Difference between revisions of "1986 AIME Problems/Problem 10"
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== Problem == | == 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, <math> | + | 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 == | ||
− | Let <math>m</math> be the number <math>100a+10b+c</math>. Observe that <math> | + | ===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>. | ||
+ | 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 | ||
− | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=1986|num-b=9|num-a=11}} | |
− | {{ | + | [[Category:Intermediate Number Theory Problems]] |
+ | {{MAA Notice}} |
Revision as of 19:03, 22 August 2019
Problem
In a parlor game, the magician asks one of the participants to think of a three digit number where , , and represent digits in base in the order indicated. The magician then asks this person to form the numbers , , , , and , to add these five numbers, and to reveal their sum, . If told the value of , the magician can identify the original number, . Play the role of the magician and determine if .
Solution
Solution 1
Let be the number . Observe that so
This reduces to one of . But also so . Of the four options, only satisfies this inequality.
Solution 2
As in Solution 1, , and so as above we get . We can also take this equation modulo ; note that , so
Therefore is mod and mod . There is a shared factor in in both, but the Chinese Remainder Theorem still tells us the value of mod , namely mod . We see that there are no other 3-digit integers that are mod , so .
Solution 3
Let then Since , we get the inequality Checking each of the multiples of from to by subtracting from each , we quickly find
~ Nafer
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.