On Twos for Twosday

At 22 seconds past 22 minutes past ten in the evening of the 22nd day of the 2nd month of the 22nd year this century (which falls on a Tuesday, or Twosday), the time and date will consist of a single repeating digit: the last occasion this will happen in our lifetimes.  In the date format dd/mm/yyyy, 22/02/2022 is also a palindromic date, i.e., is read the same backwards as it is forwards.  To mark and celebrate Twosday, here are a few curios about twos... 

2...

2 is the first (and only even) prime.

2 is the only number which, when added to itself, gives the same result as when multiplied by itself (2+2 = 2×2). 

2 is the only difference between two consecutive primes that is prime  (3 and 5, 5 and 7, 11 and 13...).

2 is the smallest prime that produces another prime seven consecutive times by adding a digit (2, 29, 293, 2939, 29399, 293999, 2939999, 29399999).

2 has a square root that is probably the first number known to be irrational, known as Pythagoras' Constant (watch this Numberphile video, and this from VSauce).

2 is the number most feared by primonumerophobics.

2 is strobogrammatic (the same upside-down) in an 8-segment display.

2 has 2 homphones (to and too).

2 is a child's age on 22/2/22 if they were born on 22/2/2020 (children born at 20:20 on 6/12/2019 will be 2 years, 2 months, 2 weeks, 2 days, 2 hours and 2 minutes old at 22:22 on 22/2/22).

Listen to 'Two - At the double' from Simon Singh's 'Numbers'.

Listen to 'Funbers 2' from Tom Rocks Maths Funbers broadcast.

22...

22 is the first self-describing number (it has 'two twos").

22 is the smallest number that can be expressed as the sum of 2 primes in more than 2 ways (3+19 = 5+17 = 11+11).

22 is the number of ways 2³ can be expressed as the sum of positive integers (8, 7+1, 6+2, 6+1+1, 5+3, 5+2+1, 5+1+1+1, 4+4, 4+3+1, 4+2+2, 4+2+1+1, 4+1+1+1+1, 3+3+2, 3+3+1+1, 3+2+2+1, 3+2+1+1+1, 3+1+1+1+1+1, 2+2+2+2, 2+2+2+1+1, 2+2+1+1+1+1, 2+1+1+1+1+1+1, 1+1+1+1+1+1+1+1). 

22 is the largest known arithmetic sequence of primes, starting with 11,410,337,850,553 and ending with 108,201,410,428,753, with a common difference of 4,609,098,694,200.

22 is the number of different ways five pentagons can be linked together (i.e., there are 22 pentahexes).

22 has a digit sum (2+2 = 4) equal to the sum of the digits of its prime factors (2, 11... 2+1+1).

22 has a digit sum (2+2 = 4) equal to the product of its digits (2×2 = 4).

22 is the maximum number of pieces that can be created when cutting a circle with six lines (i.e., 22 is the 6th term of the lazy caterer's sequence).

22 is a person's age on 22/2/22 if they were born on 22/2/2000 (people born at 00:00 on 06/12/1999 will be 22 years, 2 months, 2 weeks, 2 days, 22 hours and 22 minutes old at 22:22 on 22/2/22).

222...

222 is the product of exactly three distinct primes (2×3×37).

222 is the sum of 2 successive primes (109+113).

222 is the sum of three squares in at least three ways (1+25+196, 1+100+121, 4+49+169).

222 is the sum of all 2-digit primes formed from consecutive digits (23+43+67+89).

222 is the 22nd number that produces a prime when divided by the sum of its digits [222÷(2+2+2) = 37].

222 has factors that can be partitioned into 2 sets with equal sums (1+2+3+222 = 6+37+74+111).

222 cannot be written as a number + the sum of the number's digits, for any number.

2222...

2222 is the smallest number divisible by a 1-digit, 2-digit and 3-digit prime (2222÷2 = 1111, 2222÷11 = 202, 2222÷101 = 22).

2222 has factors that are all palindromic (1, 2, 11, 22, 101, 202, 1111, 2222).

2222 squared is the sum of 88 consecutive squares (2222² = 4,937,284 = 192²+193²+...+279²).

2222 is the sum of six cubes in exactly six ways.

22222...

22222 is a Kaprekar number because 22222² = 493,817,284 and 4,938+17,284 = 22,222.

22222 is the smallest palindromic multiple of 41 with even digits (22222 = 41×542).

 

On Descartes Numbers

A number is said to be Perfect if it is equal to the sum of its proper positive factors, i.e. the sum of its factors excluding the number itself.  Summing all of the factors of a number \(n\) — i.e. including the number \(n\) itself — is known as the sum-of-divisors function:

\[\sigma \left( n \right) = \sum\limits_{d\left| n \right.} d \]
A number \(n\) is thus said to be Perfect when \(\sigma \left( n \right)\) = 2n.  For example, the factors of 28 — 1, 2, 4, 7, 14 and 28 — sum to 28 + 28 = 56, thus \(\sigma \left( {28} \right)\) = 2 \(\times \) 28 and so 28 is a Perfect number.  In contrast, and by way of illustration, the factors of 42 — 1, 2, 3, 6, 7, 14, 21 and 42 — sum to 96, or 2.2857... \(\times \) 42, thus \(\sigma \left( {42} \right) \ne\) 2 \(\times \) 42 and so 42 is not a Perfect number.  In fact, 42 is said to be an Abundant number because the sum of its proper positive divisors is greater than the number itself, i.e. \(\sigma \left( n \right) > 2n\).  When the sum of a number's proper positive divisors is less than the number itself, i.e. \(\sigma \left( n \right) < 2n\), the number is said to be Deficient.  For example, the factors of 50 — 1, 2, 5, 10, 25 and 50 — sum to 93, or 1.86 \(\times \) 50, and thus \(\sigma \left( {50} \right)\) < 2 \(\times \) 42.  As such:

\[\begin{array}{l}\sigma \left( n \right) < 2n \Rightarrow n\;{\rm{is}}\;{\rm{deficient}}\\\sigma \left( n \right) = 2n \Rightarrow n\;{\rm{is}}\;{\rm{perfect}}\\\sigma \left( n \right) > 2n \Rightarrow n\;{\rm{is}}\;{\rm{abundant}}\end{array}\]
Perfect numbers have intrigued us since Euclid, who observed via Proposition 36 in Book IX of his Elements that a number of the form (2\(^{p - 1}\))(2\(^p\) − 1) is a perfect number whenever 2\(^p\) − 1 is prime (i.e. what subsequently became known as a Mersenne prime) [1].  The fact that every even perfect number is of this type, i.e. that every even perfect number can be written in the form (2\(^{p - 1}\))(2\(^p\) − 1) whenever 2\(^p\) − 1 is prime, was proposed by René Descartes in his famed correspondence with Marin Mersenne (specifically in his letter to Mersenne of 15 November 1638) [2], and later proven in 1849 by Leonhard Euler [3].  Descartes was, indeed, 'among the first to consider the existence of odd perfect numbers' (Greathouse and Weisstein, 2012).

As of writing, of the fifty perfect numbers that have been found, all are even, and it is not known if there are infinitely many or, indeed, whether any odd perfect numbers exist — although Pascal Ochem and Michaël Rao showed (2012) that no number up to 10\(^{1500}\) is an odd perfect number.

One number has been found, however, that would have been an odd perfect number if only one of its factors was prime rather than a 'spoof prime' (i.e. a composite number wrongly assumed to be prime).  This number — 198,585,576,189 — was found in 1638 by Descartes, documented in his 15 November letter to Mersenne, and is known as a Descartes Number, or, as an 'odd spoof perfect number' (note that the spoof prime in question is 22021, i.e. 19\(^2\) \(\times \) 61).

\[198,585,576,189 = {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\]
Descartes showed that if 22021 was prime, the proper factors of 198,585,576,189 [4] would sum to 198,585,576,189 — making it the only odd perfect number ever found.  As it is, because 22021 is not prime, or because 22021 is a spoof prime, the proper factors of 198,585,576,189 do not sum to 198,585,576,189 (they actually sum to 227,441,894,589).  This makes 198,585,576,189 an odd spoof perfect number, which in itself is the only such number ever found!

To elaborate: Descartes showed that an equivalence to the sum-of-divisors function using the prime factorisation of a number is the following [5]:

\[\sigma \left( n \right) = \prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} {\rm{ }}\]
where \(p\) is a distinct prime factor.  This means that a number's factors can be summed using only the number's prime factors.  Taking 28 again as an example, we have seen that it is a perfect number because

\[\sigma \left( 28 \right) = \sum\limits_{d\left| 28 \right.} d  = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2n\]
We can also observe this result using the prime factors of 28, i.e. 2\(^2 \times \) 7:

\[\prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}}  = \frac{{{2^{2 + 1}} - 1}}{{2 - 1}} \times \frac{{{7^{1 + 1}} - 1}}{{7 - 1}} = \frac{7}{1} \times \frac{{48}}{6} = 56 = 2n\]
And so, similarly, the sum of the divisors of Descartes' number:

\[\begin{array}{c}\begin{align}\sigma \left( 198,585,576,189 \right) &= \prod\limits_{{p^a}\left\| {\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} \\ \\&= \frac{{{3^{2 + 1}} - 1}}{{3 - 1}} \times \frac{{{7^{2 + 1}} - 1}}{{7 - 1}} \times \frac{{{{11}^{2 + 1}} - 1}}{{11 - 1}} \times \frac{{{{13}^{2 + 1}} - 1}}{{13 - 1}} \times \frac{{{{22021}^{1 + 1}} - 1}}{{22021 - 1}}\\  \\&= 13 \times 57 \times 133 \times 183 \times 22022\\ &= 13 \times \left( {3 \times 19} \right) \times \left( {7 \times 19} \right) \times \left( {3 \times 61} \right) \times \left( {2 \times 7 \times 11 \times 11 \times 13} \right)\\ &=2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times \left( {{{19}^2} \times 61} \right)\\ &= 2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\\ &= 2 \times 198,585,576,189\end{align}\end{array}\]
Hence, if 22021 were prime rather than 19\(^2\) \(\times \) 61, Descartes' number 198,585,576,189 would  be an odd perfect number because \(\sigma \left( n \right)\) = 2n.

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Notes, References & Links:

On Mountain Numbers

Cairn Toul, a beautifully wild and remote mountain in the Cairngorms massif of the eastern Scottish highlands, is notable not only for being the fourth highest mountain in the UK, but also for being the only mountain in the UK whose elevation in metres — 1291m above sea level — is also a 'mountain'.  Großglockner, by way of a further example, the highest mountain in Austria, is a mountain whose elevation in feet — 12461ft above sea level — is also a 'mountain'.

To elaborate: A number is a 'mountain' if its decimal digits start with 1, i.e. 'base camp', ascend continuously to a unique summit, i.e. to one largest digit, then descend continuously back to 1, i.e. back to base camp.  Such mountain numbers are recorded as sequence A134941 (itself a mountain) in the On-line Encyclopedia of Integer Sequences (OEIS), from which you can also find this table of all possible 21,846 mountains (including 1 itself).

Mountain primes, as the name suggests, are prime mountain numbers, prime numbers in other words whose decimal digits start with 1, ascend continuously to the summit of a single largest digit, then descend continuously back to 1.  (The area chart image above shows a mountain range of the first 17 such numbers.)  Mountain primes are recorded in the OEIS as sequence A134951 (yes, also a mountain prime), from which you can find this table of all 2620 such primes.

When climbing a mountain, it is typical to descend back to base camp, i.e. back to where you started your ascent, but it is not always the case.  Mountain numbers that ascend from one location (as an elevation) but descend to another — e.g. 3,598,432 — are considered to be generalised mountain numbers, recorded as sequence A134853 in the OEIS (there are 173,247 such mountains).

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Suggested explorations, diversions & links:

• Explain why there must be a finite number of mountain numbers.
• Construct a mountain range diagram for mountains of your choice.
• What is the Everest of mountain numbers?
• How will you define 'Everest'; how will you define 'elevation'?  (In the image above, the 'height' of mountain 1291, for example, is greater than the 'height' of mountain 1571.)  
• Explore the distribution of mountain number digits, i.e. how many mountains have 1 digit, 2 digits, 3 digits, etc. (OEIS sequence A135417).
• If we regard the number of digits in a mountain number as the horizontal distance travelled when climbing such a mountain:
• What mountain(s) has (have) the shallowest ascent?
• What mountain has the steepest ascent and descent?
• What about generalised mountains?
• Many people find the most beautiful mountains to be those most pyramidal in shape, such as for example the Matterhorn in the Alps and Machapuchare in the Himalaya.
• Define and find your most beautiful mountain numbers: 
• Palindromic mountain numbers perhaps (OEIS sequence A173070),
• Or palindromic mountain prime numbers (the largest is 123467898764321).
• Find all possible — i.e. 45 — Giza numbers.
• Giza numbers are so-called because they represent the pyramids of Giza in the sense that their first digits increase in consecutive order to a largest central digit, and their last digits decrease in the same consecutive order as they increased.  
• The largest Giza number is 12,345,678,987,654,321 (OEIS sequence A134810).
• Of two mountain numbers chosen at random, what is the probability that their numerical heights — i.e. the magnitude of the number — will be in alignment with the elevation of actual mountains?  
• For example, mountain number 1291 is 'higher' than mountain number 1571, but as actual elevations, 1291m < 1591m, so these two mountains are not in 'alignment'. 
• Find as many actual mountains whose heights are mountain numbers, or mountain primes.
• What is the highest mountain in the world whose height is also a mountain (use metres and/or feet for the height)?
• You may find PeakBagger.com and PeakList.org useful starting points to collect your data.  WolframAlpha may also help. 

On Frustrating Primes

7775353777 is a prime number.  It has ten digits and it looks like, on first inspection, that the second five digits reflect the first.  But a number cannot be prime and have such a property — and on closer inspection we can see that there is, indeed, no such reflection.  Such primes can be considered 'frustrating' because of the absence of this visually eye-catching, aesthetically pleasing pattern they appear, on the initial face of it, to promise.

A prime, therefore, is said to be frustrating if the first half of its digits occur again in the second half, but without a) the order repeating (i.e. abc‖abc, because this is divisible by 1001 [1] and thus not prime), or b) the order reflecting (i.e. abc‖cba, because every palindrome with an even number of digits is divisible by 11 and thus not prime).

As such, a frustrating prime has an even number of digits, and contains at least six digits of which at least two are distinct [2].  Of the 68,906 six-digit primes [3], 265 are frustrating (collated in the image above).  The possible arrangements of six-digits with the first three reoccurring:

aaa‖···

⇒ aaa‖aaa : Not prime ∵  divisible by 11

aab‖···

⇒ aab‖aab : Not prime ∵ divisible by 1001

⇒ aab‖aba : Possibly a frustrating prime (n = 10)

⇒ aab‖baa : Not prime ∵ divisible by 11

aba‖···

⇒ aba‖aab : Possibly a frustrating prime (n = 7)

⇒ aba‖aba : Not prime ∵ divisible by 1001

⇒ aba‖baa : Possibly a frustrating prime (n = 10)

abb‖···

⇒ abb‖abb : Not prime ∵ divisible by 1001

⇒ abb‖bab : Possibly a frustrating prime (n = 9)

⇒ abb‖bba : Not prime ∵ divisible by 11

abc‖···

⇒ abc‖abc : Not prime ∵ divisible by 1001

⇒ abc‖acb : Possibly a frustrating prime (n = 66)

⇒ abc‖bac : Possibly a frustrating prime (n = 58)

⇒ abc‖bca : Possibly a frustrating prime (n = 68)

⇒ abc‖cab : Possibly a frustrating prime (n = 37)

⇒ abc‖cba : Not prime ∵ divisible by 11

The Frustrating Primes are recorded as sequence A297994 in the On-line Encyclopedia of Integer Sequences (OEIS), from which you can find this table of Frustrating Primes a_n for n = 1 to 10,000. 

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The learning capital to be accrued from exploring the idea of frustrating primes, is especially evident in terms of the development of students' mathematical maturity.  The above sketch of the underlying premise of frustrating primes informs a 'thinking through' of how as teachers we could use the idea with (alongside) students, how we could approach the problem with students, structurally and pedagogically, and what questions can/should be asked of our students, should they not yet be able to ask them for themselves.

Considering the idea through this problem will encourage depth and maturation, and give us as teachers the opportunity, as I wrote about in a previous post, to 'model for [students] what it is to be mathematically mature, to behave in a mathematically mature manner when we are befuddled by the beautiful sixes and sevens of uncertainty.'   Students will not only deepen their sense of number and their appreciation of prime numbers, at the same time as being able to apply divisibility tests, consider prime factors, etc., but they may also get a kick out of doing something a little more mathematically novel than they are typically used to.

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Notes & (Select) Links:

[1]  \(\left( {1000 \times \overline {abc} } \right) + \left( {1 \times \overline {abc} } \right)\)

[2]  For a two-digit number to be a frustrating prime the first half of the digits would have to occur in the second, and this would thus be a number of the form a‖a, which is clearly a composite number divisible by 11.  Similarly, for a four-digit number to be a frustrating prime the first half of the digits would have to occur in the second, meaning that the number would thus be of the form ab‖ab or ab‖ba, which are both composite as the former is divisible by 101 (100×ab+1×ab) and the latter is an even palindrome and thus divisible by 11.

[3] For a full list (via Wolfram Alpha) 

Problem... Frustrating Primes

#PrimeNumbers #DivisibilityTests #Combinations

• Download pdf format.
• Download png format.
• For a post with solutions and more on Frustrating primes.
• Link to the OEIS entry for Frustrating Primes.
• For more on palprimes from Wolfram.

Some other prime types to explore with students:

• Additive primes (and list)
• Circular primes (and list)
• Dihedral primes (and list).  
• Emirps (and list)
• Factorial primes (and list for n!-1, list for n!+1)
• Double Factorial primes (and list for n!!-1, list for n!!+1)

On 73

73 is a permutable (or anagrammatic) prime, a prime that can have its digits' rearranged (in base 10) in any permutation and still be a prime number. 73 is also an emirp, a number whose reverse, 37, is also prime — a property also evident in terms of the respective ordinal positions of 73 and its emirp partner, 73 being the 21st prime [1] while 37 is the 12th prime.  73 is a Sophie Germain prime and is palindromic in binary 1001001 (interestingly, all Fermat primes and Mersenne primes are subsets of the binary palindromic primes).  73 is also an octal palindrome 111 and the only octal prime repunit.  73 is, moreover, and for some of the reasons given here, Sheldon Cooper's favourite integer in The Big Bang Theory — as was first referenced in the show’s 73rd episode.  (Jim Parsons, incidentally, the actor who plays Sheldon Cooper, was born in 1973.)

However, after listening to Charles Yang, associate professor in the University of Pennsylvania Department of Linguistics, in this episode of the superb Tell Me Something I Don't Know podcast, hosted by Stephen Dubner of Freakonomics fame, it is another little number-73-centred-nugget-of-information that has prompted this post.  Namely, that once an English speaking child has learned to count to 73, they realise they can keep going, and thus ‘understand’ the concept of infinity (NB the 1st two comments at the end of this blog).

Having delved into Yang's fascinating research, let me try and elaborate:  If there is a linguistic rule, a generalisation in other words that can be applied to a set of N words, but within this set of N words there is a subset of words, e [2], that do not follow this rule and that must therefore be memorised, then Yang (2005, cf. 2015) has shown that:

\[e < {\theta _N}\;where\;{\theta _N}: = \frac{N}{{\ln \left( N \right)}}\]

This model, 'dubbed' the Tolerance Principle by Yang (2005, cf. 2015 pp16-17), asserts that for the generalisation to be ‘productive’, i.e. for the rule that generates words to tolerate the exceptions and thus be applied instead of the pure memorisation of all N, e must not exceed N / ln(N).  It delineates, in other words, a threshold value for e (as a function of N) at which a word-forming rule ceases to be a productive mechanism for the learner to apply, and thus provides a ‘sufficiency measure’ for such linguistic generalisation.

By way of illustration, Yang (2015, pp948-949) gave this example of regular and irregular verbs: ‘If a typical English speaker knows e = 120 irregular verbs, the productivity of the -ed rule is guaranteed [my emphasis] only if there are many more regular verbs. Specifically, there must be N verbs, including both regulars and irregulars, such that θ_N = N / ln(N) ≥ 120.  The minimum value of N is 800.  In other words, if there are at least 680 regular verbs in English, the -ed rule can tolerate 120 irregular verbs’.

In his Tell Me Something I Don’t Know appearance, Yang described applying this Tolerance Principle to how we as children learn to count.  We know how many of our counting words are exceptions to the rule, those numbers that is that ‘bear no relation to their quantity, [that are] are completely arbitrary’, those numbers to put it another way that ‘won’t tell you anything about how numbers are forming' (Yang, in Berger, 2017) — in the case of children counting in English, there are 17 of these exceptions, e = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 20, 30, 50} [3].

Once we have memorised, ‘by brute force,’ the number words one to ten, we start to observe and learn rules for continuing.  For example, we start to see that fourteen is formed from ‘four and ten’, sixteen from ‘six and ten’, twenty-four from ‘two tens and four’, 'twenty-six from 'two tens and six', etc.  But there remain of course those exceptions to the rule — why is eleven not one-teen, or twelve not two-teen, etc. —  where ‘memorisation [is] the sole educational mechanism’ (Yang, in Berger 2016).  To this end, using the Tolerance Principle, we can find the least amount of words that we need to learn ‘to overcome the [17] exceptions we have to memorise’.

The smallest value of N such that θ_N = 17 is 73:

\[\begin{array}{l}\begin{align}N &= 73\;\\\because17 &= e < {\theta _N} = \frac{{73}}{{\ln \left( {73} \right)}}\\\;where\;{\rm{ }}\frac{{73}}{{\ln \left( {73} \right)}} &= 17.014...\end{align}\end{array}\]

Or in other words, once a child has learned to count to 73, they have learned the rules of the game sufficiently to overcome the cognitive need for memorisation [4].  Once children have passed this 'tipping point', when they come to a number that they are unaware has a word outside of the system they have subconsciously generalised, a thousand for example, they make up a word that fits their generalised system.  Thus, once an English speaking child can count to 73, they can count forever.

Notes & (Select) Links:

[1]  Incidentally, also the product of multiplying 7 and 3.

[2]  Not to be confused with Euler's number, e.

[3]  Yang noted that in Chinese, e = 11, meaning that a Chinese speaking child need only count up to 42 before they are able to continue and thus appreciate the concept of infinity, some one whole year before English speaking children.  

[4]  Welsh may be one of the languages with the smallest threshold at θN = 36.  The numbers 1-10 in Welsh are the only 'exceptions': Un (one), Dau (two), Tri (three), Pedwar (four), Pump (five), Chwech (six), Saith (seven), Wyth (eight), Naw (nine), and Deg (ten).  All numbers beyond this are generalised from them, for example eleven is un deg un (one ten one), twelve is un deg dau (one ten two), seventy three is saith deg tri (seven ten three), etc.

Berger, M.  (2017, October 20). PennCurrent. 'Penn linguist determines tipping point for children learning to count' [Blog post].  Retrieved from: https://penncurrent.upenn.edu/research/penn-linguist-determines-tipping-point-for-children-learning-to-count

Tell Me Something I Don't Know. (2017).  How to Count to Infinity. [podcast] Available at: http://tmsidk.com/2017/10/how-to-count-to-infinity/.

Yang, C.  (2015).  'Negative Knowledge from Positive Evidence', Language, Vol. 91, No.4, pp938-953.  Retrieved from: https://www.linguisticsociety.org/sites/default/files/07_91.4Yang.pdf

Yang, C.  (2005).  'On Productivity', in Linguistic Variation Yearbook 5, John Benjamins Publishing Company.  Retrieved from: http://www.ling.upenn.edu/~ycharles/papers/l.pdf

Problem... Vampire Numbers

#NumberProperties #Factors #VampireNumbers

• Download pdf format.
• Download png format.
• For Cliff Pickover's original post defining Vampire Numbers.
• For results, as given in the OEIS.
• Vampire Numbers on Wolfram Mathworld.
• Watch this video from @Numberphile:

Problem... Five Sets

#PropertiesOfNumber #Sequences #Venn #Sets #SetNotation

• Download pdf format.
• Download png format.