Difference between revisions of "2013 AIME II Problems/Problem 6"
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==Solution 4== | ==Solution 4== | ||
We want to find the least <math>N \in \mathbb{N}</math> such that <math>\exists m \in \mathbb{N}</math> where <math>m^2 < 1000N, 1000N + 1000 < (m+1)^2</math>. | We want to find the least <math>N \in \mathbb{N}</math> such that <math>\exists m \in \mathbb{N}</math> where <math>m^2 < 1000N, 1000N + 1000 < (m+1)^2</math>. | ||
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+ | |||
+ | Combining the two inequalities, we have, | ||
+ | |||
+ | |||
+ | <math>(m+1)^2 > m^2 + 1000</math> | ||
+ | |||
+ | |||
+ | <math>2m + 1 > 1000</math> | ||
+ | |||
+ | |||
+ | <math>m > 499.5</math> | ||
+ | |||
+ | |||
+ | Since <math>m \in \mathbb{N}, m \geq 500,</math> | ||
+ | |||
+ | |||
+ | Let <math>m = k + 500</math>, where <math>k \in \mathbb{W}</math>. | ||
+ | |||
+ | |||
+ | Then, the inequalities become, | ||
+ | |||
+ | |||
+ | <math>N > \frac{(k+500)^2}{1000} = \frac{k^2}{1000} + k + 250</math>, and | ||
+ | |||
+ | |||
+ | <math>N < \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.</math> | ||
+ | |||
+ | |||
+ | Since we want to minimize <math>N</math>, we want the minimum <math>k</math> such that there exists a | ||
+ | positive integer between <math>\frac{k^2}{1000} + k + 250</math> and <math>\frac{(k+1)^2}{1000} + k + 250.</math> Since <math>k + 250 \in \mathbb{N}</math>, we need the minimum <math>k</math> such that there | ||
+ | exists a positive integer between <math>\frac{k^2}{1000}</math> and <math>\frac{(k+1)^2}{1000}</math>. It is now trivial to see that the minimum <math>k</math> is <math>31</math>, since <math>31^2 = 961</math>, <math>32^2 = 1024</math>. This produces <math>N = 282</math> as the minimum possible. | ||
==See Also== | ==See Also== |
Revision as of 14:37, 21 July 2021
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
. The sum of the square of
and a number a little over 1000 must result in a new perfect square. 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
We want to find the least such that
where
.
Combining the two inequalities, we have,
Since
Let , where
.
Then, the inequalities become,
, and
Since we want to minimize , we want the minimum
such that there exists a
positive integer between
and
Since
, we need the minimum
such that there
exists a positive integer between
and
. It is now trivial to see that the minimum
is
, since
,
. This produces
as the minimum possible.
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) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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