= 1 /2 ×( 1+2)+ 1 /2=( 1/ 2) ×3+ 1 /2 ,

∵3? Unit 3，3？ Unit 9, 3? Unit 7, 3? The unit is 1, the 3 5 bit is 3, …

∴3 bits are 1,

The unit number of a is 1,

∫b≡a(bmod 10)，

The unit of ∴b is also 1,

∴ In a positive integer B that meets the conditions, the smallest two digits are 1 1.

So the answer is: 1 1.