Difference between revisions of "2003 AIME I Problems/Problem 8"
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We then set <math>400+x^2=y^2</math>, where <math>y</math> is an integer. Factoring using difference of squares, we have | We then set <math>400+x^2=y^2</math>, where <math>y</math> is an integer. Factoring using difference of squares, we have | ||
<cmath>400=2^4 \cdot 5^2=(y+x)(y-x).</cmath> | <cmath>400=2^4 \cdot 5^2=(y+x)(y-x).</cmath> | ||
− | Then, noticing that <math>y+x > y-x</math>, we set up several systems of equations involving the factors of <math>400</math>. The | + | Then, noticing that <math>y+x > y-x</math>, we set up several systems of equations involving the factors of <math>400</math>. The second system we set up in this manner, |
<cmath>y+x=2^3 \cdot 5^2</cmath> | <cmath>y+x=2^3 \cdot 5^2</cmath> | ||
<cmath>y-x=2,</cmath> | <cmath>y-x=2,</cmath> | ||
yields the solution <math>y=101, x=99</math>. Plugging back in, we get that <math>a=27 \implies d=9</math>, so the sequence is <math>18,</math> <math>27,</math> <math>36,</math> <math>48,</math> and the answer is <math>\boxed{129}.</math> | yields the solution <math>y=101, x=99</math>. Plugging back in, we get that <math>a=27 \implies d=9</math>, so the sequence is <math>18,</math> <math>27,</math> <math>36,</math> <math>48,</math> and the answer is <math>\boxed{129}.</math> | ||
− | *Note: we do not have to check the other systems since the <math>x</math> and <math>y</math> values obtained via this system yield integers for <math>a</math>, <math>d</math>, and this must be the only possible answer since this is an AIME problem. We | + | *Note: we do not have to check the other systems since the <math>x</math> and <math>y</math> values obtained via this system yield integers for <math>a</math>, <math>d</math>, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :) |
+ | -Fasolinka | ||
+ | |||
+ | ==Solution 3 (Guesswork)== | ||
+ | We represent the values as <math>a-d</math>, <math>a</math>, <math>a+d</math>, and <math>\frac{(a+d)^2}{a}</math> | ||
+ | Take the difference between the first and last values | ||
+ | <cmath>\frac{(a+d)^2}{a}-a+d=30</cmath> | ||
+ | Manipulating the values by expanding and then long division we see | ||
+ | <cmath>\frac{a^2+2ad+d^2}{a}-a+d=30</cmath> | ||
+ | <cmath>\frac{(a+2d)a+d^2}{a}-a+d=30</cmath> | ||
+ | <cmath>a+2d+\frac{d^2}{a}-a+d=30</cmath> | ||
+ | Combining like terms we get | ||
+ | <cmath>3d+\frac{d^2}{a}=30</cmath> | ||
+ | And looking at the values we know that since the second term must be positive (since both a and d are positive), <math>d</math> must be a maximum of 9, which offers 9 possible values (since <math>d</math> must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of <math>d=9</math> at which <math>a=27</math>. The solution is thus <math>\boxed{129}.</math> | ||
+ | * We could also lay down some further parameters for the value of <math>d</math>. as <math>a</math> must be greater than <math>d</math> (for the first term to be positive), we can substitite in a "borderline" value of a=d (which itself is just over the limit). This gives us: | ||
+ | <cmath>3d+\frac{d^2}{d}=30</cmath> | ||
+ | <cmath>3d+d=30</cmath> | ||
+ | <cmath>4d=30</cmath> | ||
+ | Thus, d must be greater than 7.5, which gives us only 2 values left (8 and 9). We can then plug in any of the two to see if an integer value is attainable. In the end, 8 doesn't work and 9 does, which gives the solution of <math>\boxed{129}.</math> | ||
+ | |||
+ | * Note: This solution is similar to Solution 2 | ||
+ | -PMaldonado13 | ||
== See also == | == See also == |
Latest revision as of 23:21, 22 January 2023
Problem 8
In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by . Find the sum of the four terms.
Solution
Denote the first term as , and the common difference between the first three terms as
. The four numbers thus are in the form
.
Since the first and fourth terms differ by , we have that
. Multiplying out by the denominator,
This simplifies to
, which upon rearranging yields
.
Both and
are positive integers, so
and
must have the same sign. Try if they are both positive (notice if they are both negative, then
and
, which is a contradiction). Then,
. Directly substituting and testing shows that
, but that if
then
. Alternatively, note that
or
implies that
, so only
may work. Hence, the four terms are
, which indeed fits the given conditions. Their sum is
.
Postscript
As another option, could be rewritten as follows:
This gives another way to prove , and when rewritten one last time:
shows that must contain a factor of 3.
-jackshi2006
EDIT by NealShrestha:
Note that once we reach this implies
since all other terms are congruent to
.
Solution 2
The sequence is of the form
. Since the first and last terms differ by 30, we have
Let
, where
is an integer. This yields the following:
We then set
, where
is an integer. Factoring using difference of squares, we have
Then, noticing that
, we set up several systems of equations involving the factors of
. The second system we set up in this manner,
yields the solution
. Plugging back in, we get that
, so the sequence is
and the answer is
- Note: we do not have to check the other systems since the
and
values obtained via this system yield integers for
,
, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :)
-Fasolinka
Solution 3 (Guesswork)
We represent the values as ,
,
, and
Take the difference between the first and last values
Manipulating the values by expanding and then long division we see
Combining like terms we get
And looking at the values we know that since the second term must be positive (since both a and d are positive),
must be a maximum of 9, which offers 9 possible values (since
must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of
at which
. The solution is thus
- We could also lay down some further parameters for the value of
. as
must be greater than
(for the first term to be positive), we can substitite in a "borderline" value of a=d (which itself is just over the limit). This gives us:
Thus, d must be greater than 7.5, which gives us only 2 values left (8 and 9). We can then plug in any of the two to see if an integer value is attainable. In the end, 8 doesn't work and 9 does, which gives the solution of
- Note: This solution is similar to Solution 2
-PMaldonado13
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.