1986 AIME Problems/Problem 10

Problem

In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if $N= 3194$ .

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 $a+b+c\geq 15$ . Of the four options, only $m = \boxed{358}$ satisfies this inequality.

Solution 2

As in Solution 1, $3194 + m = 222(a+b+c)$ . Modulo $222$ , 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 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 $m = \boxed{358}$ .

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1986 AIME Problems/Problem 10

Contents

Problem

Solution

Solution 1

Solution 2

See also