Question

a number that is between 100 and less then 200 and the number is 19 times larger than its sum

Answer 1

I assume you mean the sum of digits.

let the number be xyz
where x is 1
(1+y+z)= 1yz/19

Look at the right side. That has to be a whole number, so 1yz is a multiple of 19
I can range from 6 to 9
which means y+z can range from 5 to 8
try 5=y+z or y=5-z which means if
z; y;
2;3
3,2
4,1
1,4
5,0
0,5
but 1yz=19*6 or 1yz/6=19, which means 1yz is divisible by six, so it ends in 6, 2, 8, 4
if =2, y=4 142 /19=not six
if z=4, y=2 124/19=not six
so yz now runs from 7 to 8
If 1+y+z= then combinations
y+z=6 or 7
assume y+z=6
y, z
0,6
6,0
1,5
5,1
4,2
2,4
Now try these combinations:
1+0+6=7, 7*19=133 nope
1+1+5=7, 133 does not match
none of the y,z above will give 133

Now try y+z=7
6,1 1+6+1=8, 8*19=152 and I see immediately that y=5, z=2

152 is the number
sum digits: 8
the number is 19 times the sum