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

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

@@ Line 1: / Line 1: @@
 == Problem ==
+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 1 ===
+Let <math>m</math> be the number <math>100a+10b+c</math>. Observe that <math>3194+m=222(a+b+c)</math> so
+<cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath>
+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>.
+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 ===
+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
+<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 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>.
+=== Solution 3 ===
+Let <math>n=abc</math> then
+<cmath>N=222(a+b+c)-n</cmath>
+<cmath>N=222(a+b+c)-100a-10b-c=3194</cmath>
+Since <math>0<100a+10b+c<1000</math>, we get the inequality
+<cmath>N<222(a+b+c)<N+1000</cmath>
+<cmath>3194<222(a+b+c)<4194</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>
+~ 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 ==
-* [[1984 AIME Problems]]
+{{AIME box|year=1986|num-b=9|num-a=11}}
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

1986 AIME (Problems • Answer Key • Resources)
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