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Difference between revisions of "1986 AIME Problems/Problem 10"

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<cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath>
 
<cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath>
  
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This reduces <math>m</math> to one of 136, 358, 580, 802. But also <math>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. Only one of the values of <math>m</math> satisfies this, namely <math>\boxed{358}</math>.
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This reduces <math>m</math> to one of 136, 358, 580, 802. But also <math>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. Further, note that neither <math>b</math> nor <math>c</math> may be equal to 0 due to their locations in their respective hundred's places. Thus, only one of the values of <math>m</math> satisfies this, namely <math>\boxed{358}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:44, 25 August 2015

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 $3194+m=222(a+b+c)$ 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 $a+b+c\geq 15$. Further, note that neither $b$ nor $c$ may be equal to 0 due to their locations in their respective hundred's places. Thus, only one of the values of $m$ satisfies this, namely $\boxed{358}$.

See also

1986 AIME (Problems • Answer Key • Resources)
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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