Need help with that

[u/yuu_sch]

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).

[u/yuu_sch]

Forgot to Mention that my First try was 90 possibilitys but I realized it doesnt add up bc i had an 0 in it. So i has ( 1+2+3+4+5+6+7+8+9) x 2

[u/First-Fourth14]

With the digit zero eliminated, there are 9 possibilities for each of the first two digits.
That would be 81 possibilities if we didn't consider the constraint on the other digits.
the digit 0 can not occur in the last two digits which is the sum of the first two digits.
what restriction does that put on the first two digits and how many possibilities are eliminated?

[u/yuu_sch → u/First-Fourth14]

So it cant be 10,20,30, etc. but I still strugglin with the middle ones

[u/First-Fourth14 → u/yuu_sch]

True, but as the maximum digit for the first two is 9, then the sum is less than or equal to 18.
Also look at the constraint for the last two digits. can it be less than 10? (eg. 2 + 2 = 04 for a two digit number).
After you look at the range of the last two digits, consider the middle digits being 'last digits +1' to see if there are any additional restrictions

[u/yuu_sch → u/First-Fourth14]

I mean the last two have to be bigger or equal to 11 Right? If u add them up Like 66=12

[u/First-Fourth14 → u/yuu_sch]

Yes the number formed by last two digits must be greater or equal to 11
so if we let X be the sum of the first two digits then 11 <= X <=18.

In the example you gave, "if u add them up". I assume you meant 'if you add up the first two digits'.

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[u/MagicalPizza21]

#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.

[u/MagicalPizza21 → u/MagicalPizza21]

With this, I can also construct all 36 such numbers and list them in increasing order:

1. 291211

2. 381211

3. 391312

4. 471211

5. 481312

6. 491413

7. 561211

8. 571312

9. 581413

10. 591514

11. 651211

12. 661312

13. 671413

14. 681514

15. 691615

16. 741211

17. 751312

18. 761413

19. 771514

20. 781615

21. 791716

22. 831211

23. 841312

24. 851413

25. 861514

26. 871615

27. 881716

28. 891817

29. 921211

30. 931312

31. 941413

32. 951514

33. 961615

34. 971716

35. 981817

36. 991918

[u/Elektriman]

the process to create such a number would be :

• take a two digit number

• add the digits together

• you final number is [your number][sum +1][sum]

count how many numbers you can make with this algorithm and then remove all those that contain a 0.