NACLO 2026 - Problem OSum Times in Na

Invitational round | 10 points | N/A | Problem statement | Official solution | Tags: Number system

First analyze the structure of the number phrases. These phrases are at least 1 word and at most 3 words:

• 1 word
  • ko
  • jo
  • maka
  • muta
• 2 words
  • ko-jo
  • ko-muta
• 3 words: "X bi-jɔ Y"
  • X:
    • ko
    • jɔkɛ
    • ko-bi-jɔkɛ
    • ko-jo
    • jɔkɛ-jo
    • ko-muta
  • Y:
    • maka
    • mitikijo
    • sɔ
    • muta
    • jo

In the 3-word forms, X always starts with "ko" or "jɔkɛ", strongly suggesting that these are bases. This is supported by ko + ko = ko-jo, which tells us "-jo" means "2". We shall assume that the other 4 Y words are units digits as well, and that "bi-jɔ" performs addition. Let ko = $K$, jɔkɛ = $J$, maka = $a$, mitikijo = $i$, muta = $u$, sɔ = $s$. Hence:

$$\begin{aligned} K + K &= K\times 2 \\ a \times a &= K + a \\ (J + i)\times 2 &= \mathbf{ko‐bi‐jɔkɛ} + s \\ K + i + \mathbf{ko‐bi‐jɔkɛ} + u &= K\times u \\ (J + 2)\times u &= K\times 2 + a \\ J\times 2 + a + J\times 2 + i &= K\times u + u \end{aligned}$$

We are given very little information about possible ranges of each digit, so we must make reasonable assumptions. For starters, $K$ and $J$ are bases. Although any base may exist, bases that are not multiples of 5 are incredibly hard to deal with IRL. Also, they must be greater than the units digits, $a$, $i$, $u$, $s$. I'm going to assume that these units digits are all less than 10 (base 10 seems to be as high as you can get without inventing a subbase, because otherwise you need 19 different numerals).

From (2), $K = a\times (a-1)$. If $K$ is a multiple of 5, then $a$ is either 5 or 6, and $K$ is either 20 or 30.

In (5), RHS evaluates to 45 or 66. If the former, then $u$ is 3 or 5; if the latter, then $u$ is 2, 3, or 6 (since $u$ must be less than $J+2$). In each case:

• $a=5,K=20,u=3$: $J+2=15$, leaving $J=13$. Reject because it's not a natural base.
• $a=5,K=20,u=5$: already a contradiction because $a$ and $u$ cannot be identical.
• $a=6,K=30,u=2$: $J+2=33$, leaving $J=31$. Reject because it's not a natural base.
• $a=6,K=30,u=3$: $J+2=22$, so $J=20$.
• $a=6,K=30,u=6$: again, $a$ and $u$ cannot be identical.

So, assuming our bold assumption that $K$ and $J$ are multiples of 5 is correct, then the only viable answer is $a=6,K=30,u=3,J=20,j=2$. This gives us $i=7$. It's also nice that "jɔkɛ" and "jo", which look similar, stand for 20 and 2 respectively.

Now we only have $s$ and "ko-bi-jɔkɛ" left.

$$\begin{aligned} (20 + 7)\times 2 &= \mathbf{ko‐bi‐jɔkɛ} + s \\ 30 + 7 + \mathbf{ko‐bi‐jɔkɛ} + 3 &= 30\times 3 \end{aligned}$$

Therefore "ko-bi-jɔkɛ" = 90－40=50, which is ko+jɔkɛ. And $s=4$.

Let's recap the number system:

• jo = 2
• muta = 3
• sɔ = 4
• maka = 6
• mitikijo = 7
• jɔkɛ = 20
• jɔkɛ bi-jɔ X = 20 + X
• ko = 30
• ko bi-jɔ X = 30 + X
• ko-bi-jɔkɛ = 50
• ko-bi-jɔkɛ bi-jɔ X = 50 + X
• ko-jo = 60
• ko-jo bi-jɔ X = 60 + X

So in O2, ko-jo-bi-jɔkɛ bi-jɔ muta is (60+20) + 3 = 83.

In O3, we don't know what 40 is. It could be either (30+10) or (20×2). However we don't know what 10 is either, so the only viable answer is that jɔkɛ-jo = 40, and jɔkɛ-jo bi-jɔ sɔ = 44. As for 110, we don't know if the primary base is 20 or 30, since we have jɔkɛ and jɔkɛ-jo but also ko and ko-jo, and ko-bi-jɔkɛ. 110=30×3+20=30+20×4; the answer is given as ko-muta-bi-jɔkɛ, but I'm not certain about the acceptability of ko-bi-jɔkɛ-sɔ. Anyway, assuming the former, 117 = ko-muta-bi-jɔkɛ bi-jɔ mitikijo.