Difference between revisions of "2013 AIME II Problems/Problem 6"
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Find the least positive integer <math>N</math> such that the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer. | Find the least positive integer <math>N</math> such that the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer. | ||
+ | |||
+ | ==Solution== | ||
+ | Let us first observe the difference between <math>x^2</math> and <math>(x+1)^2</math>, for any arbitrary <math>x\ge 0</math>. <math>(x+1)^2-x^2=2x+1</math>. So that means for every <math>x\ge 0</math>, the difference between that square and the next square have a difference of <math>2x+1</math>. Now, we need to find an <math>x</math> such that <math>2x+1\ge 1000</math>. Solving gives <math>x\ge \frac{999}{2}</math>, so <math>x\ge 500</math>. Now we need to find what range of numbers has to be square-free: <math>\overline{N000}\rightarrow \overline{N999}</math> have to all be square-free. | ||
+ | Let us first plug in a few values of <math>x</math> to see if we can figure anything out. <math>x=500</math>, <math>x^2=250000</math>, and <math>(x+1)^2=251001</math>. Notice that this does not fit the criteria, because <math>250000</math> is a square, whereas <math>\overline{N000}</math> cannot be a square. This means, we must find a square, such that the last <math>3</math> digits are close to <math>1000</math>, but not there, such as <math>961</math> or <math>974</math>. 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 <math>2x+1</math>, so all we need to do is addition. After making a list, we find that <math>531^2=281961</math>, while <math>532^2=283024</math>. It skipped <math>282000</math>, so our answer is <math>\boxed{282}</math> |
Revision as of 22:00, 4 April 2013
Find the least positive integer such that the set of
consecutive integers beginning with
contains no square of an integer.
Solution
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