Question

一道小学数学奥数题，请求解答：

两个自然数，差为11，每一个的数字和都被11整除，满足要求的最小一对自然数中较小的那个为（ ）。
我现在知道答案了，是189 999 999 999

Answer 1

数字和为11的有29.38.47.56.65.74.83.92.119.128.137.146.155.164.173.182.191,209.218,227.236.245.254.263.272.281.290.308.317.326.335.344.353.362.371.380.407.416.425.434.443.452.461.470.506.515.524.533.542.551.560.605.614.623.632.641.650.704.713.722.731.740.803.812.821.830.902.911.920.1019.1028.1037.1046....
数字和为22的有(不可能为两位数):499.589.598.679.688.697.769.778.787.796.859.868.877.886.895.949.958.967.976.985.994.1399.1498.1489.1579.1588.1597,1669.1678.1687.1696.1759.1768.1777.1786.1795... 未完

Answer 2

56和65的差也不是11呀，所以推荐的答案不对

Answer 3

是56

Answer 4

什么叫每一个的数字和，是每个数的每一位数字和吗？

Answer 5

0