AMC 10A Year 2005 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"]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 

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[/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!!!!!!

[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.0"]

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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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.

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Similar Problems

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What are some facts about Leonardo Fibonacci and the Fibonacci sequence that everyone wants to know?

Italian Mathematician Leonardo Pisano( born in 1175 and died around 1250) also known as Fibonacci is mostly famous for his Fibonacci sequence. His name got originated from a misreading on a manuscript of “filius Bonacci”(son of Bonaccio).

He also played an important role in establishing the Hindu-Arabic numeral system in Europe.What Fibonacci did in his Book “Liber Abaci” in 1202,is that he played a major role in introducing the numbers which we now use to replace Roman numerals.The concept of Fibonacci sequence was referred to by him in a problem about breeding rabbits which is discussed later.He was considered to be the most talented western Mathematician of the Middle age. He introduced the mind blowing concept of Fibonacci sequence.He is also known as Leonardo Bonacci, Leonardo of Pisa, or Leonardo Bigollo Pisano (“Leonardo the Traveler from Pisa”).He had written a book known as “Liber Abaci”, which is translated as “The book of Calculation” and the book was published in 1202. He brought focus on the famous Hindu-Arabic Numeral System on his book Liber Abaci.The book has several applications related to above mentioned topic which includes conversion of weights and measures, money changing ,interest calculation and many other practical applications.The 1228 edition of the book includes methods for converting various numeral systems into Hindu-Arabic numerals.Popular Mathematical topic Abacus is also mentioned in the book .These facts have played a vital role in making calculations smoother and quicker thus aiding in the development if banking and other economic terms in Europe.Topic like prime numbers and irrational numbers are also mentioned in the book. In short he evolved the concept of Number theory .

Now discussing on his exceptional work on Fibonacci Sequence.The name “Fibonacci sequence” was first applied by Theorist Edouard Lucas in the 19th century.

In the field of Mathematics, Fibonacci numbers denoted as \(F_n\).The sequence states that that each number is the sum of the two preceding numbers starting from 0 followed by 1.

The general term of the sequence

\(F_n\) = \(F_{n-1}\) + \(F_{n−2}\) where \(F_0\) =0 and \(F_1\)=1 for all \(n>1\)

Thus the sequence becomes

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on.

Outside India, the Fibonacci sequence was first termed in Liber Abaci as mentioned above by Fibonacci.This concept was actually used to estimate the growth of rabbit population.

Fibonacci discovered a very interesting concept of the rabbit population.Rabbits usually never die and they are able to reproduce at the end of its second month.

.Now if a male and a female rabbit that is a newly born pair of rabbits are placed in a field then they will always produce a new pair at the end of each month starting from the second month.This way the following observations were made.

  1. By the end of first month there is only one pair. (\(F_1\)=1)
  2. By the end of second\d month, a new pair is born thus amounting to 2 pairs \(F_2\) =2)
  3. By the end of third month a new pair is born from the original pair thus amounting to 3 pairs ( \(F_3\) = \(F_2\) + \(F_1\) = 2+1 = 3)
  4. By the end of fourth month again a new pair is born from the original pair and another pair is born from the first female produced by the original female amounting to 5 pairs ( \(F_4\) = \(F_3\) + \(F_2\) = 3+2 = 5)

We can conclude from the above mentioned facts that by the end of n month, the number of pairs will be

\(F_n\) = \(F_{n-1}\) + \(F_{n−2}\) , which is the Mathematical generalised expression of the Fibonacci Sequence.

Now mentioning a few applications of the Fibonacci Sequence.

  1. The Fibonacci numbers are vital in analyzing Euclid’s Algorithm to determine the greatest common factor of two integers.
  2. Every positive integer can be expressed as a sum of Fibonacci numbers provided any one number is used at most once thus resulting in a complete sequence.
  3. Sculpture and Painter, Mario Merz included Fibonacci sequence in his works in 1970s.
  4. Fibonacci numbers also has its applications in Physics. In optics the number of different beam paths when a ray of light shines at an angle through two different transparent plates of different refractive index and material,there are k reflections ,for k>1 and k is the Fibonacci number.
  5. This sequence plays a very essential role in Computer Programming as as well.
  6. It is widely used in the filed of Botany.

Another very interesting fact about the Fibonacci number is that number of petals on flower Daisy is always a Fibonacci number (21, 34, 55 being most common numbers).

As recorded 1597, was the last year that was a Fibonacci number and the next will be 2584.

Number Theory - AMC 10A 2013 Problem 21 Sequential Hints

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Understand the problem

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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 ! 

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]

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 ?  

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

 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 ?      

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]

 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.                        

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]

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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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://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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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 ?

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]

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 ?  

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

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 ?    

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]

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 ?                      

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]

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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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://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"]

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Number Theory - Germany MO 2019, Problem 4

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Understand the problem

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Show that for each non-negative integer $n$ there are unique non-negative integers $x$ and $y$ such that we have
\[n=\frac{(x+y)^2+3x+y}{2}.\]
[/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="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0" hover_enabled="0"]Germany MO, 2019, Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]4/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="on"]Challenges and Thrills of Pre College Mathematics [/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

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]If you observe the expression \( \frac{(x+y)^2 + 3x + y}{2} \). It has two free variables. Now, if let's play with the expression to make it a little more symmetric and easy to work with and make some meaning out of it. Try to remove the asymmetry of x and write it in a much more symmetric form in x and y.

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Observe that  \( \frac{(x+y)^2 + 3x + y}{2} = \frac{(x+y)^2 + (x+y)}{2} + x\) is the expression what we get after bringing in the symmetry. Now, factorize it and see what we are looking for is \( n = \frac{(x+y)(x+y+1)}{2} + x\).  Can you guess anything about the expression \( \frac{(x+y)(x+y+1)}{2} \).

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]

\( \frac{(k)(k+1)}{2} \)  is the sum of the first k natural numbers.  So, now the idea is that somehow you are taking the first k natural numbers and adding another number x to it to make any number. Can you get the final logic?   [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Now, observe that sum of the first k numbers is increasing with k.  Now, take any number say 17. Now observe that 17 lies between 15 and 21.  This means that 17 = 15 + 2 = ( 1 +2 + 3 +4 + 5) +  2. So, this is the idea that k = x+y is the largest number such that n is greater than or equal to the ( 1 + 2 + ... + k ) and observe that we may need something like 2 in case of n = 17. Call that x and obviously \( k \geq x \). Hence, define y = k - x. So, for n = 33 = ( 1 + 2 + ... +  7 ) + 5.  Hence k = x + y = 7, x = 5, y = 2.  This is the whole idea. QED.  [/et_pb_tab][/et_pb_tabs][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"]

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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://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"]

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What are the first three Ramanujan numbers?

The world of mathematics is renowned for a number of interesting and fascinating numbers. Now Ramanujan Number also makes such a place in the list. Ramanujan Numbers (preciously termed as Hardy-Ramanujan Numbers) are those numbers that are the smallest positive integers that can be represented or expressed as a sum of 2 positive integers in n ways. Before discussing what are the first three Ramanujan numbers and how I found them in an amazing way, let's understand why is 1729 a special number.

Why is 1729 a special number and why is it called Hardy-Ramanujan number?

Now this Ramanujan’s n-way solution is a way in which a positive integer that can be expressed as a sum of 2 cubes in n different ways.

Like

  1. Ramanujan’s 1-way solution
  2. Ramanujan’s 2-way solution
  3. Ramanujan’s 3-way solution
  4. Ramanujan’s 4-way solution
  5. Ramanujan’s 5-way solution
  6. Ramanujan’s 6-way solution

Now let’s discuss the above ways in a mathematical way.

Ramanujan’s 1-way solution

Integers that are expressed as the sum of 2 cubes (in at least one way). Some of these numbers include :

{2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 152, 189, 217, 224, 243, 250, 280, 341, 344, 351, 370, 407, 432, 468, 513, 520, 539, 559, 576, 637, 686, 728, 730, 737, ...}

——Verifying——

2=\(1^3\)+\(1^3\)

9=\(2^3\)+\(1^3\)

16=\(2^3\)+\(2^3\)

Here all these numbers can be expressed as a sum of 2 cubes in a single way and so all these numbers from the above set can be expressed in this way.

Ramanujan’s 2-way solution

Integers that can be expressed as sum of 2 cubes in more than 1 way(at least in 2 ways) . Some of these numbers includes

{1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, ...}

——Verifying——-

1729=\(12^3\)+\(1^3\)=\(10^3\)+\(9^3\)

4104=\(16^3\)+\(2^3\)=\(15^3\)+\(9^3\)

13832=\(24^3\)+\(2^3\)=\(20^3\)+\(18^3\)

20683=\(27^3\)+\(10^3\)=\(24^3\)+\(19^3\)

Here all these numbers can be expressed as a sum of 2 cubes in 2 different ways . All of these numbers form the set can be expressed in these ways.

Ramanujan’s 3-way solution

Integers that can be expressed as sum of 2 cubes in at least 3 ways. Some of these numbers includes

{87539319, 119824488, 143604279, 175959000, 327763000, 700314552, 804360375, 958595904, 1148834232, 1407672000, 1840667192, 1915865217, 2363561613, 2622104000, ...}

——Verifying——-

87539319=\(167^3\)+\(436^3\)=\(228^3\)+\(423^3\)=\(255^3\)+\(414^3\)

Here all the numbers from the above set can be expressed as a sum of 2 cubes in 3 ways.

Ramanujan’s 4-way solution

Integers that can be expressed as sum of 2 cubes in at least 4 ways. Some of these numbers includes

{6963472309248, 12625136269928, 21131226514944, 26059452841000, 55707778473984, 74213505639000, 95773976104625, 101001090159424, 159380205560856, ...}

——Verifying——

6963472309248=\(2421^3\)+\(19083^3\)=\(5436^3\)+\(18948^3\)=\(10200^3\)+\(18072^3\)=\(13322^3\)+\(16630^3\)

Here all the numbers from the above set can be expressed as a sum of 2 cubes in 4 ways.

Ramanujan’s 5-way solutions

Integers that can be expressed as sum of 2 cubes in at least 5 ways. Some of these numbers are

{48988659276962496, 391909274215699968, 490593422681271000, 1322693800477987392, 3135274193725599744, 3924747381450168000, 6123582409620312000, 6355491080314102272, ...}

—-Verifying——

48988659276962496=\(38787^3\)+\(365757^3\)=\(107839^3\)+\(363753^3\)=\(205292^3\)+\(342952^3\)=\(221424^3\)+\(336588^3\)=\(231518^3\)+\(331954^3\)

Here this numbers can be expressed as a sum of 2 cubes in 5 different ways . All of these numbers of the above set can be expressed in these ways.

Ramanujan’s 6-way solutions

Integers that can be expressed as a sum of 2 cubes in at least 6 ways. Example

24153319581254312065344,…….

So these are all the numbers, now for the first three Ramanujan’s numbers, one can check for the type first, then can see the first 3 numbers.

——Verifying——

24153319581254312065344=\(582162^3\)+\(28906206^3\)=\(3064173^3\)+\(28894803^3\)=\(8519281^3\)+\(28657487^3\)=\(16218068^3\)+\(27093208^3\)=\(17492496^3\)+\(26590452^3\)=\(18289922^3\)+\(26224366^3\)

Here these numbers can be expressed as a sum of 2 cubes in 6 different ways . All of these numbers of the above set can be expressed in these ways.

All this numbers were also known as taxicab numbers and also denoted by Ta(n)

Some Useful Links:

Number Theory - Italy MO 2019, Problem 2

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Understand the problem

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Let $p,q$ be prime numbers$.$ Prove that if $p+q^2$ is a perfect square$,$ then $p^2+q^n$ is not a perfect square for any positive integer $n.$
[/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="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Italy MO 2019, Problem 2 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]5/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills of Pre College Mathematics [/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

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Let us first write the condition and use the idea that p and q are prime. Let $a \in \mathbb{N}$ such that \[ p + q^2 = a^2 \] =>   \[ p = a^2 - q^2 = (a - q)(a + q) \]. As $p$ is a prime and $a + q > a - q$, then $a - q = 1$. We must have $p = 2q + 1$. Now, try to use this condition to rewrite the expression we are interested in. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]p = 2q+1. Suppose that there exists $b \in \mathbb{N}$ such that for some positive integer $n$, then
\[ p^2 + q^n = b^2 \] => \[ q^n = b^2 - p^2 \] => \[ q^n = (b - p)(b + p) \]. It implies that both (b-p) and (b+p) must be powers of q. Hence gcd(b-p, b+p) = (b-p). b-p | b+p => b-p | 2p = 4q + 2. As, b-p is a power of q, it implies that b-p = 1 and q can be a prime or q = 2 and b-p = 2 as gcd (b-p, 4q+2) = 1 if q is an odd prime or 2 if q = 2. Thus, we get that ( b-p = 1 ) or ( b-p =2, q = 2 ) and p = 2q+1.   [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Case 1 ( b-p = 1 ): b -p = 1,  (b - p) = 1, (b+p) = \(q^n = 1 + 2p = 3 + 4q\). Now, you see that LHS will grow much more fast as q increases than the RHS.  Mathematically, \( q^n \geq q^2 \geq 4q+3 \) as \( q \geq 5, n \geq 2 \). So, the only possibility for q is q = 2, 3, 5. But it has no solution as you can see. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2 ( q = 2 ):  It implies p = 2q+1 = 5. \( 2^n = (b-5)(b+5) \). It implies that the both b-5 and b+5 must be powers of 2.  Let \( b-5 = 2^a \leq b+5 = 2^b \rightarrow 2^b - 2^a = 10 \rightarrow a = 1 and 2^(b-1) - 1 = 5 \). This has no solution. 

QED.

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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://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"]

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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 ? 

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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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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://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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Number Theory - Dutch MO 2015, Problem 4

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]Find all pairs of prime numbers $(p, q)$ for which $7pq^2 + p = q^3 + 43p^3 + 1$ [/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="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Dutch MO 2015 Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]5/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills in Pre College Mathematics [/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" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="4.0" hover_enabled="0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]This Diophantine Equation may seem a bit difficult to handle and will force you to try various techniques like making modulo 7, modulo p, modulo q as p and q are given as primes. But, let's go through the basic techniques for handling it. So what is it? Checking the parity of p and q in the given equation. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]Som check that if both p and q are odd primes, then the LHS will be even but the RHS will be odd, which is a contradiction.  Hence the only way it can happen that one of them must be even i.e. 2. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Now, things seem to be under control. We have two cases, p = 2 and q = 2. For p = 2, we get the equation \( q^3 - 14q^2 = -343 \). This implies that q must divide \( 343 = 7^3\). Hence q can be only 7. This gives rise to the solution (2,7). The next hint offers the other case. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]For  q = 2. we get \(43p^3 - 29p + 9 = 0\). How to solve this? Clearly apply the same idea. Observe that if p = odd the LHS will be odd which can't be 0. Hence, p must be 2, but it doesn't satisfy the equation. The only solution is (2,7). [/et_pb_tab][/et_pb_tabs][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"]

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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://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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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

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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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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://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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