Difference between revisions of "1986 AIME Problems/Problem 10"
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== Solution == | == 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 | 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> | <cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath> | ||
− | This reduces <math>m</math> to one of 136, 358, 580, 802. 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 136, 358, 580, 802. 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 = 222(a+b+c)</math>. | ||
+ | Modulo <math>222</math>, 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 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>. | ||
== See also == | == See also == |
Revision as of 21:31, 1 September 2016
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, . If told the value of
, the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if
.
Solution
Solution 1
Let be the number
. Observe that
so
This reduces to one of 136, 358, 580, 802. But also
so
.
Of the four options, only
satisfies this inequality.
Solution 2
As in Solution 1, .
Modulo
, 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 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
.
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.