a three digit number that has more tens than ones and less hundreds than tens

Question

how do you know thats the correct number and how did you get that number

Answer 1

The 3-digit number can be represented by 100A+10B+C.

More tens than ones: B>C.

Fewer hundreds than tens: A<B so B>A.

We only know that the tens digit is bigger than the ones and the hundreds.

The smallest B can be is 2, A<2, C<2, so because it's 3 digits A can't be zero, so C=0 and A=1 or C=A=1.

The smallest number is 120. The next possibility is 121.

If B=3, A<3, C<3 so A can be 1 or 2 and C can be 0, 1 or 2:

130, 131, 132, 230, 231, 232.

If B=4, C and A < 4, so A can be 1, 2 or 3 and C can be 0, 1, 2, 3:

140, 141, 142, 143, 240, 241, 242, 243, 340, 341, 342, 343.

...

If B=9, A can be 1, 2, 3, 4, 5, 6, 7, 8 and C can be 0, 1, 2, 3, 4, 5, 6, 7, 8:

190, 191, 192, 193, 194, 195, 196, 197, 198, ..., 890, 891, ..., 898.

So there are many solutions (2+6+12+20+30+42+56+72=240 solutions).