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AMC 10 USA Math Olympiad

AMC 10A Year 2005 Problem 22 Sequential Hints

A beautiful problem form American Math Contest (AMC 10A) 2005, Problem 22 solution. Key idea here is to apply the greatest integer function.

[et_pb_section fb_built=”1″ _builder_version=”4.0″][et_pb_row _builder_version=”3.25″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][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″]

Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Let S be the set of the 2005 smallest positive multiples of 4, and let T be the set of the 2005 smallest positive multiples of 6. How many elements are common to S and T? (a) 166.       (b)333.      (c)500.      (d)668.      (e)1001

[/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=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0″]American Mathematical Contest 2005 10A Problem 22

[/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Number Theory 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]5/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

[/et_pb_accordion_item][/et_pb_accordion][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”4.0″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” min_height=”148px” custom_padding=”||24px|20px||”][et_pb_tab title=”Hint 0″ _builder_version=”4.0″]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]Step 1. First use the concept of lcm in order to find the type of common elements in both the sets.

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]Step 2. After getting the gcd as 12 you can easily see that the common elements in S and T must be in the form 12k (Where k is positive integers) . Now try to find out the number of elements in S and T.

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]Step 3.  Here lies the main concept of this problem as S has 8020 elements and T has 12030 (That is \(T>S\)) elements so you can see that many multiples of 12 are in T but not in S. As T has more elements than S so it’s obvious that multiples of 12 in S will definitely present in the set T but the converse is not true . So you have to only concentrate in the set S . Now find out the multiples of 12 in S. Think and give it a try!!!!!!!!!!

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]Step 4 Now you can see that as the lcm(4,6) is 12

In S you can see 4*3=12 so the set becomes 4,8,12,16,20,24,28,32,36,40,…… .Now you can see the 3rd , 6th , 9th , …. element are multiples of 12 ,so the multiples of 12 in S are in 3p ( values of p will determine the position of multiples of 12 in set S)  positions. Now here comes the interesting part of the problem. Think carefully what we can do with this information and hint!!!!!!

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Step 5 . So now you know that every 3rd element in S will be a multiple of 12 .So you have to calculate [\(\frac{2005}{3}\)] where [.] is the greatest integer function (Which means as many time this 3rd position or the 3p form appears in the set you will get a multiple of 12 so we will use the greatest integer function in order to find out the multiples of 12 in S). Therefore  [\(\frac{2005}{3}\)]=668. which is your required answer 

 

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Start with hints

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Watch video

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