$1-90 C_{10}^{1}+90^{2} C_{10}^{2}-90^{3} C_{10}^{3}+\cdots+90^{10} C_{10}^{10}$ 除以 88 的余数是 ( )
答案:B

【解析】因为 $1-90 C_{10}^{1}+90^{2} C_{10}^{2}-90^{3} C_{10}^{3}+\cdots+90^{10} C_{10}^{10}=(1-90)^{10}=(1+88)^{10}$

$=1+88 \mathrm{C}_{10}^{1}+88^{2} \mathrm{C}_{10}^{2}+88^{3} \mathrm{C}_{10}^{3}+\cdots+88^{10} \mathrm{C}_{10}^{10}$

$=1+88\left(\mathrm{C}_{10}^{1}+88 \mathrm{C}_{10}^{2}+88^{2} \mathrm{C}_{10}^{3}+\cdots+88^{9} \mathrm{C}_{10}^{10}\right)$,

所以 $1-90 C_{10}^{1}+90^{2} C_{10}^{2}-90^{3} C_{10}^{3}+\cdots+90^{10} C_{10}^{10}$ 除以 88 的余数是 1 , 故选 B.