How many positive three digit even numbers lesser than 700 are there with all three different digits and where the sum of the three digits is 12 or less ?

Am I supposed to write solve the expression for each of the 12 cases like solution for a + b + C=12, 11, 10...1 and then add them together?

Is there a better approach to this problem?

  • $\begingroup$ I'd start by drawing a tree of cases that begins with the least significant digit, knowing it has to be even. You can then work out the possible choices for upper two digits, taking into account the restriction that all digits are different as well as having sum less than (or equal to??) 12. Note how your title states the last restriction. $\endgroup$ – hardmath Aug 9 '11 at 16:40
  • 1
    $\begingroup$ Less than 12, or 12 or less? $\endgroup$ – Daniel R Hicks Aug 9 '11 at 17:42
  • $\begingroup$ @Daniel R Hicks : 12 or less $\endgroup$ – user118102114 Aug 9 '11 at 18:49

I would just enter the numbers 2 through 698 into a spreadsheet, use the mod function to extract the digits, and if statements to determine whether the digits are different and sum to less than 12. Then ask it to count the results. The copy function means you only write the things once.


I believe that there are 115 numbers that meet the requirements. We only need to consider even numbers so you can make a table with the possible hundreds digits and even ones digits and easily compute the number of possible tens for each one like this:

1t0---8 possibilities 2,3,4,5,6,7,8,9

2t0---8 possibilities 1,3,4,5,6,7,8,9

3t0---8 possibilities 1,2,4,5,6,7,8,9

4t0---7 possibilities 1,2,3,5,6,7,8

5t0---6 possibilities 1,2,3,4,6,7

6t0---5 possibilities 1,2,3,4,5

1t2---8 possibilities 0,3,4,5,6,7,8,9


Remembering to skip repeated the digits like 2t2, 4t4, and 6t6 ie.

4t4--- 0 possibilities

8+2+2 (error)
7+2+2 (error)
6+3+3 (error)

Each can be permuted ABC, ACB, BAC, BCA, CAB, CBA, so 13x6 if I counted right.

[Oops, I failed to eliminate the numbers above 700. We'll leave that as an exercise for the student. Easily done by breaking the above list into two ranges.]

[Gotta learn to read more carefully! Sum of digits LESS than 12! Oh well...]

[I suspect the above scheme would work, though, with a little adjustment.]

  • $\begingroup$ 921, 831, 741, 651, 633, and 543 are not even. And you can't permute digits in a way that results in an odd. $\endgroup$ – anon Aug 9 '11 at 17:40
  • $\begingroup$ Ah, missed that point too. $\endgroup$ – Daniel R Hicks Aug 9 '11 at 17:44
  • $\begingroup$ (I never was that good at math.) $\endgroup$ – Daniel R Hicks Aug 9 '11 at 17:44

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