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

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== Problem ==
 
== Problem ==
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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 <tt>(acb)</tt>, <tt>(bca)</tt>, <tt>(bac)</tt>, <tt>(cab)</tt>, and <tt>(cba)</tt>, to add these five numbers, and to reveal their sum, <math>N</math>. If told the value of <math>N</math>, the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if <math>N= 3194</math>.
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In a parlor game, the magician asks one of the participants to think of a three digit number <math>(abc)</math> where <math>a</math>, <math>b</math>, and <math>c</math> represent digits in base <math>10</math> in the order indicated. The magician then asks this person to form the numbers <math>(acb)</math>, <math>(bca)</math>, <math>(bac)</math>, <math>(cab)</math>, and <math>(cba)</math>, to add these five numbers, and to reveal their sum, <math>N</math>. If told the value of <math>N</math>, the magician can identify the original number, <math>(abc)</math>. Play the role of the magician and determine <math>(abc)</math> if <math>N= 3194</math>.
  
 
== Solution ==
 
== Solution ==
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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>.
+
This reduces <math>m</math> to one of <math>136, 358, 580, 802</math>. But also <math>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>.
 
Of the four options, only <math>m = \boxed{358}</math> satisfies this inequality.
 
Of the four options, only <math>m = \boxed{358}</math> satisfies this inequality.
  
 
===Solution 2 ===
 
===Solution 2 ===
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As in Solution 1, <math>3194 + m = 222(a+b+c)</math>.
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As in Solution 1, <math>3194 + m \equiv 222(a+b+c) \pmod{222}</math>, and so as above we get <math>m \equiv 136 \pmod{222}</math>.
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Modulo <math>222</math>, as above we get <math>m \equiv 136 \pmod{222}</math>.
 
 
We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so
 
We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so
  
 
<cmath>3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.</cmath>
 
<cmath>3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.</cmath>
  
−
Therefore <math>m</math> is <math>7</math> mod <math>9</math> and <math>136</math> mod <math>222</math>. There is a shared factor in <math>3</math> in both, but Chinese remainder theorem still tells us the value of <math>m</math> mod <math>666</math>, namely <math>m \equiv 358</math> mod <math>666</math>. We see that there are no other 3-digit integers that are <math>358</math> mod <math>666</math>, so <math>m = \boxed{358}</math>.
+
Therefore <math>m</math> is <math>7</math> mod <math>9</math> and <math>136</math> mod <math>222</math>. There is a shared factor in <math>3</math> in both, but the Chinese Remainder Theorem still tells us the value of <math>m</math> mod <math>666</math>, namely <math>m \equiv 358</math> mod <math>666</math>. We see that there are no other 3-digit integers that are <math>358</math> mod <math>666</math>, so <math>m = \boxed{358}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:56, 2 April 2018

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 $(abc)$ 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, $3194 + m \equiv 222(a+b+c) \pmod{222}$, 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 $m = \boxed{358}$.

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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