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AMC 10 USA Math Olympiad

Number System ISI Entrance , 2012 Problem 8

Try this beautiful problem from AMC 10A. It involves finding the remainder when anumber is divided by another digit. We provide sequential hints so that you can try the problem.

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What are we learning ?

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 Competency in Focus: Number System

This problem from Indian Statistical Institute (ISI Entrance 2012) is based on Number System. It includes finding the remainder when a number is divided by another digit.

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First look at the knowledge graph.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/02/isi-2012-p8.png” align=”center” force_fullwidth=”on” _builder_version=”4.2.2″ min_height=”388px” height=”198px” max_height=”207px”][/et_pb_image][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″ inline_fonts=”Aclonica”]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ 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″]The last digit of  \(9!+3^{9966}\) is (A) 3 (B) 9 (C) 7 (D) 1[/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=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”on”]Indian Statistical Institute (ISI) 2012 Problem 8.[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″]

Number system

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]4/10[/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 

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Start with hints 

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][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=”4.2.2″]Unit digit of the whoe expression will be sum of unit digit of the first term and unit digit of the second term. So if the first term gives last digit 5 and 2nd terms gives 2 then unit digit of whole expresion is (5+2) or 7.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]when we expand \(9!\) there will be 5 and 2 in between that when multiplied will give 10 as a factor so the term \(9!\) will have \(0!\) as last digit.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]\(3^{1}\) has last digit as 3 \(3^{2}\) has last digit as 9 \(3^{3}\) has last digit as 7 \(3^{4}\) has last digit as 1 \(3^{5} \) has last digit as 3  and the pattern repeats with the power(indxe) at an interval of 4.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]9966 when divided by 4 gives 2 as remainder. So, in \(3^{9964}  3^{2} \) , 9 will be the last digit.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ saved_tabs=”all” locked=”off”][et_pb_fullwidth_header title=”I.S.I. & C.M.I. Program” button_one_text=”Learn more” button_one_url=”https://www.cheenta.com/isicmientrance/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/ISI.png” _builder_version=”4.2.2″ title_level=”h2″ title_font=”Acme||||||||” background_color=”#220e58″]

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