Difference between revisions of "1993 AIME Problems/Problem 6"
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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? | 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? | ||
− | == | + | ==Solutions== |
=== Solution 1 === | === Solution 1 === | ||
Denote the first of each of the series of consecutive integers as <math>a,\ b,\ c</math>. Therefore, <math>n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55</math>. Simplifying, <math>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>. | Denote the first of each of the series of consecutive integers as <math>a,\ b,\ c</math>. Therefore, <math>n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55</math>. Simplifying, <math>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>. | ||
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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}} |
Revision as of 18:06, 9 December 2023
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?
Solutions
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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