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## Perfect square Problem | AIME I, 1999 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.

## Perfect square Problem – AIME I, 1999

Find the sum of all positive integers n for which $n^{2}-19n+99$ is a perfect square.

• is 107
• is 38
• is 840
• cannot be determined from the given information

### Key Concepts

Perfect Square

Integers

Inequalities

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

## Try with Hints

$(n-10)^{2}$ $\lt$ $n^{2}-19n+99$ $\lt$ $(n-8)^{2}$ and $n^{2}-19n+99$ is perfect square then $n^{2}-19n+99$=$(n-9)^{2}$ that is n=18

and $n^{2}-19n+99$ also perfect square for n=1,9,10

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# What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Competency in Focus: Perfect square numbers  This problem from American Mathematics contest (AMC 10A, 2014) is based on the concept that when a number is a perfect square . [/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

# Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” hover_enabled=”0″ box_shadow_style=”preset2″]Which of the following numbers is a perfect square? $\textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.1″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.1″ hover_enabled=”0″]American Mathematical Contest 2014, AMC 10A  Problem 8 [/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.1″ hover_enabled=”0″ open=”off”]This number theory problem is based on the concept that when a number is a perfect square  [/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.1″ hover_enabled=”0″ open=”off”]5/10 [/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.1″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.1″ hover_enabled=”0″]First of all look at the examples , see that   for all positive $n$, we have$$\dfrac{n!(n+1)!}{2}$$.Now what we have to do with this ? [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.0.9″]Now we have to find which member has what uniform numbers from the given  conversation .[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.1″ hover_enabled=”0″]After some simple manipulations , we have$$\dfrac{n!(n+1)!}{2}$$$$\implies\dfrac{(n!)^2\cdot(n+1)}{2}$$$$\implies (n!)^2\cdot\dfrac{n+1}{2}$$ . Thus now the problem reduces to  finding  a value of $n$ such that $(n!)^2\cdot\dfrac{n+1}{2}$ is a perfect square. [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.1″ hover_enabled=”0″]Since $(n!)^2$ is a perfect square, we must also have $\frac{n+1}{2}$ be a perfect square. In order for $\frac{n+1}{2}$ to be a perfect square,  $n+1$ must be twice a perfect square.

[/et_pb_tab][et_pb_tab title=”HINT 5″ _builder_version=”4.1″]

Now check the options and see for what value of n , $n+1$ must be twice a perfect square.   $n+1=18$ works, thus, $n=17$ and our desired answer is $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$ [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]