|
Question 1110788: Find the six-digit number in which the first digit is 2 more than the second, the second digit is 2 more than the third, the third is 2 less than the second, the fourth is 2 less than the third, the fifth is 1 less than the fourth, and the last two digits are the sum of the first four. The sum of all six digits is 30.
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
Eliminating redundant information, and keeping all the comparisons of the digits "going in the same direction", we can re-state the problem like this:
the 1st digit is 2 more than the 2nd
the 2nd digit is 2 more than the 3rd
the 3rd digit is 2 more than the 4th
the 4th digit is 1 more than the 5th
So the 1st digit is the largest of the first five digits, and probably the largest of all the digits.
Now you could probably finish solving the problem using formal algebra; but it seems logical reasoning will get you to the answer faster.
So let's see what happens if we let the first digit be the largest it can be: 9. Then the 6-digit number is 97532x; and since the sum of all 6 digits is 30, the 6-digit number would be 975324.
And that number satisfies the remaining condition of the problem -- the sum of the first four digits is 24, and the last two digits are 24. So
Answer: the 6-digit number is 975324.
|
|
|
| |
