Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 11"
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First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of <math>4</math> when divided by <math>5</math>. | First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of <math>4</math> when divided by <math>5</math>. | ||
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− | Finally, the sequence of such squares is | + | Suppose a perfect square <math>m^2</math> is in this arithmetic progression. Observe that the remainders when <math>0^2</math>, <math>1^2</math>, <math>2^2</math>, <math>3^2</math>, and <math>4^2</math> are divided by <math>5</math> are <math>0</math>, <math>1</math>, <math>4</math>, <math>4</math>, and <math>1</math>, respectively. Furthermore, for any integer <math>m</math>, <cmath>(m+5)^2 = m^2 + 10m + 25 = m^2 + 5(2m + 5),</cmath> and so <math>(m+5)^2</math> and <math>m^2</math> leave the same remainder when divided by <math>5</math>. It follows that the perfect squares in this arithmetic progression are exactly the numbers of the form <math>(5k+2)^2</math> and <math>(5k+3)^2</math>, respectively. |
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+ | Finally, the sequence of such squares is <cmath>(5\cdot 0 + 2)^2, (5\cdot 0 + 3)^2, (5\cdot 1 + 2)^2, (5\cdot 1 + 3)^2,\cdots.</cmath> | ||
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+ | In particular, the first and second such squares are associated with <math>k=1</math>, the third and fourth are associated with <math>k=2</math>, and so on. It follows that the <math>37^{\text{th}}</math> such number, which is associated with <math>k=18</math>, is <cmath>(5\cdot 18 + 2)^2 = 92^2 = 9409.</cmath> | ||
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+ | Therefore the arithmetic progression must not reach <math>8464</math>. This means the desired answer is <math>\boxed{8459}.</math> ~djmathman | ||
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+ | ==Solution 2== | ||
+ | We examine all perfect squares ending in <math>4</math> or <math>9</math> are part of our sequence, so for every cycle of <math>10</math> perfect squares, exactly <math>4</math> are included. This means <math>9</math> cycles are included, which goes until <math>90^2=8100</math>. Now, note <math>92^2=8464</math> is not part of our sequence, but is the <math>37</math>th perfect square. Therefore, <math>5</math> below this yields <math>\boxed{8459}</math>, which is the answer. | ||
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+ | ~Geometry285 | ||
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+ | ==Solution 3== | ||
+ | Since <math>9-4=5,</math> this arithmetic progression has common ratio 5. Thus, all terms in it are in the form <math>4+5n.</math> Taking the modulo 5, all have either <math>1^2\equiv1\pmod5,2^2\equiv4\pmod5,3^2\equiv4\pmod5,4^2\equiv1\pmod5</math> or <math>5^2\equiv0\pmod5.</math> Thus all integers of the form <math>(2k+2)^2,(2k+3)^2</math> are in the arithmetic progression and are perfect squares. This means that the 37 perfect square in the progression is <math>92^2.</math> This also implies that the maximum value of <math>n</math> is <math>92^2-5=\boxed{8459}</math> | ||
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+ | ~pinkpig | ||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||
+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 21:15, 11 July 2021
Problem
For some , the arithmetic progression
has exactly
perfect squares. Find the maximum possible value of
Solution
First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of when divided by
.
Suppose a perfect square is in this arithmetic progression. Observe that the remainders when
,
,
,
, and
are divided by
are
,
,
,
, and
, respectively. Furthermore, for any integer
,
and so
and
leave the same remainder when divided by
. It follows that the perfect squares in this arithmetic progression are exactly the numbers of the form
and
, respectively.
Finally, the sequence of such squares is
In particular, the first and second such squares are associated with , the third and fourth are associated with
, and so on. It follows that the
such number, which is associated with
, is
Therefore the arithmetic progression must not reach . This means the desired answer is
~djmathman
Solution 2
We examine all perfect squares ending in or
are part of our sequence, so for every cycle of
perfect squares, exactly
are included. This means
cycles are included, which goes until
. Now, note
is not part of our sequence, but is the
th perfect square. Therefore,
below this yields
, which is the answer.
~Geometry285
Solution 3
Since this arithmetic progression has common ratio 5. Thus, all terms in it are in the form
Taking the modulo 5, all have either
or
Thus all integers of the form
are in the arithmetic progression and are perfect squares. This means that the 37 perfect square in the progression is
This also implies that the maximum value of
is
~pinkpig
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.