1993 AIME Problems/Problem 6
Denote the first of each of the series of consecutive integers as . Therefore, . Simplifying, . The relationship between suggests that is divisible by . Also, , so is divisible by . We find that the least possible value of , so the answer is .
Let the desired integer be . From the information given, it can be determined that, for positive integers :
This can be rewritten as the following congruences:
Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is
Let be the desired integer. From the given information, we have here, and are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have as the 4th term of the sequence. Since, is a multiple of and it is also a multiple of Hence, for some So, we have It follows that is the smallest integer that can be represented in such a way.
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