1986 AIME Problems/Problem 10
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 .
Let be the number . Observe that so
This reduces to one of . But also so . Of the four options, only satisfies this inequality.
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 .
Let then Since , we get the inequality Checking each of the multiples of from to by subtracting from each , we quickly find
|1986 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|