AMC 10 USA Math Olympiad

Number Theory – AMC 10A 2013 Problem 21 Sequential Hints

AMC 10A 2013, Problem 21 needed a clever trick to play with numbers. See the solution with sequential hints.

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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″]A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?

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Number Theory

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Elementary Number Theory by David M. Burton

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Start with hints

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Well, just give the problem a good read. Probably, with a little bit of thought, you can even get this done without a hint ! 

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We could start this the traditional way, be assuming the number of coins to be x.  Now, ask yourself after the k’th pirate has taken his share, what is the remanant number of coins ? This is  ( 12-k / 12 ) of what was originally there. [ Why ? Because each pirate takes k/12 of the coins, remember ? ]  Now, could you try taking things up from here…by yourself ?  

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 Let’s understand the next thing the problem is trying to focus on. “Each pirate receives a whole number of coins” Now, this should actually help us conclude   x. ( ( / 12 ) is supposed to be an integer. Since this actually implies divisibility.   Cancellation of terms leads us to : x. ( ( / ( ) )  Can you try and approach the solution by yourself now ?      

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 Now, this tells us the intuition of the problem. We make sure that the quotient should be an integer ! Also, recall that the 12’th pirate definitely takes the entirety of what is left, practically unity since it is exactly divisible.                        

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So basically, we just realized that the denominator is entirely multiplied out…cancelled !  And since we know that the denominator cancels out, the number of gold coins received by the 12th pirate is just going to be the product of the numerators !!! That evaluates to : = 1925 And that completes our solution !


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Watch video

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Connected Program at Cheenta

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url=”” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”3.23.3″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” background_layout=”dark” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″][/et_pb_button][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_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Similar Problems

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