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AMC 10 USA Math Olympiad

Number Theory – AMC 10A 2013 Problem 21 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?

[/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” _builder_version=”4.0″ open=”off”]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]

Number Theory

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7/10

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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. ( (11.10.9.8.7.6.5.4.3.2.1) / 12 ) is supposed to be an integer. Since this actually implies divisibility.   Cancellation of terms leads us to : x. ( (11.5.1.7.1.5.1.1.1.1) / ( 12.6.2.12.2.12.3.4.6.12 ) )  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 : 11.5.7.5 = 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=”https://www.cheenta.com/matholympiad/” 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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Categories
AMC 10 USA Math Olympiad

Number Theory – AMC 10A 2014 Problem 24 Sequential Hints

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Understand the problem

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A sequence of natural numbers is constructed by listing the first $4$, then skipping one, listing the next $5$, skipping $2$, listing $6$, skipping $3$, and, on the $n$th iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,\!000$th number in the sequence ? 

[/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 Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]

Number Theory, Sequences

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7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Challenges and Thrills of Pre-College Mathematics

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Start with hints

[/et_pb_text][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” custom_padding=”||153px|25px||”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]You could give it a thought first…are you sure you really need a hint ?

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Stuck…? Well, don’t worry. The key to solving this problem is not thinking too much about the skips. We can start with natural numbers, from 1 to 500,000. So, a useful strategy could be to find how many numbers we have actually skipped, n and then add them back accordingly.  So, now could you take things on from here ?  

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If you’re a tad bit doubtful of where we’re heading even now, well no problem. Clearly, we can say 999.(1000) / 2   < 500,000 < 1000.(1001) / 2 So, now can you find out how many blocks of gaps we have in the sequence ?    

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Now see, finding the blocks of gaps easy ! There’s just one small thing you would have to recall. We began the count from 4…so now, the number of skipped blocks in the sequence = 999 – 3 = 996.  Now to find n, from the number of blocks , we have =  (996.997) / 2 = 496,506 This stands for the number of numbers we skipped. Now concluding this is fairly easy…could you try that out yourself ?                      

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What remains for us to do is to add back those skipped numbers to the count, 500,000 to obtain the final answer. That gives us = 500000 +496506 = 996506

And we are done !

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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=”https://www.cheenta.com/matholympiad/” 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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AMC 10 USA Math Olympiad

AMC 10A Year 2007 Problem 20 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″]Suppose that the number \(a\) satisfies the equation \(a\)+\(\frac{1}{a}\)=\(4\) What is the value of \(a^4\)+\(\frac{1}{a^4}\)?  

(a) 164.        (b)172.       (c)192.       (d)194.       (e)212

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American Mathematical Contest 2007 10 A Problem 20

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Basic Algebra 

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4/10

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Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

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Start with hints

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Step 1. Look’s hard , lets see we are given with \(a\)+\(\frac{1}{a}\)=\(4\) and we need to find \(a^4\)+\(\frac{1}{a^4}\). So first we will square both sides in order to get  \(a^2\)+\(\frac{1}{a^2}\) and find its value. Give it a try!!!!.

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Step 2. We reached the power 2 now aim is to get to power 4. Now After getting the the value of  \(a^2\)+\(\frac{1}{a^2}\) , again square both sides to get the value of the expression \(a^4\)+\(\frac{1}{a^4}\) . Very close to the solution !!!!!!.

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Step 3 . Now we get the equation \(a^4\)+\(\frac{1}{a^4}\) after solving  \((a^2+\frac{1}{a^2})^2\). Now I hope u can see the answer by your self just a last step!!!!.

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Step 4 By solving we get the results that is   \(a^2\)+\(\frac{1}{a^2}\)=14

and \(a^4\)+\(\frac{1}{a^4}\) =194 which is our required value to find ,thats it!!!

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AMC 10 USA Math Olympiad

Probability – AMC 10A 2015 Problem 22 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″]Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

[/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 Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Probability / Combinatorics

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]’7/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Challenges and Thrills of Pre-College Mathematics

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Start with hints

[/et_pb_text][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” custom_padding=”||150px|18px||”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]You could give it a thought first…are you sure you really need a hint ?

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Let’s not get worried by the idea of having to find out the probability. For, we may simply recall : Probability = Number of favorable events / Total number events in the sample space So, the denominator is easy to find out here, 8 people, all of whom are either standing or sitting. So, 2 possibilities for every person. And that makes our denominator, \(2^8 = 256 \) Now, with a little bit of focus, it is not at all hard to see that the numerator, “number of favorable events” is essentially a Combinatorics arrangement problem. 

So, would you like to have a go at this by yourself now ?

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Now, as we go ahead with the solution, let us develop the idea of the strategy we could use here. You must be familiar with recursion, but let’s just recall its philosophy. Recursion essentially meant that you could express a variable or a term in terms of its own parameters. As in, when a function calls itself. If you want a brief look-up, catch the short story in the paragraph below [ Much like imagine, you’re eating a large cake. You won’t really eat the whole thing at once, would you ? You would possibly enjoy it in parts. Imagine you’d have 1/3 rd of it, to begin with. Then if your problem initially is, EAT THE CAKE. Even, after you have had a first bite, as I just said, what’s the name of your problem ? It’s still EAT THE CAKE , isn’t it ? Yes, you’re still eating the same cake, only your problem is reduced by 1/3-rd, somewhat like a miniature version. Something we could call, ( EAT THE CAKE ) / 3…That’s pretty much, an intuitive idea of recursion. ] How could we model our problem in that fashion ? Imagine we talk of 8 persons standing in the line, to begin with. Let us try to develop a recurrence relation for the problem. Let’s talk of a person, N in the queue. See the catch here, if this person N is sitting, our problem size reduces by 1 – which means it is now equivalent to 7 people standing in the queue. On the other hand if this person N is standing, it is implied by the statement of the problem, that the adjacent people on his either side must be sitting ! That reduces our problem size by 2… Hey, so now that we have kind of linguistically framed it, could you try out a mathematical interpretation ?     

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Let, the answer we are looking for is a_n. ( That is, the number of ways to arrange n people in a line such that no two of them, who are adjacent, are standing ) So for a_n >= 2,
See there can be only two broad-spectrum cases : Case I : Person #1 in the line is standing If Person #1 in the line is standing, the second one is sitting. So the rest can be arranged in a_n-2 ways.  Case II : Person #1 in the line is sitting
This means the rest can be arranged in a_n-1 ways. So, Case I + Case II, adds up to the relation we are looking for ? That would be something like :  a_n = a_n-2 + a_n-1 Wait ! Does this, by any chance seem familiar ? Does it ? Well yes, this is nothing but the Fibonacci Recurrence Relation. The problem of rabbits in the field, if you need some off-beat reference !  [ The Fibonacci Sequence goes like :
   1,1,2,3,5,8,13, 21, 34,….
It’s quite easy to see that from the third time onwards, every term is numerically equal to the sum of its two previous terms ] So, do you think you could conclude this…and find the required probability ?                  

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]

So we just saw, we can map the given problem at hand to the Fibonacci problem.  Our starting cases here would be a_0 = 1 and a_1 = 2. ( By the problem ). This means our ans would be the (n+2)th Fibonacci number for a_n F_n stands for the n’th Fibonacci number, of the sequence you just came across in the last hint. So that gives us a_5 + a_7 = F_7 + F_9 = 13 + 34 = 47. So, that’s our numerator ! Hence, the required probability =   47 / 256 

And that’s pretty much of all we need ! 

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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=”https://www.cheenta.com/matholympiad/” 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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Categories
AMC 10 USA Math Olympiad

Number Theory – AMC 10A 2016 Problem 22

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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″] For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have ? 

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.25″][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 Mathematics Competition [/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”]8/10

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Elementary Number Theory by David M. Burton

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Start with hints

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Check the problem out…give its statement a thorough read. Might appear a bit daunting on the first couple of reads. Think for some time, you could be on to something without any help whatsoever ![/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

Okay, now let’s think about what our first thoughts could be, on the problem. It’s definitely about the n in the problem, which acts as our unknown here.  Can you somehow try finding the n ? Let’s take the first step in that direction. How could we prime factorize 110 ? That’s easy 110 = 2.5.11. Could you take things from hereon to find more about the n ?

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]

However interestingly the problem says, the number 110. (n^3)  has 110 factors. Just as we saw, 110. (n^3) = 2.5.11.(n^3) Now, let’s use some basic number theoretic knowledge here. How many divisors would 110. (n^3) have then ?  If n=1 Clearly it would have, (1+1). (1+1). (1+1) = 8 divisors.  So see, that’s the idea isn’t it ? Pretty much of plug and play. We actually get to control how many divisors the number has, once we adjust (n^3).  Now you could try some advances…        

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]

 Okay, so as we just understood we need to achieve a count of 110 divisors.  If we have 110.(n^3) = 2^(10). 5^(4). 11 which actually conforms to :  (10+1).(4+1).(1+1) = 11.5.2 = 110  So, that implies :   (n^3) = 2^(9). 5^(3), which means, n = 2^(3). 5 Now that we have found out n…the rest dosen’t seem really a big deal. You could do it…try !                  

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]

Well, it’s pretty straightforward now.  Let’s call 81.(n^4) equal to some X. First let’s prime factorize 81. That would be 81 = (3^4). So, finally X = (3^4). (2^12). (5^4) How many divisors does that make ? Yes, (4+1).(12+1). (4+1) = 13.25 = 325.

That’s our answer. 325 factors

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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=”https://www.cheenta.com/matholympiad/” 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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Categories
AMC 10 USA Math Olympiad

Number Theory – AMC 10A 2014 Problem 20 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″]The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.25″][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=”off” _builder_version=”4.0″]American Mathematics Competition [/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”]7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”on”]

Elementary Number Theory by David M, Burton 

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Start with hints

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So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

Think you could do with some help ? Okay let’s get ourselves a headstart. As you might have guessed, sometimes the most fruitful thing to be done is to observe what’s going on in these kind of problems. The intuition is clear, there are too many instances of ‘8‘-s for someone to account for them manually.  So, yes, that’s something we can expect. Try working out with a few simple cases like, k=1, k=2, and so on…

 

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]

Okay, so let’s see what’s happening for a few small values of k…
k=1
8 * 8 = 64 k=2
8 * 88 = 704 k=3
8 * 888 = 7104 k=4 
8 * 8888 = 71104 …
So, Wait ! Look closely…Do you see something ?    

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]

A really nice pattern is evolving. If you can see it, and yet find it a bit difficult to articulate it mathematically, don’t worry. See for k=4, the product gives us the result 71104. That means, we have the starting digit to be 7. And the ending suffix is 04. What varies are the ones. See, for k=4, we have ( k-2 = 2 ) ones. That’s it ! The generalization should be fairly simple for us to do now… For every k >=2, the product result consists of a 7 to start with, exactly k-2 1’s follow, and we conclude with single occurrences of 0 and 4 each. Now, think…can you take this till the end ?          

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]

Once we’ve seen the pattern, it’s easy to get this done.  What we have noted, we need that to sum up to 1000.  In simple mathematical terms, 7 + (k-2).1 + 0 + 4 = 1000
11 + k = 1002
k = 991

So, we need ‘8’ to come 991 times in the multiplicand, so that the digits sum up to 1000. So, that seals the deal !

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Watch video

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Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title=”Math Olympiad Program” url=”https://www.cheenta.com/matholympiad/” url_new_window=”on” image=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”3.23.3″ header_font=”||||||||” header_text_color=”#e02b20″ header_font_size=”48px” link_option_url=”https://www.cheenta.com/matholympiad/” link_option_url_new_window=”on”]

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=”https://www.cheenta.com/matholympiad/” 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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Categories
AMC 10 USA Math Olympiad

Enumerative Combinatorics- AMC 10A 2017 Problem 25 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″]How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. )

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.25″][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 Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Enumerative Combinatorics 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Introductory Combinatorics by Richard Brualdi

[/et_pb_accordion_item][/et_pb_accordion][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=”48px||48px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Start with hints

[/et_pb_text][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”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. 
What is the largest number in the range divisible by 11 ? Easy, 990.
How about the smallest such number ? Yeah, 110.
So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81.
Now, how about you try taking things ahead from here onwards ?

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]

Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If you find out the answer to this “why”, you can might as well say you’ve gone far enough to solve the problem…  

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]

    Let’s answer the “why” here. Say, we have a 3-digit number, abc. Now, ideally we would have had 6 permutations ( 3 ! simply ) for each such abc. But here’s a catch ! If abc is divisible by 11, so is cba.…!!! Yeah, that’s it. So, basically if we multiply by 6, we are accounting for same kind of permutations twice. So basically, each multiple of 11 in the range has it’s ( 6/2 ) = 3 permutations, that we are bothered about. This clearly justifies the fact that we can at maximum have 81 x 3 = 243 numbers in our desired solution set. But wait ? Why do I say, at maximum ? So…have we overcounted ? Yeah,we have. Why don’t you think about it…?          

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Well, as you might have felt, we did overcount. We did not account for the numbers where 0 could be one of the digits. We overcounted cases where the middle digit of the number is 0 and the last digit is 0. So, what are these ? Let’s find them out. In how many of the numbers is the last digit 0 ? That’s easy…they have a pattern. It goes like 110, 220,….990. That makes it 9. Now, in how many of those 243 numbers that we are bothered about, does ‘0’ occur as a middle digit ? With a little bit of insight, you’d find out they are 803, 902, 704, 605. And well their permutations too. So that makes it 4 x 2 = 8.  So, in total, 9+8 = 17 elements have been overcounted. 

 Subtract that from 243, ( 24317 ) = 226, that’s your answer.

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