两位不可能 ,设三位数是100x+10y+z 则 x+y+z=13,100x+10y+z=99x+11y+x-y+z是11的倍数
从而可设 x-y+z=11m 2y=13-11m =10-10m+3-m 所以只能 m=1 ,y=1,x+z=12
满足条件的数有913,814,715,616,517,418,319 共7个
四位的仿此