93
3x + 12 < 20
3
6(x+12)
the number is 11
3(x + 12)
3(x=12)
i am a two digit # my tens digit# is 3 times my ones digit #and the sum of my digit is 12 what am i
The larger number is 12.
93
3x + 12 < 20
3
6(x+12)
the number is 11
Where n = any number, n(12+3)
3(x + 12)
Nine hundred thirty (930) is the highest three digit number whose digits have a sum of 12. 9+3+0=12. Any numbers higher than that will give you an sum greater than 12.