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

Revision as of 18:03, 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 $N=222(a+b+c)-n$ $N=222(a+b+c)-100a-10b-c=3194$ 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

@@ Line 25: / Line 25: @@
 <cmath>N=222(a+b+c)-n</cmath>
 <cmath>N=222(a+b+c)-100a-10b-c=3194</cmath>
-Since <math><100a+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>3194<222(a+b+c)<4194</cmath>

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
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