According to the meaning of the question, the new three digits are 100x+ 10y+z+3, because the sum of the new three digits is 1/3 of the original three digits, that is, (x+y+z)/3.
Re-analysis, carry situation:
1. If there is no carry, the sum of the new three digits is (x+y+z+3), which can't be 1/3 of x+y+z, so there must be carry.
2. Obviously, if carry occurs, it must be caused by "plus 3", so one digit must be carried to the tenth digit, so the new three-digit unit is z+3- 10=z-7 (at this time, z≥7, so we can just analyze whether the tenth digit is carried to the hundredth digit:
1) is rounded up, so the digit of 100 is x+ 1, and the digit of 10 is.
Y+ 1- 10=y-9, unit number is z-7,
List the equations:
x+ 1+y-9+z-7=(x+y+z)/3
That is 2(x+y+z)=45.
Impossible (even number on the left, odd number on the right)
2) there is no carry, so the hundred digits are x and the ten digits are X.
Y+ 1, unit number is z-7,
Column equation:
x+y+ 1+z-7=(x+y+z)/3
That is, x+y+z=9.
While z≥7 and x≥ 1 are rare:
Only:
x= 1，y= 1，z = 7；
x=2，y=0，z = 7；
x= 1，y=0，z = 8；
In other words, it may be:
1 17,207, 108,
1 17+207+ 108=432