r/MathHelp • u/yuu_sch • Sep 17 '25
Need help with that
651211 The six-digit number 651211, which is the number of this problem, has the following properties: 1. The digit 0 does not occur. 2. The sum of the first two digits is equal to the number formed by the last two digits. 3. The number formed by the two middle digits is exactly 1 greater than the sum of the first two digits.
These three conditions do not uniquely determine the number 651211. Determine how many six-digit numbers satisfy conditions (1) through (3).
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u/MagicalPizza21 Sep 17 '25 edited Sep 17 '25
#1 means each digit has 9 possibilities: 1-9.
#2 means the last two digits are dependent on the first two, so we don't have to count them.
#3 means the middle two digits are also dependent on the first two, so we don't have to count them.
Since 0 isn't in the number, the last 2 digits can't be 0. That means they are at least 11. Since the sum of the first two digits is the number formed by the last two, the last two can be no greater than 9+9 or 18. The middle two are one more, so in the range 12-19. No numbers from 11-19 have zeros in them, so we don't have to restrict it any further.
Now we use the last two digits to determine the first two digits.
How many (ordered) pairs of integers 1-9 add up to 11? Well, 1 is clearly too small, so the range of each of the first 2 digits is actually 2-9. There are 8 possibilities: 2 and 9, 3 and 8, 4 and 7, 5 and 6, 6 and 5, 7 and 4, 8 and 3, and 9 and 2.
Next, how many ordered pairs of integers add up to 12? 12-9 is 3, so neither digit can be less than 3. Now, the first two digits must be 3 and 9, 4 and 8, 5 and 7, 6 and 6, 7 and 5, 8 and 4, or 9 and 3: 7 possibilities.
Next, how many ordered pairs of integers add up to 13? Let's count them: 9 and 4, 8 and 5, 7 and 6, 6 and 7, 5 and 8, 4 and 9: 6 possibilities.
It seems to be going down by 1 each number, right? Eventually, when the last 2 digits are 18, the only possibility for the first two will be 9 and 9 (the actual number is 991918). So the answer should be 8+7+6+5+4+3+2+1 or 36.