23 thoughts on “PnC Examples – Numbers with Fixed Digit-sums

    • I’ve assumed you’ve already read (and understood!) my previous posts on PnC – specifically the ones on the Theory of Partitioning. If you haven’t please go through them, it will be clear I guess 🙂


  1. sir, in 1st question if (a – 1) will be 9, then “a” will be equal to 10 which, which is not possible….please tell me where am i going wrong??

  2. Seema, if you are considering those solutions you will also need to arrange (1, 5, 8) in 6 ways and (1, 1, 40) in 3 ways as the question asks for (a, b, c) so we need ordered triplets. I will also suggest you consider the next case where two numbers are divisible by 2 and one not. See if that adds anything to your set of solutions.


    • yes, (2,20,1) will also be there, but do we need to arrange them? u mean it matters weather a =2, b=20, c=1 and a=20, b=2 and c=1?

      • OK let me clarify.

        If the question asks “how many pairs of natural numbers multiply to give 30” then the answer will be 4 i.e. (1, 30), (2, 15), (3,10) and (5,6) [here order is not important]

        If the question asks “how many natural number solutions exist for (a, b) such that ab = 30”, then the answer will be 8 [here order is important as a = 15 and b = 2 is different from a = 2 and b = 15]

        The latter question could also be asked as “how many ordered pairs of natural numbers exists such that their product is 30”


  3. Dear sir,
    Please correct me if wrong but won’t the answer to 2nd question be 2002-1=2001 as there will one case where all the digits are zero resulting in the number 0?
    After all, we are looking for natural numbers.

  4. Sir, In an earlier post where we checked how many 2 digit numbers exist whose sum of digits is 13, we had checked such that a < 10 as digits are from 0-9 and we assumed a = a'+10 . But here in Qn.1 why are we not checking if each digit value has upper limit 9. We are ensuring only that the first digit is not zero. Could you please help me understand

    • In this case, the sum of the digits is only 9. So, none of the individual digits can cross 9. Had the question been, say, “How much numbers of 6 digits have their sum of digits less than or equal to 15”, then indeed we would need to take that extra step. However, in that case the question would have too much complexity to make sense in something like the CAT. So, it is unlikely.


  5. sir, in the 4th ques there will be other cases too where all the 6 digits are 0 and dummy got all 1,2,3,4,5,6,7,8 as we have to find upto 9?

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