Difference between revisions of "1993 AIME Problems/Problem 6"
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=== Solution 5 === | === Solution 5 === | ||
− | 9 x | + | First note that the integer clearly must be divisible by <math>9</math> and <math>11</math> since we can use the "let the middle number be x" trick. Let the number be <math>99k</math> for some integer <math>k.</math> Now let the <math>10</math> numbers be <math>x,x+1, \cdots x+9.</math> We have <math>10x+45 = 99k.</math> Taking mod <math>5</math> yields <math>k \equiv 0 \pmod{5}.</math> Since <math>k</math> is positive, we take <math>k=5</math> thus obtaining <math>99 \cdot 5 = \boxed{495}</math> as our answer. |
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=5|num-a=7}} | {{AIME box|year=1993|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:28, 21 September 2020
Contents
Problem
What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?
Solution
Solution 1
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 .
Solution 2
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
Solution 3
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.
Solution 4
By the method in Solution 1, we find that the number can be written as for some integers . From this, we can see that must be divisible by 9, 5, and 11. This means must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that cannot be divisible by 10, so must equal . Solution by Zeroman.
Solution 5
First note that the integer clearly must be divisible by and since we can use the "let the middle number be x" trick. Let the number be for some integer Now let the numbers be We have Taking mod yields Since is positive, we take thus obtaining as our answer.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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