Find all of the six-digit numbers in which the first digit is one less than the second, the third digit is half the second, the fourth digit is three times the third and the last two digits are the sum of the fourth and fifth

Let abcdef denote any such six-digit number and convert each requirement in the problem statement into an
equation.
a = b - 1
c =1/2b
d = 3c
e + f = d + e
24 = a + b + c + d + e + f
In a more standard form this becomes
a - b = -1
-b + 2c = 0
-3c + d = 0
-d + f = 0
a + b + c + d + e + f = 24
Using equation operations (or the techniques of the upcoming Section RREF [23]), this system can be converted
to the equivalent system
a -2/3f = -1
b - 2/3f = 0
c -1/3f = 0
d - f = 0
e +11/3f = 25
Clearly, we must choose f to be a multiple of 3, and of the choices f = 0; 3; 6; 9 only f = 6 results in a sensible
(positive, single-digit) value for e. So with f = 6 we have
e = 3 d = 6 c = 2 b = 4 a = 3
So the only such number is 342636. Notice that the question casts the numbers as digits, but their role as place
values is not relevant.