Difference between revisions of "1986 AIME Problems/Problem 10"
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<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 <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>. | + | 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 === | ===Solution 2 === | ||
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=== Solution 3 === | === Solution 3 === | ||
− | Let <math>n=abc</math> then | + | Let <math>n=abc</math> then |
− | <cmath> | + | <cmath>N=222(a+b+c)-n</cmath> |
− | N | + | <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 == | == See also == |
Latest revision as of 16:51, 1 October 2023
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
.
Recall that
refer to the digits the three digit number
, 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
Solution 4
The sum of the five numbers is We can see that
(mod
) and
(mod
) so we need to make sure that
(mod
) by some testing. So we let
Then, we know that so only
lie in the interval
When we test , impossible
When we test
When we test , well, it's impossible
The answer is then
~bluesoul
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.