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 $61$
