Difference between revisions of "1993 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | + | What is the smallest [[positive]] [[integer]] than 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 == | ||
| − | + | Denote the first of each of the series of consecutive integers as <math>a,\ b,\ c</math>. Therefore, <math>\displaystyle n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55</math>. Simplifying, <math>\displaystyle 9a = 10b + 9 = 11c + 19</math>. The relationship between <math>a,\ b</math> suggests that <math>b</math> is divisible by <math>9</math>. Also, <math>10b -10 = 10(b-1) = 11c</math>, so <math>b-1</math> is divisible by <math>11</math>. We find that the least possible value of <math>b = 45</math>, so the answer is <math>10(45) + 45 = 495</math>. | |
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=5|num-a=7}} | {{AIME box|year=1993|num-b=5|num-a=7}} | ||
Revision as of 20:34, 27 March 2007
Problem
What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?
Solution
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 
.
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 | ||
