DeHayward,
Method 1: trial and error
Make a guess that the first digit is 1, so the number is 1_ _ _ _ _
(the spaces represent the remaining digits).
The first digit is one less than the second, so 12_ _ _ _
The third is half the second, so 121_ _ _
The fourth is three times the third, so 1213_ _
The sum of the digits is 24, so sum of last two gitis is 247 =17,
which means the last two digits must be 9 and 8 (9+8=17).
So, the number is either 121398 or 121389. But neither of thesesatisfy the remaining condition: the last 2 digits are the sum of the fourth and fifth digits. Thus, the initial guess of 1 as the first digit is incorrect.
Make a guess that the first digit is 2, so the number is 2_ _ _ _ _
The first digit is one less than the second, so 23_ _ _ _
The third is half the second, so 23?_ _ _ (can't work since half of 3 is 1.5 and digits are whole numbers). Thus, the initial guess of 2 as the first digit is incorrect.
Make a guess that the first digit is 3, so the number is 3_ _ _ _ _
The first digit is one less than the second, so 34_ _ _ _
The third is half the second, so 342_ _ _
The fourth is three times the third, so 3426_ _
The sum of the digits is 24, so sum of last two digits is 2415 =9. The fourth and fifth digits must also have a sum of 9, because the last 2 digits are the sum of the fourth and fifth digits. The fourth digit is 6, so the fifth digit must be 3 (6+3=9). The fifth digit is 3 and the sum of the last two digits 9, so the last digit is 6 (3+6=9). Therefore, the number is 342636.
Method 2  Using algebra
Suppose f is the first digit
The first digit is one less than the second, so the second digit is f+1
The third is half the second, so the third digit is (f+1)/2
The fourth is three times the third, so the fourth digit is 3*(f+1)/2
The sum of the firts four digits is:
f + (f+1) + (f+1)/2 + 3*(f+1)/2
= f + f + f/2 + 3f/2 + 1 + 1/2 + 3/2
= 2f/2 + 2f/2 + f/2 + 3f/2 + 2/2 + 1/2 + 3/2
= 8f/2 + 6/2
= 4f + 3
The sum of all the digits is 24, so sum of last two digits is:
24(4f+3)
= 214f
The fourth and fifth digits must also have a sum of 214f, because the last 2 digits are the sum of the fourth and fifth digits.
The fourth digit is 3*(f+1)/2, so the fifth digit must be:
21  4f  3*(f+1)/2
= 21  4f  3f/3  3/2
= 42/2  3/2  8f/2  3f/3
= 39/2  11f/2
The fifth digit is 39/2  11f/2 and the sum of the last two digits
214f, so the last digit is:
214f  (39/2  11f/2)
= 42/2  39/2  8f/2 + 11f/2
= 3f/2 + 3/2
= 3*(f+1)/2
= fourth digit
So the digits in order are:
f, f+1, (f+1)/2, 3*(f+1)/2, 39/2  11f/2, 3*(f+1)/2
(f+1)/2 must be a whole number, so f+1 must be even. Hence, f is odd.
39/2  11f/2 = (39  11f)/2 must be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9
So, 3911f must be 0, 2, 4, 6, 8, 10, 12, 14, 16, or 18
(respectively).
Thus, 11f must be 39, 37, 35, 33, 31, 29, 27, 25, 23, or 21
(respectively).
But f must be a whole number, so 11f must be a multiple of 11. The only mulitiple of 11 in the last list above is 33. Hence, 11f=33, and f = 3.
Therefore, the number is 342636.
Note  there is a potential for two interpretations of this problem
Interpretation 1:
"The last 2 digits are the sum of the fourth and fifth" means "fifth digit + sixth digit = fourth digit + fifth digit"
This is how I interpreted the problem in the solutions above.
Interpretation 2:
"The last 2 digits are the sum of the fourth and fifth" means "fifth digit = fourth digit + fifth digit"
AND
"sixth digit = fourth digit + fifth digit"
"fifth digit = fourth digit + fifth digit" means fourth digit is 0
The fourth digit is three times the third digit, so the third digit is 0. The third digit is half the second digit, so the second digit is 0. The first digit is one less than the second digit, so the first digit is 1. But digits must be positive. Therefore, interpretation 2 is not possible.
Paul
