Difference between revisions of "2013 AIME II Problems/Problem 6"
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− | The difference between consecutive integral squares must be greater than 1000. <math>(x+1)^2-x^2\geq1000</math>, so <math>x\geq\frac{999}{2}\implies x\geq500</math>. <math>x=500</math> does not work, so <math>x>500</math>. Let <math>n=x-500</math> | + | The difference between consecutive integral squares must be greater than 1000. <math>(x+1)^2-x^2\geq1000</math>, so <math>x\geq\frac{999}{2}\implies x\geq500</math>. <math>x=500</math> does not work, so <math>x>500</math>. Let <math>n=x-500</math>. By inspection, <math>n^2</math> should end in a number close to but less than 1000 such that there exists <math>1000N</math> within the difference of the two squares. Examine when <math>n^2=1000</math>. Then, <math>n=10\sqrt{10}</math>. One example way to estimate <math>\sqrt{10}</math> follows. |
<math>3^2=9</math>, so <math>10=(x+3)^2=x^2+6x+9</math>. <math>x^2</math> is small, so <math>10=6x+9</math>. <math>x=1/6\implies \sqrt{10}\approx 19/6</math>. This is 3.16. | <math>3^2=9</math>, so <math>10=(x+3)^2=x^2+6x+9</math>. <math>x^2</math> is small, so <math>10=6x+9</math>. <math>x=1/6\implies \sqrt{10}\approx 19/6</math>. This is 3.16. |
Revision as of 09:29, 1 January 2023
Contents
Problem 6
Find the least positive integer such that the set of
consecutive integers beginning with
contains no square of an integer.
Solutions
Solution 1
The difference between consecutive integral squares must be greater than 1000. , so
.
does not work, so
. Let
. By inspection,
should end in a number close to but less than 1000 such that there exists
within the difference of the two squares. Examine when
. Then,
. One example way to estimate
follows.
, so
.
is small, so
.
. This is 3.16.
Then, .
, so
could be
. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are
and
. Checking,
and
.
straddles the two squares, which have a difference of 1063. The difference has been minimized, so
is minimized
~BJHHar
Solution 2
Let us first observe the difference between and
, for any arbitrary
.
. So that means for every
, the difference between that square and the next square have a difference of
. Now, we need to find an
such that
. Solving gives
, so
. Now we need to find what range of numbers has to be square-free:
have to all be square-free.
Let us first plug in a few values of
to see if we can figure anything out.
,
, and
. Notice that this does not fit the criteria, because
is a square, whereas
cannot be a square. This means, we must find a square, such that the last
digits are close to
, but not there, such as
or
. Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are
, so all we need to do is addition. After making a list, we find that
, while
. It skipped
, so our answer is
.
Solution 3
Let be the number being squared. Based on the reasoning above, we know that
must be at least
, so
has to be at least
. Let
be
. We can write
as
, or
. We can disregard
and
, since they won't affect the last three digits, which determines if there are any squares between
. So we must find a square,
, such that it is under
, but the next square is over
. We find that
gives
, and so
. We can be sure that this skips a thousand because the
increments it up
each time. Now we can solve for
:
, while
. We skipped
, so the answer is
.
Solution 4
The goal is to find the least such that
where
.
Combining the two inequalities leads to .
Let , where
, then the inequalities become,
, and
For , one can verify that
is the unique integer satisfying the inequalities.
For ,
,
i.e., , a contradiction.
Note leads to larger
(s).
Hence, the answer is .
~yuxiaomatt
See Also
Very similar to 2016 AMC 12 A Problem 25: https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25
2013 AIME II (Problems • Answer Key • Resources) | ||
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