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 === |
Latest revision as of 15:51, 1 October 2023
Contents
[hide]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.