1986 AIME Problems/Problem 10

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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

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

$m\equiv -3194\equiv -86\equiv 136\pmod{222}$

This reduces $m$ to one of 136, 358, 580, 802. But also $a+b+c=\frac{N+m}{222}>\frac{N}{222}>14$ so $a+b+c\geq 15$ . Only one of the values of $m$ satisfies this, namely $358$ .

1986 AIME (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1986 AIME Problems/Problem 10

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Problem

Solution

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