1986 AIME Problems/Problem 10

Revision as of 20:45, 26 March 2007 by Scorpius119 (talk | contribs) (solution added)


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, $\displaystyle N$. If told the value of $\displaystyle N$, the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if $\displaystyle N= 3194$.


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 $358$.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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