Difference between revisions of "1986 AIME Problems/Problem 10"
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=== Solution 3 === | === Solution 3 === | ||
− | Let <math>n=abc</math> then < | + | 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> | <cmath>N=222(a+b+c)-100a-10b-c=3194</cmath> | ||
Since <math><100a+10b+c<1000</math>, we get the inequality | Since <math><100a+10b+c<1000</math>, we get the inequality | ||
Line 28: | Line 29: | ||
<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> | ||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Revision as of 19:02, 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.