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

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(→‎Solution 3)
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Let <math>n=abc</math> then  
 
Let <math>n=abc</math> then  
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<cmath>N=222(a+b+c)-n=222(a+b+c)-100a-10b-c</cmath>
+
<cmath>\begin{align*}
 +
N&=222(a+b+c)-n\\
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&=222(a+b+c)-100a-10b-c\\
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\end{align*}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 17:51, 22 August 2019

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 $m = \boxed{358}$.


Solution 3

Let $n=abc$ then \begin{align*} N&=222(a+b+c)-n\\ &=222(a+b+c)-100a-10b-c\\ \end{align*}

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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