Difference between revisions of "1993 AIME Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
| − | Let the desired integer be <math>n</math>. From the information given, it can be determined that, for integers <math>a, \ b, \ c</math>: | + | Let the desired integer be <math>n</math>. From the information given, it can be determined that, for positive integers <math>a, \ b, \ c</math>: |
<math>n = 9a + 36 = 10b + 45 = 11c + 55</math> | <math>n = 9a + 36 = 10b + 45 = 11c + 55</math> | ||
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This can be rewritten as the following congruences: | This can be rewritten as the following congruences: | ||
| − | <math>n \equiv 0 \pmod{9}</math> | + | <math>n \equiv 0 \pmod{9}</math> and <math>n \geq 36</math> |
| − | <math>n \equiv 5 \pmod{10}</math> | + | <math>n \equiv 5 \pmod{10}</math> and <math>n \geq 45</math> |
| − | <math>n \equiv 0 \pmod{11}</math> | + | <math>n \equiv 0 \pmod{11}</math> and <math>n \geq 55</math> |
| − | 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 interpretted from the 2nd congruence) is <math>\boxed{495}</math> | + | 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 interpretted from the 2nd congruence) is 495. This also satisfies the size restrictions, and is thus the answer. <math>\boxed{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 22:27, 10 July 2008
Contents
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 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:
and 
and 
and 
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 interpretted from the 2nd congruence) is 495. This also satisfies the size restrictions, and is thus the 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 | ||
