Difference between revisions of "2021 AIME I Problems/Problem 5"
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===Solution 2=== | ===Solution 2=== | ||
Proceed as in solution 1, until we reach <cmath>3x^2+2d^2=xd^2,</cmath>. Write | Proceed as in solution 1, until we reach <cmath>3x^2+2d^2=xd^2,</cmath>. Write | ||
− | <math>d^2=\frac{3a^2}{a-2}</math>, it follows that <math>a-2=3k^2</math> for some (positive) integer k and <math>k \mid a</math>. \ | + | <math>d^2=\frac{3a^2}{a-2}</math>, it follows that <math>a-2=3k^2</math> for some (positive) integer k and <math>k \mid a</math>. \par |
− | Taking both sides modulo <math>k</math>, <math>-2 \equiv 0 \pmod{k}</math>, so <math>k \mid 2 \rightarrow k=1,2</math>. | + | Taking both sides modulo <math>k</math>, <math>-2 \equiv 0 \pmod{k}</math>, so <math>k \mid 2 \rightarrow k=1,2</math>. \par |
When k=1, x=5 and d=5. When k=2, x=14 and d=7. | When k=1, x=5 and d=5. When k=2, x=14 and d=7. | ||
Revision as of 23:41, 11 March 2021
Problem
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
Solution
Let the terms be ,
, and
. Then we want
, or
. Rearranging, we get
. Simplifying further,
. Looking at this second equation, since the right side must be an integer,
must equal
. Looking at the first equation, we see
since
is positive. This means we must test
. After testing these, we see that only
and
work which give
and
respectively. Thus the answer is
.
~JHawk0224
Solution 1
Let the common difference be and let the middle term be
. Then, we have that the sequence is
This means that the sum of the sequence is
We know that this must be equal to
so we can write that
and it follows that
Now, we can treat as a constant and use the quadratic formula to get
We can factor pull
out of the square root to get
Here, it is easy to test values of
. We find that
and
are the only positive integer values of
that make
a positive integer.
gives
and
, but we can ignore the latter.
gives
, as well as a fraction which we can ignore.
Since and
are the only two solutions and we want the sum of the third terms, our answer is
. -BorealBear
Solution 2
Proceed as in solution 1, until we reach . Write
, it follows that
for some (positive) integer k and
. \par
Taking both sides modulo
,
, so
. \par
When k=1, x=5 and d=5. When k=2, x=14 and d=7.
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.