SOLUTION: all of 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

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Question 1136522: all of 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 from a number that is equals the sum of fourth and fifth, and the sum of all digits is 24
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


According to the given information, let the digits be the following:

1st digit: x
2nd digit: x+1
3rd digit: (x+1)/2
4th digit: 3(x+1)/2
5th digit: y
6th digit: z

The requirement for the 5th and 6th digits is that the 2-digit number they form must be the sum of the 4th and 5th digits: 10y+z = 3(x+1)/2+y

Since the third digit is (x+1)/2, we know the first digit must be odd.

Armed with only that knowledge, we can plunge into trial and error:

1st digit 1 --> 2nd digit 2 --> 3rd digit 1 --> 4th digit 3
The 5th digit y has to be such that the 2-digit number with first digit y is the sum of y and 3 -- clearly impossible.

1st digit 3 --> 2nd digit 4 --> 3rd digit 2 --> 4th digit 6
Again the same problem -- the 5th digit y has to be such that the 2-digit number with first digit y is the sum of y and 6 -- clearly impossible.

1st digit 5 --> 2nd digit 6 --> 3rd digit 3 --> 4th digit 9
Now if the 5th digit is 1, the sum of the 4th and 5th digits is 10, which means the last digit has to be 0. So there is one number that satisfies the given conditions.

And clearly if the first digit is greater than 5, the 4th "digit" will not be a single digit.
So the number 563910 is the only number that satisfies the given conditions.

Note that the statement that the sum of all the digits is 24 is not necessary in the problem. The number 563910 is the ONLY number that satisfies all the other requirements.

The sum of the digits is in fact 24, so the statement that the sum of all the digits is 24 is superfluous. If the statement of the problem has said that the sum of the digits was anything OTHER THAN 24, then there would have been no solution.

Note further that someone with greater insight into the problem would have jumped immediately to the solution, knowing that the 5th digit has to be 1, in order to make the last two digits form a 2-digit number that is the sum of some two digits. That would then lead to the 4th digit being 9, and finishing the problem from there would be easy.