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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 <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>.
+
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.
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Recall that <math>a, b, c</math> refer to the digits the three digit number <math>(abc)</math>, so of the four options, only <math>m = \boxed{358}</math> satisfies this inequality.
  
 
===Solution 2 ===
 
===Solution 2 ===
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=== Solution 3 ===
 
=== Solution 3 ===
  
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Let <math>n=abc</math> then  
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Let <math>n=abc</math> then
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<cmath>N=222(a+b+c)-n</cmath>
 
<cmath>N=222(a+b+c)-100a-10b-c=3194</cmath>
 
<cmath>N=222(a+b+c)-100a-10b-c=3194</cmath>
−
Since <math><0100a+10b+c<1000</math>, we get the inequality
+
Since <math>0<100a+10b+c<1000</math>, we get the inequality
 
<cmath>N<222(a+b+c)<N+1000</cmath>
 
<cmath>N<222(a+b+c)<N+1000</cmath>
 
<cmath>3194<222(a+b+c)<4194</cmath>
 
<cmath>3194<222(a+b+c)<4194</cmath>
 
<cmath>14<a+b+c<19</cmath>
 
<cmath>14<a+b+c<19</cmath>
 +
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
 +
 +
== Solution 4 ==
 +
 +
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>
 +
 +
Then, we know that <math>1\leq a+b+c \leq 27</math> so only <math>7,16,25</math> lie in the interval
 +
 +
When we test <math>a+b+c=25, 10b+11c=16</math>, impossible
 +
 +
When we test <math>a+b+c=16, 10b+11c=138, b=5,c=8,a=3</math>
 +
 +
When we test <math>a+b+c=7, 10b+11c=260</math>, well, it's impossible
 +
 +
The answer is <math>\boxed{358}</math> then
 +
 +
~bluesoul
  
 
== See also ==
 
== See also ==

Latest revision as of 15:51, 1 October 2023

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 $a+b+c\geq 15$. Recall that $a, b, c$ refer to the digits the three digit number $(abc)$, so 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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