equiv -28\pmod {61}\)

...........

\(60\equiv -1\pmod {61}\)

$\Rightarrow 31.32....60\equiv (-30)(-29)(-28)..(-1)\pmod {61}$

Hay $B\equiv A\pmod {61}$

Hay $B-A\equiv 0\pmod {61}$

Tức là $B-A$ chia hết cho $61$

Tham khảo thêm tại đây:

https://hoc24.vn/cau-hoi/cho-a-1232930-b-3132335960a-chung-minh-b-chia-het-cho-230b-chung-minh-b-a-chia-het-cho-61.1506388592636

\(A=10\cdot11+11\cdot12+12\cdot13+...+29\cdot30\)
\(\Leftrightarrow A=\left(1\cdot2+2\cdot3+3\cdot4+...+29\cdot30\right)-\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow A=\frac{29\cdot30\cdot31}{3}-\frac{9\cdot10\cdot11}{3}\)
\(\Leftrightarrow A=29\cdot10\cdot31-3\cdot10\cdot11\)
\(\Leftrightarrow A=8990-330\)
\(\Leftrightarrow A=8660\)

\(B=1\cdot2\cdot3+2\cdot3\cdot4+...+49\cdot50\cdot51\)
\(\Leftrightarrow B=\frac{1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+49\cdot50\cdot51\cdot\left(52-48\right)}{4}\)
\(\Leftrightarrow B=\frac{1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+49\cdot50\cdot51\cdot52-48\cdot49\cdot50\cdot51}{4}\)
\(\Leftrightarrow B=\frac{49\cdot50\cdot51\cdot52}{4}\)
\(\Leftrightarrow B=49\cdot50\cdot51\cdot13\)
\(\Leftrightarrow B=1624350\)

mình không biết làm bài 2