1
$\begingroup$

In decimal representation, we call an integer k-carboxylic if and only if it can be represented as a sum of k distinct integers, all of them greater than 9, whose digits are the same. For instance, 2008 is 5-carboxylic because $2008 = 1111 + 666 + 99 + 88 + 44$. Find, with an example, the smallest integer k such that 8002 is k-carboxylic.

I started by writing $8002 = 1111a + 111b + 11c$. To minimize k, we want to maximize a, then b and c. But it seems there are still a lot of combinations. Help?

|cite|improve this question
$\endgroup$
  • $\begingroup$ You are implicitly assuming you can do it with just 3 parameters $a,b,c$. This need not be true, as suggested by the example they gave for 2008. You may need several distinct contributions from multiples of 1111; and 111; and 11. $\endgroup$ – zibadawa timmy Nov 12 '13 at 19:35
  • $\begingroup$ There is not really any difference so long as you step back and allow $a,b,c$ to be any integer between $0$ and $9$ inclusive. Obviously, you can't have a series of more than $4$ $1$'s. If the smallest number $k$ you seek for example where to be $2$, then this is the case where $b=0$ or $c=0$. So approaching from this problem is not necessarily bad. $\endgroup$ – mathematics2x2life Nov 12 '13 at 19:53
  • $\begingroup$ @mathematics2x2life Except as Daniel's answer and my statement say, in general you will require several distinct multiples of 11, and possibly even 111. It's not a 3-carboxylic number, so you're going to need several of at least one of them. $\endgroup$ – zibadawa timmy Nov 12 '13 at 21:11
4
$\begingroup$

Since we have only multiples of $11$, $111$, and $1111$ to consider, it is relatively easy. $1111$ is a multiple of $11$, so we can determine the required number of $111$s (modulo $11$) immediately, $b \equiv 8002 \pmod{11}$, which tells us $b \equiv 5 \pmod{11}$.

For the remaining $8002 - 555 = 7447$, we use $6666$, and for the remainder $781 = 71\cdot 11$, we use $7$ times $99$ plus $88$ for a total of $10$ summands.

It is then easy to check that we don't get fewer summands if we try $b = 16$, which requires two three-digit summands, or $b = 27$, which requires three three-digit summands, or $b = 38, \, 49,\, \dotsc$, so indeed $10$ is the minimal $k$ for which $8002$ is $k$-carboxylic.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.