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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 (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum, <math>\displaystyle N</math>. If told the value of <math>\displaystyle N</math>, the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if <math>\displaystyle 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>.
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== Solution ==
 
== Solution ==
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{{solution}}
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===Solution 1 ===
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Let <math>m</math> be the number <math>100a+10b+c</math>. Observe that <math>3194+m=222(a+b+c)</math> so
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<cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath>
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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>.
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Of the four options, only <math>m = \boxed{358}</math> satisfies this inequality.
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===Solution 2 ===
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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>.
 +
We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so
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<cmath>3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.</cmath>
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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>.
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 +
 
 +
=== Solution 3 ===
 +
 
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Let <math>n=abc</math> then
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<cmath>N=222(a+b+c)-n</cmath>
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<cmath>N=222(a+b+c)-100a-10b-c=3194</cmath>
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Since <math>0<100a+10b+c<1000</math>, we get the inequality
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<cmath>N<222(a+b+c)<N+1000</cmath>
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<cmath>3194<222(a+b+c)<4194</cmath>
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<cmath>14<a+b+c<19</cmath>
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Checking each of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> from each <math>222(a+b+c)</math>, we quickly find <math>n=\boxed{358}</math>
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 +
~ Nafer
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== Solution 4 ==
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The sum of the five numbers is <math>222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194</math> We can see that <math>3194 \equiv 8 </math> （mod <math>9</math>） and <math>122 \equiv 5</math> （mod <math>9</math>） so we need to make sure that <math>a+b+c \equiv 7</math> （mod <math>9</math>） by some testing. So we let <math>a+b+c=9k+7</math>
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Then, we know that <math>1\leq a+b+c \leq 27</math> so only <math>7,16,25</math> lie in the interval
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When we test <math>a+b+c=25, 10b+11c=16</math>, impossible
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When we test <math>a+b+c=16, 10b+11c=138, b=5,c=8,a=3</math>
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When we test <math>a+b+c=7, 10b+11c=260</math>, well, it's impossible
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The answer is <math>\boxed{358}</math> then
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 +
~bluesoul
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== See also ==
 
== See also ==
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* [[1986 AIME Problems]]
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{{AIME box|year=1986|num-b=9|num-a=11}}
  
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{{AIME box|year=1986|num-b=9|num-a=11}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 20:41, 5 July 2022

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 \[N=222(a+b+c)-n\] \[N=222(a+b+c)-100a-10b-c=3194\] Since $0<100a+10b+c<1000$, we get the inequality \[N<222(a+b+c)<N+1000\] \[3194<222(a+b+c)<4194\] \[14<a+b+c<19\] Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$, we quickly find $n=\boxed{358}$

~ Nafer

Solution 4

The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$） and $122 \equiv 5$ （mod $9$） so we need to make sure that $a+b+c \equiv 7$ （mod $9$） by some testing. So we let $a+b+c=9k+7$

Then, we know that $1\leq a+b+c \leq 27$ so only $7,16,25$ lie in the interval

When we test $a+b+c=25, 10b+11c=16$, impossible

When we test $a+b+c=16, 10b+11c=138, b=5,c=8,a=3$

When we test $a+b+c=7, 10b+11c=260$, well, it's impossible

The answer is $\boxed{358}$ then

~bluesoul

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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