The search for answers to simple questions like “why is the sky blue? ” and “why are there selfish acts” is a nice way to bite into a fair amount of science and philosophy, respectively. In math the same is true. Why do even-perfect numbers only end in 6 or 8? If you’re fortunate enough to have a child asking you such a question , all you need is about 1200 words to answer it—only 1/9 th of the number of words in a sitcom script. 🙂 And the discussion of perfect numbers begins with a venture into a special type of prime number.

As of March 2017, of all the known primes, only 49 of them are Mersenne primes. A Mersenne prime *M _{n}* can be expressed as

*M*= 2

_{n}^{p}− 1, where

**itself is prime. The largest prime to date is 2**

*p*^{74,207,281}− 1, known as M49. To put things into perspective, our universe only has 10

^{80}atoms, a number with only 81 digits. But M49 has 44 677 235 digits, meaning that it is bigger than the inconceivably large 10

^{ 107}. Given that the number of primes less than 10

^{ 107 }is approximately equal to 10

^{ 107}/log(10

^{ 107}), in comparison, 49 (plus the possibly few smaller Mersennes yet to be discovered), divided by that gargantuan number is an unimaginably small fraction indeed.

**Why***p*must be prime if 2− 1 is prime^{p}

An interesting matter is why ** p **must be prime if 2

*− 1 is prime. To convince ourselves, we will use a proof by contradiction—in other words we will make an initial assumption that states the opposite condition, logically follow it to a contradiction, and in the process reveal the assumption to be invalid.*

^{p}We know from a geometric series that :

x^{n} − 1 = (x − 1)(x^{n − 1 }+ x^{n − 2} + · · · + a + 1)

For example x^{3} – 1 = (x-1) ( x² + x + 1)

Let’s assume that the exponent was composite, such that the Mersenne prime would be expressed as 2* ^{an}* – 1. In other words we assume its exponent to be composite. Let x = 2ª. from the above geometric series; then

2

^{an}− 1 = (2ª − 1)(2ª

^{(n − 1)}+ 2ª

^{(n − 2)}+ · · · + 2ª + 1)

But now we have expressed a prime number as two factors, neither of which is equal to one, which is impossible. The original assumption about ** p** being composite was therefore false, so

*must be prime.*

**p**2. ** Getting a perfect number from a Mersenne Prime**

Next we will multiply the Mersenne prime by a power of 2 whose exponent is one less than the prime used to generate the Mersenne prime. In other words:

(2* ^{p-1 }*)(2

*− 1).*

^{p}Interestingly this resulting number will always be a ** perfect number**, a number that is the sum of all of its factors except itself. For example 28 is perfect because ( 2

*)(2*

^{3-1 }*− 1) = 1 + 2+ 4+7 + 14 = 28.*

^{3}To reveal the “perfection” of any number of the form (2* ^{p}* − 1) ( 2

*) where p is prime, we remind you of the function σ(x). σ(x) is equal to the sum of all the factors of x*

^{p-1 }*including x*. Adding a perfect number, N= ( 2

*)(2*

^{p-1 }*− 1), to a sum that already equals N, implies that in order for a number to be perfect,*

^{p}σ(N) = 2N = 2 ( 2* ^{p-1 }*)(2

*− 1)= ( 2*

^{p}*)(2*

^{p }*− 1) … (equation 1)*

^{p}We start with the theorem σ(ab) = σ(a)σ(b), where a and b’s greatest common divisor =1. The reason this theorem is valid is that multiplying the sum of factors of two individual factors a and b is equivalent to getting all the *non-redundant* combinations of the product’s factors. For example, consider ab = 12 where σ(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28. That is equivalent to doing the following : σ(4)σ(3) = ( 1 + 2 + 4)(1 + 3) = 1(1) + 1(2) + 1(3) +2(3) + 4(1) +4(3) = 28.

Applying the theorem to our expression:

σ[( 2* ^{p-1 }*)(2

*− 1)] =*

^{p}**σ( 2**σ(2

*) ∗*^{p-1 }*− 1). … equation (2)*

^{p}Beginning with **σ( 2^{p-1 }), **we have here simply the sum of the factors of a power of two up to that power.

**σ( 2**

*) = 2*^{p-1 }^{0}+ 2^{1}+ 2^{2}+ 2^{3}+ …..2^{p-2}+ 2^{p-1}.We add 1 to each side:

**1+ σ( 2^{p-1 }) =( 1 + 2^{0} )+ 2^{1} + 2^{2} + 2^{3} + …..2^{p-2} + 2^{p-1}. **

We notice that the first terms’ sum is equal to the next power such that:

**1+ σ( 2^{p-1 }) = (2^{1} + 2^{1} )+ 2^{2} + 2^{3} + …..2^{p-2} + 2^{p-1}. ** But this keeps happening:

**1+ σ( 2^{p-1 }) = (2^{2} + 2^{2} )+ 2^{3} + …..2^{p-2} + 2^{p-1}. ** When we get to the second last term, we will have

**1+ σ( 2^{p-1 }) = (2^{p-2} +2^{p-2} )+ 2^{p-1}.**

**1+ σ( 2^{p-1 }) = (2^{p-1} )+ 2^{p-1 }= 2^{p} ,**

or **σ( 2^{p-1 }) = 2^{p} – 1 ……equation (2b)**

We substitute this result into equation (2) and obtain

σ[ ( 2* ^{p-1 }*)(2

*− 1)] = (*

^{p}**2**σ(2

*– 1 ) **^{p}*− 1). … equation(3)*

^{p}But σ(2* ^{p}* − 1) is the sum of the factors of a prime number which means that we merely have to add 1 to itself = 1 + 2

*− 1 =*

^{p}**2**

*.*^{p}Substituting this result into equation (3) we obtain:

σ[( 2* ^{p-1 }*)(2

*− 1)] = (*

^{p}**2**, which meets the condition we had outlined for a perfect number in equation (1)!

*– 1 )2*^{p}^{p}**3. Why multiply a Mersenne by ( 2^{p-1 })?**

But where does the idea of multiplying a Mersenne prime (2* ^{p}* − 1) by ( 2

*) come from? To get the sum, S, of all the natural numbers from 1 to 100 in other words (1 + 2+ 3+ 4 ….100), we simply multiply the largest number by the next natural number and divide by 2, so S(100) = 100*101/2 = 5050. The formula is S(n)= n(n+1)/2 and I derived it informally and partly pictorially in the adjacent drawing.*

^{p-1 }Applying the formula to the Mersenne prime we get (2* ^{p}* − 1)(2

*− 1 + 1)/2 = (2*

^{p}*− 1)(2*

^{p}*)/2¹ = (2*

^{p}*− 1)(2*

^{p}*)*

^{p-1}**4. But does an even perfect number have to be of the form ****(2*** ^{p}* − 1)(2

*) ?*

^{p-1}The answer is yes, and here is my summary of a clearer version of the Euclid-Euler Theorem found here. Let the even perfect number = 2^{k} *a, where a is an odd number.

As seen earlier, if the number is perfect then

σ(2^{k} *a) = σ(2^{k}) σ(a) , which applies since a power of 2 and an odd number will never have a common factor greater than 1

As we demonstrated to get to equation 2(b), the sum of the factors of a power of 2 is equal to the next power of 2 minus 1, so

σ(2^{k} *a) = 2^{k+1} -1 * σ(a). ….equation (4)

Since 2^{k} *a is perfect, the sum of all of its factors will also equal twice itself, so:

σ(2^{k} *a) = 2*2^{k} *a = 2^{k+1} *a …equation (5)

Equating equations (4) and (5) we obtain,

(2^{k+1} -1)* σ(a) = 2^{k+1} *a …equation (6)

or rearranging σ(a) = 2^{k+1} *a /(2^{k+1} -1)

We can see from this expression that σ(a) is the product of 2^{k+1 }and another expression which we’ll simplify as** b**.

so σ(a) = 2^{k+1} *b , where b = a/ (2^{k+1} -1) …equation (7)

Rearranging the expression for b,

a = (2^{k+1} -1) * b and

σ(a) = σ( (2^{k+1} -1) * b) …..equation (8)

From equation (7)

σ(a) = 2^{k+1} *b , which we can rewrite as σ(a) =2^{k+1} *b -b + b . If we then factor the first two terms to get:

σ(a) = b(2^{k+1} -1) + b . ….equation (9)

Equating (8) and (9):

σ( (2^{k+1} -1) * b) = b(2^{k+1} -1) + b . The only way that the factors of (2^{k+1} -1) * b won’t exceed the discovered sum of b(2^{k+1} -1) + b is if *b* =1, so our original odd number ** a **only has two factors, one of which is 1, revealing that

**is prime. Moreover**

*a***is a Mersenne prime since it is of the form 2**

*a*^{k+1}– 1

7. Finally we return to the question in the title. **Why do all even perfect numbers (no one yet found an odd-numbered one ) have to end in 6 or 8?**

A perfect number = ( 2* ^{p-1 }*)(2

*− 1). The first factor of that expression, 2*

^{p}*, will always create an even power of 2, except for 2*

^{p-1}^{2-1}which will still satisfy our overall generalisation by creating 6. All even powers of 2 end either with the digit 6 or 4. The next power 2

*, which follows 2*

^{p}*, ends in either 2 or 8, respectively. But we subtract 1 from each of those numbers because we are multiplying ( 2*

^{p-1 }*) by (2*

^{p-1 }*− 1). That translates into either multiplying a number ending in 6 by (1=2-1), which will end in 6 or a number ending in 4 by (7= 8-1), whose result will end in 8.*

^{p}Here’s a table from Wikipedia of all known perfect numbers as of March 4, 2017. To date, thirty of the discovered perfect numbers end in 6 and nineteen end in 8.

Rank | p |
Perfect number | Digits | Year | Discoverer |
---|---|---|---|---|---|

1 | 2 | 6 | 1 | 4th century B.C.^{[5]} |
Euclid |

2 | 3 | 28 | 2 | 4th century B.C. | Euclid |

3 | 5 | 496 | 3 | 4th century B.C. | Euclid |

4 | 7 | 8128 | 4 | 4th century B.C. | Euclid |

5 | 13 | 33,550,336 | 8 | 1456 | First seen in a medieval manuscript, Munich, Bayerische Staatsbibliothek, CLM 14908, fol. 33^{[6]} |

6 | 17 | 8,589,869,056 | 10 | 1588 | Cataldi^{[1]} |

7 | 19 | 137,438,691,328 | 12 | 1588 | Cataldi^{[1]} |

8 | 31 | 2,305,843,008,139,952,128 | 19 | 1772 | Euler |

9 | 61 | 265845599156…615953842176 | 37 | 1883 | Pervushin |

10 | 89 | 191561942608…321548169216 | 54 | 1911 | Powers |

11 | 107 | 131640364585…117783728128 | 65 | 1914 | Powers |

12 | 127 | 144740111546…131199152128 | 77 | 1876 | Lucas |

13 | 521 | 235627234572…160555646976 | 314 | 1952 | Robinson |

14 | 607 | 141053783706…759537328128 | 366 | 1952 | Robinson |

15 | 1,279 | 541625262843…764984291328 | 770 | 1952 | Robinson |

16 | 2,203 | 108925835505…834453782528 | 1,327 | 1952 | Robinson |

17 | 2,281 | 994970543370…675139915776 | 1,373 | 1952 | Robinson |

18 | 3,217 | 335708321319…332628525056 | 1,937 | 1957 | Riesel |

19 | 4,253 | 182017490401…437133377536 | 2,561 | 1961 | Hurwitz |

20 | 4,423 | 407672717110…642912534528 | 2,663 | 1961 | Hurwitz |

21 | 9,689 | 114347317530…558429577216 | 5,834 | 1963 | Gillies |

22 | 9,941 | 598885496387…324073496576 | 5,985 | 1963 | Gillies |

23 | 11,213 | 395961321281…702691086336 | 6,751 | 1963 | Gillies |

24 | 19,937 | 931144559095…790271942656 | 12,003 | 1971 | Tuckerman |

25 | 21,701 | 100656497054…255141605376 | 13,066 | 1978 | Noll & Nickel |

26 | 23,209 | 811537765823…603941666816 | 13,973 | 1979 | Noll |

27 | 44,497 | 365093519915…353031827456 | 26,790 | 1979 | Nelson & Slowinski |

28 | 86,243 | 144145836177…957360406528 | 51,924 | 1982 | Slowinski |

29 | 110,503 | 136204582133…233603862528 | 66,530 | 1988 | Colquitt & Welsh |

30 | 132,049 | 131451295454…491774550016 | 79,502 | 1983 | Slowinski |

31 | 216,091 | 278327459220…416840880128 | 130,100 | 1985 | Slowinski |

32 | 756,839 | 151616570220…600565731328 | 455,663 | 1992 | Slowinski & Gage |

33 | 859,433 | 838488226750…540416167936 | 517,430 | 1994 | Slowinski & Gage |

34 | 1,257,787 | 849732889343…028118704128 | 757,263 | 1996 | Slowinski & Gage |

35 | 1,398,269 | 331882354881…017723375616 | 841,842 | 1996 | Armengaud, Woltman, et al. |

36 | 2,976,221 | 194276425328…724174462976 | 1,791,864 | 1997 | Spence, Woltman, et al. |

37 | 3,021,377 | 811686848628…573022457856 | 1,819,050 | 1998 | Clarkson, Woltman, Kurowski, et al. |

38 | 6,972,593 | 955176030521…475123572736 | 4,197,919 | 1999 | Hajratwala, Woltman, Kurowski, et al. |

39 | 13,466,917 | 427764159021…460863021056 | 8,107,892 | 2001 | Cameron, Woltman, Kurowski, et al. |

40 | 20,996,011 | 793508909365…578206896128 | 12,640,858 | 2003 | Shafer, Woltman, Kurowski, et al. |

41 | 24,036,583 | 448233026179…460572950528 | 14,471,465 | 2004 | Findley, Woltman, Kurowski, et al. |

42 | 25,964,951 | 746209841900…874791088128 | 15,632,458 | 2005 | Nowak, Woltman, Kurowski, et al. |

43 | 30,402,457 | 497437765459…536164704256 | 18,304,103 | 2005 | Cooper, Boone, Woltman, Kurowski, et al. |

44 | 32,582,657 | 775946855336…476577120256 | 19,616,714 | 2006 | Cooper, Boone, Woltman, Kurowski, et al. |

45 | 37,156,667 | 204534225534…975074480128 | 22,370,543 | 2008 | Elvenich, Woltman, Kurowski, et al. |

46 | 42,643,801 | 144285057960…837377253376 | 25,674,127 | 2009 | Strindmo, Woltman, Kurowski, et al. |

47 | 43,112,609 | 500767156849…221145378816 | 25,956,377 | 2008 | Smith, Woltman, Kurowski, et al. |

48 | 57,885,161 | 169296395301…626270130176 | 34,850,340 | 2013 | Cooper, Woltman, Kurowski, et al. |

49 | 74,207,281 | 451129962706…557930315776 | 44,677,235 | 2016 | Cooper, Woltman, Kurowski, Blosser, et al. |

Thanks sir for this maths one….. I’m also a maths lover……follow u,….. looking forward for your next posts..!!!☺☺☺☺

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Glad you enjoyed it!

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Loved it!

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Hello, I see actually that all known perfect numbers end in either 6 or 28. Can your explanation be easily extended to explain the ‘2’ in 28? I suppose it can by considering the last 2 digits in 2^(p-1) and (2^p)-1.

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Yes, I’m sure it can. I’ll get around to it eventually— unless you get to it first and let me know. 🙂

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