AIME 2018 I · 第 2 题

Solution 1

We have these equations: $196a+14b+c=225a+15c+b=222a+37c$. Taking the last two we get $3a+b=22c$. Because $c \neq 0$ otherwise $a \ngtr 0$, and $a \leq 5$, $c=1$.

Then we know $3a+b=22$. Taking the first two equations we see that $29a+14c=13b$. Combining the two gives $a=4, b=10, c=1$. Then we see that $222 \times 4+37 \times1=\boxed{925}$.

Solution 2

We know that $196a+14b+c=225a+15c+b=222a+37c$. Combining the first and third equations give that $196a+14b+c=222a+37c$, or

$$7b=13a+18c$$ The second and third gives $222a+37c=225a+15c+b$, or

$$22c-3a=b$$ $$154c-21a=7b=13a+18c$$ $$4c=a$$ We can have $a=4,8,12,16,20$, but only $a=4$ falls within the possible digits of base $6$. Thus $a=4$, $c=1$, and thus you can find $b$ which equals $10$. Thus, our answer is $4\cdot225+1\cdot15+10=\boxed{925}$. ~SHEN KISLAY KAI 2023

Solution 3 (Official MAA)

The problem is equivalent to finding a solution to the system of Diophantine equations $196a+14b+c=225a+15c+b$ and $225a+15c+b=216a+36c+6a+c,$ where $1\le a\le 5,\,0\le b\le 13,$ and $0\le c\le 5.$ Simplifying the second equation gives $b=22c-3a.$ Substituting for $b$ in the first equation and simplifying then gives $a=4c,$ so $a = 4$ and $c = 1,$ and the base-$10$ representation of $n$ is $222 \cdot 4 + 37 \cdot 1 = 925.$ It may be verified that $b=10\le 13.$

Solution 4 (Simple Modular Arithmetic)

We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\pmod{13}$, we receive $3a+c \equiv 0 \pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We can just check these two, and find that $(a,b,c)=(4,10,1),$ and just plugging in $(a,c)$ into the third expression we receive $888+37=\boxed{925}.$

~SirAppel

Video Solution

https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S