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# AMC 10A Year 2005 Problem 21 Sequential Hints

American Mathematical Olympiard 10A Problem 21 Solutuon. The main idea here in this problem is to use some formulae of induction and finding factors.

# 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″]For how many positive integers $n$ does 1+2+3+4+….+n evenly divide from 6n? (a)3.       (b)5.       (c)7.       (d)9.       (e)11

[/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 21

[/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”]6/10

[/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. So after having a deep look into this problem you can see that if 1+2+3+…..+n evenly divides 6n that is $\frac{6n}{1+2+3+….+n}$ now to think about formula of the sum of 1+2+3+…..+n.

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Step 2. After getting the formula as 1+2+3+4+….+n=$\frac{n(n+1)}{2}$ substitute it in the equation $\frac{6n}{1+2+3+….+n}$ and simplify it. Give it a try!!!!!!

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Step 3 Now by simplifying you will get $\frac{12}{n+1}$. Now here lies the main concept of this problem as you have to find integer n so you must see that if (n+1) is a factor of 12 then only $\frac{12}{n+1}$ will become an integer. Now find out the factors of 12 and try to build up some logic how to make this $\frac{12}{n+1}$ an integer.

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Step 4 So you can easily say that the factors of 12 are 1,2,3,4,6 and 12 respectively now try to think who you can use this information here in this $\frac{12}{n+1}$. Like what are the values of n (from the factors of 12) in order to make it a (n+1) factor of 12.

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Step 5 . Here n can take values 0,1,2,3,5 and 11 respectively as n+1 must be a factor of  12 . But here 0 is not a positive integer so you have to exclude 0 so you are left with 5 different values of n . So your answer is 5

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