A number between 100 and 200 is 19 times the sum of its digits. The tens digit is 3 more than the units digit. Find the number.

Given that the number is between 100 and
200,

Then the number is a three digit
number.

Then, the hundreds digit is
1.

==> Let the number be  x such that "a" is the
tens and "b" is the
units.

Then,

x = 100 + 10a + b
..............(1)

Given that 19 times the sum of the digits
= x

==>19( 1 + a + b) =
x...............(2)

==> 19(1+ a + b) = 100 + 10a +
b

==> 19 + 19a + 19b = 100 + 10a
+b

==> 9a + 18b =
81

Divide by 9:

==> a +
2b = 9................(3)

Also, given that the tens digit =
3 more that the units.

==> a = 3 +
b

We will substitute into
(3)

==> (3+b) + 2b =
9

==> 3 + 3b =
9

==> 3b = 6

==>
b = 2

==> a = 3+ b = 3+2 =
5

==> a =
5

Then the number is: x =
152

To
check:

152 = 19 * (1+5
+2)

152 = 19 * 8

152 = 52