AIME 1986 · 第 10 题

Solution

Solution 1

Let $m$ be the number $100a+10b+c$. Observe that $3194+m=222(a+b+c)$ so

$$m\equiv -3194\equiv -86\equiv 136\pmod{222}$$ This reduces $m$ to one of $136, 358, 580, 802$. But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$. Recall that $a, b, c$ refer to the digits the three digit number $(abc)$, so of the four options, only $m = \boxed{358}$ satisfies this inequality.

Solution 2

As in Solution 1, $3194 + m \equiv 222(a+b+c) \pmod{222}$, and so as above we get $m \equiv 136 \pmod{222}$. We can also take this equation modulo $9$; note that $m \equiv a+b+c \pmod{9}$, so

$$3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.$$ Therefore $m$ is $7$ mod $9$ and $136$ mod $222$. There is a shared factor in $3$ in both, but the Chinese Remainder Theorem still tells us the value of $m$ mod $666$, namely $m \equiv 358$ mod $666$. We see that there are no other 3-digit integers that are $358$ mod $666$, so $m = \boxed{358}$.

Solution 3

Let $n=abc$ then

$$N=222(a+b+c)-n$$ $$N=222(a+b+c)-100a-10b-c=3194$$ Since $0<100a+10b+c<1000$, we get the inequality

$$N<222(a+b+c)<N+1000$$ $$3194<222(a+b+c)<4194$$ $$14<a+b+c<19$$ Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$, we quickly find $n=\boxed{358}$

~ Nafer

Solution 4

The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$） and $122 \equiv 5$ （mod $9$） so we need to make sure that $a+b+c \equiv 7$ （mod $9$） by some testing. So we let $a+b+c=9k+7$

Then, we know that $1\leq a+b+c \leq 27$ so only $7,16,25$ lie in the interval

When we test $a+b+c=25, 10b+11c=16$, impossible

When we test $a+b+c=16, 10b+11c=138, b=5,c=8,a=3$

When we test $a+b+c=7, 10b+11c=260$, well, it's impossible

The answer is $\boxed{358}$ then

~bluesoul